RBSE Solutions Class 11 Physics Chapter 6 Gravitation

RBSE Class 11 Physics Chapter 6 Textbook Exercises with Solutions

RBSE Class 11 Physics Chapter 6 Very Short Answer Type Questions

 

Question 1. Weightlessness is experienced in artificial satellite but why not on the Moon?
Answer: Weightlessness is not felt on the Moon because it has a large mass, which creates its own significant gravitational pull. This pull causes a noticeable gravitational acceleration. In contrast, artificial satellites experience almost no gravitational acceleration, making things inside them feel weightless. This difference in mass and resulting gravity is key.
In simple words: Things feel weightless in a satellite because there is almost no gravity. The Moon is big enough to have its own gravity, so you don't feel weightless there.

🎯 Exam Tip: Clearly distinguish between the presence of gravity (Moon) and the sensation of weightlessness due to constant freefall (satellite).

 

Question 2. How much energy is required by a satellite to revolve in its orbit?
Answer: A satellite does not require any additional energy to keep revolving in its orbit once it has reached the correct speed and altitude. This is because the centripetal force, which keeps it in orbit, always acts at a right angle to the satellite's direction of motion. When force is perpendicular to displacement, the work done is zero, meaning no energy is needed to maintain its orbital speed. The initial energy used to launch the satellite is what gets it into orbit.
In simple words: A satellite needs no energy to stay in orbit because the force pulling it (gravity) works sideways to its motion, so it doesn't slow it down.

🎯 Exam Tip: Remember that work is only done when a force acts in the direction of displacement. For a satellite in a stable orbit, the gravitational force is always perpendicular to its velocity.

 

Question 4. Why are high tides and low tides in the sea?
Answer: High and low tides in the sea happen because of the Moon's gravitational pull on Earth. The Moon's gravity pulls on the oceans, creating bulges of water on both the side of Earth facing the Moon and the side opposite to it. These bulges are the high tides, while the areas in between have lower water levels, creating low tides. The Sun also affects tides, but the Moon has a stronger influence due to its closer proximity.
In simple words: The Moon's gravity pulls on the ocean water, making the water level go up in some places (high tide) and down in others (low tide).

🎯 Exam Tip: When explaining tides, always mention the Moon's gravitational effect and the concept of bulges on both sides of the Earth.

 

Question 5. Why is the Earth flat at the poles?
Answer: The Earth is slightly flattened at the poles and bulges at the equator because of its rotation. As the Earth spins on its axis, the centrifugal force pushes matter outwards at the equator. This force is strongest at the equator and weakest at the poles, causing the Earth to take on an oblate spheroid shape rather than a perfect sphere. This phenomenon is known as the Earth's oblateness.
In simple words: The Earth spins, which makes it bulge out in the middle and flatten a little at the top and bottom.

🎯 Exam Tip: Link the Earth's rotation directly to the centrifugal force acting outwards, causing the equatorial bulge and polar flattening.

 

Question 6. If the Earth revolves around the Sun in circular orbit then what would be the work done by the gravitational force?
Answer: If the Earth revolves around the Sun in a perfectly circular orbit, the work done by the gravitational force would be zero. This is because the gravitational force (centripetal force) pulling the Earth towards the Sun is always directed towards the center, which is at a right angle to the Earth's instantaneous velocity. According to the formula for work done \( W = Fd \cos \theta \), where \( \theta \) is the angle between force and displacement, when \( \theta = 90^\circ \), \( \cos 90^\circ = 0 \).
Therefore, if \( W = Fd \cos 0 \)
\( \implies W = Fd \cos 90 = 0 \).
Hence, no work is done by the gravitational force in this ideal scenario. This is a fundamental concept in rotational motion.
In simple words: When Earth moves in a circle around the Sun, the Sun's pull (gravity) is always sideways to Earth's movement. Because of this, gravity does no work, meaning it doesn't speed up or slow down the Earth.

🎯 Exam Tip: Emphasize that work done by a centripetal force in uniform circular motion is always zero because the force is perpendicular to the displacement.

 

Question 7. How many Newtons are there in 1 kg weight?
Answer: The weight of an object is the force exerted on it by gravity. On Earth, 1 kg of mass has a weight of approximately 9.8 Newtons. This value comes from the formula \( W = mg \), where \( m \) is mass (1 kg) and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, m/s^2 \)). This means that the Earth pulls a 1 kg mass with a force of 9.8 Newtons. Therefore, for 1 kg weight:
\( W = mg \)
\( \implies 1 \, \text{kg weight} = 1 \, \text{kg} \times g = 1 \times 9.8 = 9.8 \, \text{N} \).
In simple words: One kilogram of weight is equal to about 9.8 Newtons on Earth. This is because weight is the force of gravity, and Earth pulls a 1 kg mass with 9.8 Newtons of force.

🎯 Exam Tip: Clearly state the formula \( W=mg \) and the value of \( g \) (\( 9.8 \, m/s^2 \)) to calculate weight in Newtons.

 

Question 8. The value of escape velocity for any object from the surface of the Earth is 11.2 km/s. If the object is thrown at an angle of 30° then what would be the value of escape velocity?
Answer: The escape velocity for an object from the surface of the Earth is 11.2 km/s, and this value does not change regardless of the angle at which the object is thrown. Escape velocity is a scalar quantity, which means it only depends on the magnitude of the speed, not its direction. Once an object reaches this speed, its trajectory will be an open path, allowing it to escape Earth's gravitational field. Therefore, throwing it at a 30° angle will still require the same escape speed.
In simple words: Escape velocity is always 11.2 km/s from Earth, no matter what angle you throw the object. It's about how fast you throw it, not the direction.

🎯 Exam Tip: Remember that escape velocity is independent of the projection angle; it's a minimum speed required to overcome gravity.

 

Question 10. The weight of 10 g gold is more at the poles in comparison to the equator. Why?
Answer: The weight of an object, like 10 grams of gold, is slightly more at the Earth's poles than at the equator. This is because weight is determined by the formula \( W = mg \), where \( m \) is mass and \( g \) is the acceleration due to gravity. The value of \( g \) is maximum at the poles and minimum at the equator. This variation occurs because the Earth is an oblate spheroid (flattened at the poles and bulging at the equator), making the poles closer to the Earth's center. Also, the centrifugal force due to Earth's rotation reduces the effective gravity more at the equator than at the poles. The relation for \( g \) is approximately \( g = \frac{GM}{R^2} \), and since \( R \) is smaller at the poles, \( g \) is larger there. That's why the gold weighs more at the poles.
In simple words: Gold weighs more at the poles because gravity is stronger there. The poles are closer to the Earth's center, and the Earth's spin pulls less away from gravity there.

🎯 Exam Tip: Explain both reasons for the variation in \( g \): the Earth's oblate shape (shorter radius at poles) and the effect of centrifugal force (stronger at equator).

 

Question 11. Name the first satellite launched by India.
Answer: The first satellite launched by India was Aryabhatta. It was successfully launched on 19th April, 1975. This marked a significant milestone in India's space program, laying the foundation for future advancements. Aryabhatta was named after the ancient Indian astronomer.
In simple words: India's first satellite was called Aryabhatta, launched on April 19, 1975.

🎯 Exam Tip: State the name and launch date accurately. Mentioning its significance as a milestone can earn extra points.

 

Question 12. Give the dimensional formula of gravitational potential.
Answer: The dimensional formula for gravitational potential can be derived from the concept of gravitational field intensity \( E_g \), which is force per unit mass. Gravitational potential is work done per unit mass, or the negative of the integral of gravitational field intensity with respect to distance. The dimensional formula of force \( F \) is \( [M L T^{-2}] \) and mass \( m \) is \( [M] \).
So, intensity of gravitational field,
\( E_g = \frac{F}{m} \)
The dimensional formula of \( E_g \) is \( \frac{[M^1 L^1 T^{-2}]}{[M^1]} \)
\( \implies E_g = [L^1 T^{-2}] \).
Gravitational potential is \( V = E_g \times \text{distance} \).
Therefore, the dimensional formula for gravitational potential is \( [L^1 T^{-2}] \times [L^1] = [M^0 L^2 T^{-2}] \). Gravitational potential is a scalar quantity.
In simple words: The dimensional formula for gravitational potential tells us its basic units. It is \( [M^0 L^2 T^{-2}] \), meaning it has no mass unit, two length units, and minus two time units.

🎯 Exam Tip: Derive the dimensional formula step-by-step from fundamental quantities like force and mass, or work and mass, to ensure accuracy.

RBSE Class 11 Physics Chapter 6 Short Answer Type Questions

 

Question 1. Give differences between weight and mass.
Answer:
Mass: The mass of an object shows how much substance it contains. It is a fundamental property that doesn't change with location. Mass is measured using a physical balance in units of kilograms (kg). An object's mass remains constant whether it is on Earth, the Moon, or in space.
Weight: The weight of an object is the force with which Earth (or any other celestial body) pulls the object towards its center. It is a vector quantity and its value is found using the formula \( W = mg \), where \( m \) is the object's mass and \( g \) is the acceleration due to gravity at that location. Weight is measured using a spring balance. The weight changes depending on the value of \( g \), so it varies with location on Earth and is different on other planets.
In simple words: Mass is how much stuff is in an object, and it never changes. Weight is how hard gravity pulls on that object, and it can change depending on where you are.

🎯 Exam Tip: Clearly define mass as a measure of inertia and a scalar quantity, and weight as the gravitational force and a vector quantity. Highlight their differing dependence on location.

 

Question 2. If, during revolving around the Earth's orbit, the mass of a satellite is doubled due to any reason, then what will happen to its time period?
Answer: The time period \( T \) of a satellite revolving in orbit around a celestial body (like Earth) does not depend on the mass \( m \) of the satellite itself. The formula for the time period of a satellite is primarily dependent on the mass of the central body (Earth in this case), the radius of the orbit, and the gravitational constant. Therefore, if the mass of the satellite is doubled for any reason, its time period will remain unchanged. This means the satellite will continue to complete its orbit in the same amount of time. This is why satellites of different masses can share similar orbits.
In simple words: If a satellite's mass doubles, its time to go around Earth stays the same. The speed and orbit only depend on the Earth's mass, not the satellite's mass.

🎯 Exam Tip: State clearly that the orbital period is independent of the satellite's mass, and focus on the factors it *does* depend on (central body mass, orbital radius).

 

Question 3. Can any artificial satellite be placed at any orbit so that it seen always above Rajasthan's capital Jaipur? Explain clearly.
Answer: No, an artificial satellite cannot be placed in an orbit so that it is always seen directly above Rajasthan's capital, Jaipur. For a satellite to appear stationary over a specific point on Earth (like a geostationary satellite), it must orbit directly above the Earth's equator. Jaipur, however, does not lie on the equatorial plane; it is located north of the equator. Therefore, it is not possible to have an artificial satellite that constantly remains visible over Jaipur. Geostationary satellites are crucial for communication and weather monitoring because they appear fixed in the sky from the ground.
In simple words: No, a satellite cannot always stay over Jaipur. Satellites that stay still in the sky must be above the Earth's equator, and Jaipur is not on the equator.

🎯 Exam Tip: Emphasize the requirement for geostationary satellites to be in an equatorial orbit and explain why locations off the equator cannot have a permanently overhead satellite.

 

Question 4. Generally why are rockets launched at the equator surface from west to east?
Answer: Rockets are generally launched from the equator surface and in a west-to-east direction to take advantage of the Earth's rotation. The Earth spins from west to east, providing a natural "boost" to rockets launched in the same direction. At the equator, the rotational velocity of the Earth is highest. By launching from the equator and moving east, rockets gain an initial speed boost, which helps to minimize fuel consumption. This added velocity helps rockets achieve orbital speed more efficiently, especially for missions aiming for geostationary orbits, which must also be in the equatorial plane. This method conserves a significant amount of fuel.
In simple words: Rockets launch from the equator towards the east to use Earth's spin for a speed boost. This saves fuel and helps them reach orbit faster.

🎯 Exam Tip: Highlight the two key reasons: utilizing Earth's maximum rotational velocity at the equator and gaining an initial speed boost by launching eastward.

 

Question 5. If a watch based on the simple pendulum is kept at the center of the Earth, then what would be its time period? Will the watch work?
Answer: If a simple pendulum watch is kept at the center of the Earth, its time period would become infinite, meaning the watch would stop working. At the Earth's center, the acceleration due to gravity (\( g' \)) is zero. This can be understood using the formula for gravity at a depth \( d \) below the Earth's surface: \( g' = g\left(1-\frac{d}{R}\right) \), where \( R \) is the Earth's radius and \( g \) is gravity at the surface. For the center of the Earth, \( d = R \).
So, \( g' = g\left(1-\frac{R}{R}\right) = g (1 - 1) \)
\( \implies g' = 0 \).
The time period of a simple pendulum is given by \( T = 2\pi\sqrt{\frac{L}{g'}} \). Since \( g' = 0 \), the time period \( T \) would be infinite. A pendulum needs gravity to swing, so without gravity, it cannot function. Therefore, the watch would not work.
In simple words: At the Earth's center, there is no gravity, so a pendulum watch would stop. Its swing time would be endless because there's nothing to pull it back.

🎯 Exam Tip: Explain that at the Earth's center, effective gravity is zero, which makes the pendulum's time period infinite and causes it to stop working.

 

Question 7. Different countries establish different communication satellites in the orbits for their communication system. Are these orbits different? Explain.
Answer: Although different countries establish their own communication satellites, these satellites generally occupy the same type of orbit: the geostationary orbit. The time period for each geostationary satellite revolving around the Earth is 24 hours, matching Earth's rotation, and they are typically positioned at a height of about 36,000 km from the Earth's surface. This allows them to appear stationary from the ground. While the *type* of orbit is the same, each country's satellite is placed at a different *specific longitude* along this orbit to avoid interference and cover its designated service area. So, the orbits are identical in altitude and period but differ in their angular positions. This ensures global coverage without physical collision.
In simple words: Different countries use the same type of orbit for communication satellites, called geostationary orbit. But each country puts its satellite in a different spot in that orbit so they don't crash and can cover different areas.

🎯 Exam Tip: Clarify that while the *type* of orbit (geostationary, 24-hour period, specific height) is the same, the *positions* (longitudes) of the satellites within that orbit are different.

 

Question 8. If the Earth stops rotation on its axis, then what would be the change in the value of gravitational acceleration at equator and poles?
Answer: If the Earth stops rotating on its axis, the value of gravitational acceleration (\( g' \)) would change, specifically at the equator, but not at the poles. The value of gravitational acceleration at any latitude \( \lambda \) is given by the formula: \( g' = g - \omega^2 R \cos^2 \lambda \), where \( g \) is the acceleration due to gravity without rotation, \( \omega \) is the angular velocity of Earth, \( R \) is Earth's radius, and \( \lambda \) is the latitude. If Earth stops rotating, \( \omega = 0 \).
(i) For poles: At the poles, \( \lambda = 90^\circ \), so \( \cos^2 90^\circ = 0 \).
Thus, \( g' = g - 0 = g \).
Therefore, even if the Earth stops its rotation, there will be no change in gravitational acceleration at the poles.
(ii) For equator: At the equator, \( \lambda = 0^\circ \), so \( \cos^2 0^\circ = 1 \).
Thus, \( g' = g - \omega^2 R \). If Earth stops rotation, then \( \omega = 0 \), so \( g' = g \).
This means the value of gravitational acceleration would increase at the equator when Earth stops its rotation, becoming equal to the true gravitational acceleration without the rotational effect. Without rotation, the centrifugal force effect at the equator would be gone. This small increase makes the force of gravity stronger there.
In simple words: If Earth stops spinning, gravity at the poles stays the same. But at the equator, gravity would get stronger because there's no outward push from the Earth's spin anymore.

🎯 Exam Tip: Use the formula \( g' = g - \omega^2 R \cos^2 \lambda \) and substitute the appropriate values for \( \lambda \) (\( 0^\circ \) for equator, \( 90^\circ \) for poles) to show the effect of \( \omega = 0 \).

 

Question 9. Are there some celestial bodies for which the value of acceleration due to gravity is infinity?
Answer: Theoretically, for the value of acceleration due to gravity to be infinite, two conditions would need to be met. Firstly, the mass \( M \) of the celestial body would need to be infinite, or secondly, its radius \( R \) would need to be zero. This is based on the formula \( g = \frac{GM}{R^2} \). Neither of these conditions is possible for any real, physical body in the universe. Therefore, there are no known celestial bodies for which the value of acceleration due to gravity is truly infinite. However, extremely dense objects like black holes have a gravitational field so intense that nothing, not even light, can escape once it crosses the event horizon, which is sometimes conceptually linked to extreme gravity, though not mathematically infinite at the "surface."
In simple words: No, there are no real objects with infinite gravity. For gravity to be endless, an object would need to have infinite mass or zero size, which is impossible.

🎯 Exam Tip: Explain the conditions (infinite mass or zero radius) required for infinite gravity based on the formula \( g = \frac{GM}{R^2} \), and conclude why this is not possible for real bodies.

 

Question 10. A satellite revolves around a planet in elliptical orbit. If F is gravitational force then what would be the central force?
Answer: When a satellite revolves around a planet in an elliptical orbit, the gravitational force \( F \) acting between the planet and the satellite *is* the centripetal force. The gravitational attraction always pulls the satellite towards the center of the planet, providing the necessary force to keep the satellite in its curved path. So, for the satellite, the required centripetal force is precisely equal to the gravitational attraction between the planet and the satellite. This gravitational force continuously changes in magnitude and direction as the satellite moves along its elliptical path, always pointing towards the central body. This is a clear demonstration of gravity acting as the central force.
In simple words: When a satellite orbits a planet, the gravitational pull between them acts as the central force. This force makes the satellite move in its curved path.

🎯 Exam Tip: Identify gravitational force as the source of the centripetal force in any orbiting system, especially in elliptical orbits where its magnitude changes.

 

Question 11. Define gravitational potential and give its units in S.I.
Answer: Gravitational potential at any point in a gravitational field is defined as the amount of work done to bring a unit mass (a mass of 1 kg) from infinity to that specific point without any acceleration. It is a scalar quantity, meaning it has magnitude but no direction. Gravitational potential is usually denoted by \( V_G \). The work done is typically negative, indicating that the gravitational field does work on the mass, and potential energy decreases as the mass gets closer to the gravitational source. The S.I. unit of gravitational potential is Joules per kilogram (\( J \cdot kg^{-1} \)). It represents the potential energy per unit mass.
So, \( V_G = \frac{W}{M} \), where \( W \) is the work done in bringing mass \( M \) from infinity to that point.
Thus, the S.I. unit of \( V_G = \frac{J}{\text{kg}} = J \cdot kg^{-1} \).
In simple words: Gravitational potential is the energy needed to move a 1 kg mass from very far away to a certain point in space. Its unit is Joules per kilogram.

🎯 Exam Tip: Provide a precise definition of gravitational potential, emphasizing "work done," "unit mass," and "from infinity," along with its correct S.I. unit.

 

Question 12. If due to any reason the kinetic energy of a satellite increases 100% then what would be its behaviour?
Answer: The kinetic energy of a satellite in orbit is given by \( K = \frac{1}{2} m v^2 \), which is also \( K = \frac{1}{2} \frac{GMm}{r} \), where \( G \) is the gravitational constant, \( M \) is the mass of the central body, \( m \) is the satellite's mass, and \( r \) is the orbital radius. If the kinetic energy of a satellite increases by 100%, it means its new kinetic energy \( K' \) would be twice its original value, so \( K' = 2K \).
Therefore, if the kinetic energy is increased by 100%, then
\( K' = 2K = 2 \times \frac{1}{2} \frac{GMm}{r} \)
\( \implies K' = \frac{GMm}{r} \).
The total energy of a satellite in orbit is \( E = -\frac{1}{2} \frac{GMm}{r} \). The binding energy, or the energy needed for the satellite to escape, is \( E_b = \frac{1}{2} \frac{GMm}{r} \). If the kinetic energy becomes \( \frac{GMm}{r} \), which is twice the binding energy, the satellite will gain enough energy to escape the planet's gravitational field. It will then move away from the planet in an unbound trajectory. This is because its total energy will become positive, signifying an escape.
In simple words: If a satellite's movement energy doubles, it will get enough speed to break free from the planet's gravity and fly away into space.

🎯 Exam Tip: Relate the increase in kinetic energy to the total energy and escape energy. An increase of 100% in kinetic energy means the satellite gains enough energy to escape its orbit.

 

Question 1. Derive the formulae for kinetic energy and binding energy of a satellite.
Answer:
**Energy of a Satellite**
When a satellite orbits around a planet, it possesses two types of energy: kinetic energy due to its orbital motion and potential energy due to its position in the gravitational field. The total energy of the satellite is the sum of these two energies.
**Kinetic Energy (\( E_k \)):**
For a satellite of mass \( m \) moving with orbital velocity \( v_0 \) in an orbit of radius \( r \) around a planet of mass \( M \), its kinetic energy is given by:
\( E_k = \frac{1}{2} m v_0^2 \)
The orbital velocity \( v_0 \) is determined by balancing the gravitational force with the centripetal force, giving:
\( v_0 = \sqrt{\frac{GM}{r}} \)
Substituting this into the kinetic energy formula:
\( E_k = \frac{1}{2} m \left(\frac{GM}{r}\right) \)
\( \implies E_k = \frac{1}{2} \frac{GMm}{r} \)

**Potential Energy (\( U \)):**
The gravitational potential energy of a satellite of mass \( m \) at a distance \( r \) from the center of a planet of mass \( M \) is given by:
\( U = -\frac{GMm}{r} \)
The negative sign indicates that the satellite is bound to the planet by gravity.

**Total Energy (\( E_t \)):**
The total energy of the satellite is the sum of its kinetic and potential energies:
\( E_t = E_k + U \)
\( E_t = \frac{1}{2} \frac{GMm}{r} + \left(-\frac{GMm}{r}\right) \)
\( \implies E_t = -\frac{1}{2} \frac{GMm}{r} \)
The total energy being negative means the satellite is gravitationally bound to the planet. To free the satellite from orbit, an external energy input equal to the magnitude of its total energy is required.

**Binding Energy (\( E_b \)):**
Binding energy is the minimum amount of energy required for a satellite to completely escape its orbit and move out of the planet's gravitational influence. It is represented by \( E_b \). For a satellite to escape, its total energy must become zero (or positive). Since the total energy is \( E_t = -\frac{1}{2} \frac{GMm}{r} \), the binding energy needed is the positive equivalent of this total energy:
\( E_b = -E_t \)
\( E_b = -\left(-\frac{1}{2} \frac{GMm}{r}\right) \)
\( \implies E_b = \frac{1}{2} \frac{GMm}{r} \)
This means that the binding energy is equal to the magnitude of the satellite's total energy, and also equal to its kinetic energy.
In simple words: A satellite has kinetic energy from moving and potential energy from its position. Total energy is the sum, and it's negative because the satellite is "stuck" to the planet. Binding energy is how much energy you need to give it to break free from the planet's pull.

🎯 Exam Tip: Clearly define each energy type (kinetic, potential, total, binding) and show their derivations. Ensure the negative sign for potential and total energy is explained correctly.

Chandrashekhar Limit.

 

The Chandrashekhar Limit is an important concept in astrophysics. According to this limit, the maximum mass a stable white dwarf star can have is about 1.44 times the mass of our Sun. If a star's core, after using up its fuel, has a mass greater than this limit, it cannot remain a white dwarf. Instead, it will collapse further due to its own gravity, potentially leading to a supernova and then forming either a neutron star or a black hole. In honor of Dr. S. Chandrashekhar's work, NASA established an X-ray laboratory in 1999 dedicated to the study of stars.

Pioneers like Dr. C.V. Raman and Shri Meghnad Saha also contributed significantly to spaceworks. Dr. Vikram Sarabhai, often considered the father of the Indian space program, established the Physical Research Laboratory in Ahmedabad, which became a cornerstone for space research in India.

India's journey into space began with its first satellite, Aryabhatta, which was built by ISRO and launched by the Soviet Union on April 19, 1975. Later, the Rohini satellite was launched by India's own SLV-3 vehicle in 1980 from Sriharikota, marking India's first indigenous satellite launch. In 2014, a heavy satellite, GSAT-14, was successfully put into orbit by GSLV-D5, using an Indian cryogenic engine developed by ISRO. This achievement made India the sixth country to master this complex technology. Following this, the GSAT-6 vehicle was launched on August 27, 2015, using GSLV-D6. In July 2012, former President of India and the "Missile Man," Dr. A.P.J. Abdul Kalam, highlighted the collaborative efforts of ISRO and DRDO to make space techniques more affordable and efficient. This dedication to cost-effectiveness is partly why the Mangalyaan mission's successful launch was notably less expensive than the Hollywood film 'Gravity'.

Indian Space Research achievements are as following :

• 1. In the year 1963, India launched its first rocket from Thumba.
• 2. In the year 1967, an Experimental Satellite Communication Earth Station was established at Ahmedabad.
• 3. The Department of Space was established in the year 1972.
• 4. The first Indian satellite, Aryabhatta, was successfully launched on 19th April 1975.
• 5. Bhaskara-1 satellite was launched in the year 1979.
• 6. Rohini-1 satellite was launched in the year 1980.
• 7. The INSAT-1A satellite was launched, but it was not successful.
• 8. INSAT-1B was launched in the year 1983.
• 9. Shri Rakesh Sharma was the first Astronaut of India in the year 1984.
• 10. India's first operational remote sensing satellite was launched in the year 1988.
• 11. INSAT-ID was launched in the year 1990.
• 12. INSAT-2C was launched in the year 1995.
• 13. INSAT-2D was not launched successfully in June 1997, but in September 1997, IRS-1D was successfully launched.
• 14. GSLV-D1 was partly successful in the year 2001.
• 15. In the year 2004, GSLV-EDUSAT was successfully launched.
• 16. On 22nd October 2008, Chandrayaan-1 was successfully launched.
• 17. On 5th November 2013, Mangalyaan was successfully launched.
• 18. On 24th September 2014, Mangalyaan (298 days after launching) successfully established itself in the orbit of Mars.
• 19. On 5th January 2014, GSLV-D5 was successfully launched.
• 20. On 18th December 2014, the flight of GSLV-MKIII was successful.
• 21. On 28th September 2015, India's first dedicated multi-wavelength space observatory was launched, named 'Astrosat.'

Main Achievements of the Century

Chandrayaan-1 made an important discovery by detecting the presence of water molecules on the Lunar surface. Chandrayaan-1 was sent to the Moon on October 22, 2008. It was launched by PSLV-C11 from the Satish Dhawan Space Center. Its main goal was to find water and helium gas. For this achievement, India became the sixth country in the world to reach the Moon. After it disconnected from its orbit, it was eventually closed. According to ISRO's Head, Shri Madhwan Nayar, Chandrayaan-1 was officially closed on August 30, 2009.

According to Times Magazine, Mangalyaan was selected in the list of Best Inventions in the year 2014. Mangalyaan's successful launch demonstrated the talent of Indian scientists, establishing their capabilities globally.

 

Question 3. What do you mean by orbital velocity and escape velocity? Give formulae and relationship for these.
Answer:
**Orbital Velocity:**
Orbital velocity is the speed at which an object revolves in a stable orbit around a larger celestial body. This velocity is just enough for the object to stay in orbit, balancing the gravitational pull with the centrifugal force. It is the velocity needed for a satellite to continuously fall around a planet without hitting it or flying off into space. The orbital velocity \( v_0 \) of a satellite revolving in a circular orbit of radius \( r \) around a planet of mass \( M \) is given by:
\( v_0 = \sqrt{\frac{GM}{r}} \)
where \( G \) is the universal gravitational constant.

**Escape Velocity:**
Escape velocity is the minimum speed an object needs to completely escape from the gravitational influence of a massive body, like a planet, and never return. Unlike orbital velocity, escape velocity does not depend on the direction of projection. It is often referred to as escape speed for this reason. If an object is launched with this speed, its kinetic energy will be sufficient to overcome its gravitational potential energy.
The formula for escape velocity \( v_e \) from the Earth's surface (radius \( R \), mass \( M \)) is:
\( v_e = \sqrt{\frac{2GM}{R}} \)
Since \( g = \frac{GM}{R^2} \), we can write \( GM = gR^2 \). Substituting this into the escape velocity formula:
\( v_e = \sqrt{\frac{2gR^2}{R}} \)
\( \implies v_e = \sqrt{2gR} \)
It is clear that escape velocity does not depend upon the mass of the object being launched. For Earth, with \( g = 9.8 \, m/s^2 \) and \( R = 6.4 \times 10^6 \, m \), the escape velocity is approximately 11.2 km/s.

**Relationship between Orbital Velocity and Escape Velocity:**
Comparing the formulas for orbital velocity and escape velocity near the Earth's surface (\( r \approx R \)):
Orbital velocity: \( v_0 = \sqrt{gR} \)
Escape velocity: \( v_e = \sqrt{2gR} \)
From these, we can see that:
\( v_e = \sqrt{2} \times \sqrt{gR} \)
\( \implies v_e = \sqrt{2} v_0 \)
This means the escape velocity is \( \sqrt{2} \) times the orbital velocity for an object orbiting very close to the Earth's surface. If an object's orbital velocity increases by \( \sqrt{2} \) times, it will escape its orbit.
In simple words: Orbital velocity is the speed to stay in orbit, like a satellite. Escape velocity is the speed needed to fly away from a planet's gravity forever. Escape velocity is always about 1.414 times faster than orbital velocity near the surface.

🎯 Exam Tip: Provide clear, distinct definitions and formulas for both velocities. The relationship \( v_e = \sqrt{2} v_0 \) is crucial and must be derived or stated clearly.

 

Question 4. Derive relationship between g and G.
Answer:
**Relation between Acceleration due to Gravity (\( g \)) and Universal Gravitational Constant (\( G \))**
The acceleration produced in an object due to the gravitational force of the Earth is called acceleration due to gravity (\( g \)). This value of \( g \) is independent of the mass and shape of the object. It is represented by \( g \), and its S.I. unit is meters per second squared (\( m/s^2 \)).
Suppose the Earth has a mass \( M \) and radius \( R \). Consider an object of mass \( m \) on the surface of the Earth. The gravitational force \( F \) acting on this object, according to Newton's law of universal gravitation, is:
\( F = \frac{GMm}{R^2} \) ... (1)
where \( G \) is the universal gravitational constant. This force pulls the object towards the Earth's center. According to Newton's second law of motion, the force acting on an object is equal to its mass times its acceleration. In this case, the acceleration is the acceleration due to gravity, \( g \).
So, \( F = mg \) ... (2)
By equating the two expressions for the force \( F \) (from equations 1 and 2):
\( mg = \frac{GMm}{R^2} \)
We can cancel out the mass \( m \) of the object from both sides:
\( g = \frac{GM}{R^2} \)
This is the fundamental relationship between \( g \) and \( G \). It shows that the acceleration due to gravity depends on the mass of the planet \( M \) and its radius \( R \), but not on the mass of the object experiencing the gravity. This formula can be extended to find \( g \) at a height \( h \) above the surface:
\( g' = \frac{GM}{(R+h)^2} \)
If we consider the average density \( \rho \) of the Earth, then its mass \( M \) can be expressed as:
\( M = \text{Volume} \times \text{Density} = \frac{4}{3} \pi R^3 \rho \)
Substituting this into the formula for \( g \):
\( g = \frac{G \left(\frac{4}{3} \pi R^3 \rho\right)}{R^2} \)
\( \implies g = \frac{4}{3} \pi GR \rho \)
This equation shows the relationship between \( g \), \( G \), Earth's radius \( R \), and its average density \( \rho \).
In simple words: The formula \( g = \frac{GM}{R^2} \) links the small 'g' (gravity's pull on Earth) to the big 'G' (universal gravity constant). It shows that gravity depends on how heavy the Earth is and how far you are from its center.

🎯 Exam Tip: Start by defining both \( g \) and \( G \). Clearly show the derivation by equating Newton's second law (\( F=mg \)) with the law of universal gravitation (\( F=\frac{GMm}{R^2} \)).

 

Question 6. Calculate the height of Geo-stationary satellite above the Earth's surface. How can it be used for communication?
Answer: A geostationary satellite is placed at a specific height directly above the Earth's equator. It moves in the same direction as the Earth rotates, so its orbital time period is exactly 24 hours. This makes the satellite appear stationary to an observer on Earth, which is why it's also called a geo-synchronous satellite.
We calculate the orbital radius \( r \) using the formula for a satellite's time period:
\( T = \frac{2 \pi r^{3 / 2}}{\sqrt{G M}} \)
Rearranging this formula to find \( r \):
\( r = \left[\frac{G M T^{2}}{4 \pi^{2}}\right]^{1 / 3} \)
Using the given values:
Gravitational constant \( G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \)
Mass of Earth \( M = 6 \times 10^{24} \text{ kg} \)
Time period \( T = 24 \text{ hours} = 86400 \text{ seconds} \)
\( \pi = 3.14 \)
Plugging in the values, we get the orbital radius from the Earth's center:
\( r \approx 4.2 \times 10^4 \text{ km} = 42000 \text{ km} \)
To find the height \( h \) above the Earth's surface, we subtract the Earth's radius \( R \). Assuming Earth's radius \( R = 6400 \text{ km} \):
\( h = r - R = 42000 \text{ km} - 6400 \text{ km} = 35600 \text{ km} \)
Geostationary satellites are crucial for communication because their fixed position relative to the Earth allows antennas on the ground to stay pointed at them. This ensures continuous, uninterrupted signal transmission for television broadcasting, internet services, and telephone communication across large areas. Communication satellites transmit signals to about one-third of the Earth's surface, so at least three satellites are needed to cover the entire globe. These satellites cannot be placed over locations like New Delhi because it is not on the equatorial line where geostationary orbits are located.
In simple words: Geostationary satellites stay in one spot above Earth's equator and take 24 hours to go around. We find how high they are using special formulas, which comes out to about 35,600 km above the surface. They are used for TV and phone signals because they always stay in the same place in the sky.

🎯 Exam Tip: Remember that geostationary satellites are always located above the equator and have a 24-hour orbital period, which are key facts for answering questions about their use in communication.

 

Question 7. Give reasons for the change in the intensity of the gravitational field of the Earth with respect to; (1) Height, (2) Depth, (3) Due to rotation of the Earth.
Answer: The intensity of the gravitational field, often represented by the acceleration due to gravity (\( g \)), changes based on several factors:
(1) Change with Height:
The acceleration due to gravity decreases as we move to a higher altitude above the Earth's surface. At the Earth's surface, \( g = \frac{GM}{R^2} \). At a height \( h \) above the surface, the effective distance from the Earth's center becomes \( (R+h) \).
So, the acceleration due to gravity \( g' \) at height \( h \) is:
\( g' = \frac{GM}{(R+h)^2} = g \left(\frac{R}{R+h}\right)^2 = g \left(1+\frac{h}{R}\right)^{-2} \)
From this formula, it is clear that as \( h \) increases, \( g' \) decreases because the object is farther from the Earth's center. For small heights (\( h \ll R \)), this can be approximated as \( g' \approx g \left(1-\frac{2h}{R}\right) \).
(2) Change with Depth:
The acceleration due to gravity also decreases as we go deeper inside the Earth. At a depth \( d \) below the Earth's surface, the acceleration due to gravity \( g' \) is given by:
\( g' = g \left(1-\frac{d}{R}\right) \)
Here, \( R \) is the Earth's radius. This shows that \( g' \) decreases with increasing depth \( d \). At the center of the Earth (\( d=R \)), the acceleration due to gravity becomes zero. This happens because the net gravitational force from the surrounding mass cancels out at the center.
(3) Change due to Rotation of the Earth:
The Earth's rotation causes a variation in the value of \( g \) with latitude. Because the Earth is not a perfect sphere but rather slightly flattened at the poles and bulging at the equator, it affects gravity.
At the equator, the centrifugal force due to rotation is maximum and acts outwards, reducing the effective gravity. At the poles, there is no centrifugal force component, and the radius is slightly smaller, making gravity stronger.
The effective acceleration due to gravity \( g' \) at a latitude \( \lambda \) is given by:
\( g' = g - \omega^2 R \cos^2 \lambda \)
where \( \omega \) is the angular velocity of Earth's rotation and \( R \) is Earth's radius.
- At the poles (\( \lambda = 90^\circ \), \( \cos \lambda = 0 \)): \( g' = g \). Gravity is maximum.
- At the equator (\( \lambda = 0^\circ \), \( \cos \lambda = 1 \)): \( g' = g - \omega^2 R \). Gravity is minimum.
Therefore, gravity is greater at the poles than at the equator.
In simple words: Gravity changes in three ways: It gets weaker as you go higher up because you're farther from Earth. It also gets weaker as you go deeper inside Earth, becoming zero at the very center. Lastly, Earth's spinning makes gravity a little weaker at the middle (equator) and stronger at the top and bottom (poles).

🎯 Exam Tip: Remember the three main factors affecting gravity (height, depth, rotation) and understand the general trend for each: decreases with height, decreases with depth, and is maximum at poles, minimum at the equator.

 

Question 8. Define gravitational potential energy. Calculate the change in the potential energy in sending a particle to height h from the Earth's surface. When h << R discuss about the changes in potential energy.
Answer:
Gravitational Potential Energy:
Gravitational potential energy is the energy an object possesses due to its position within a gravitational field. It is equal to the work done in bringing a mass from infinity to a specific point inside the gravitational field, without any acceleration. It is a scalar quantity and is generally considered to be zero at infinity, becoming negative as an object gets closer to a massive body.
Change in Potential Energy:
Let's calculate the change in potential energy when a particle of mass \( m \) is moved from the Earth's surface to a height \( h \).
The gravitational potential energy at the Earth's surface (\( r = R \)) is:
\( U_1 = -\frac{GMm}{R} \)
The gravitational potential energy at height \( h \) from the Earth's surface (i.e., at a distance \( r = R+h \) from the center) is:
\( U_2 = -\frac{GMm}{R+h} \)
The change in potential energy (\( \Delta U \)) is the difference between the final and initial potential energies:
\( \Delta U = U_2 - U_1 \)
\( \Delta U = -\frac{GMm}{R+h} - \left(-\frac{GMm}{R}\right) \)
\( \Delta U = GMm \left(\frac{1}{R} - \frac{1}{R+h}\right) \)
\( \Delta U = GMm \left(\frac{(R+h) - R}{R(R+h)}\right) \)
\( \Delta U = \frac{GMmh}{R(R+h)} \)
This change in potential energy is equal to the work done (\( W \)) against gravity to raise the object.
Discussion when \( h \ll R \):
When the height \( h \) is much smaller than the Earth's radius \( R \) (\( h \ll R \)), we can simplify the expression for \( \Delta U \).
In the denominator, \( R(R+h) \approx R(R) = R^2 \).
Also, we know that the acceleration due to gravity on the Earth's surface is \( g = \frac{GM}{R^2} \), which means \( GM = gR^2 \).
Substituting these into the formula for \( \Delta U \):
\( \Delta U = \frac{(gR^2)mh}{R^2} \)
\( \Delta U = mgh \)
This shows that for small heights, the change in potential energy is approximately \( mgh \), which is the familiar formula for potential energy close to the Earth's surface. This approximation simplifies calculations for objects moving short distances.
In simple words: Gravitational potential energy is the energy an object has because of its position in Earth's pull. When you lift something, its potential energy goes up. To find how much it changes when lifted to height \( h \), we calculate the difference between its energy at that height and at the ground. If the height is very small compared to Earth's size, this change in energy is simply \( mgh \), which is mass times gravity times height.

🎯 Exam Tip: Remember that gravitational potential energy is always negative, and its value increases (becomes less negative) as an object moves farther from the Earth. For small heights, the \( mgh \) approximation is very useful.

 

Question 9. Explain Kepler's laws.
Answer: Johannes Kepler studied the detailed observations of planetary motion made by Tycho Brahe and formulated three fundamental laws that describe how planets orbit the Sun. These laws, known as Kepler's Laws of Planetary Motion, are:
1. First Law (Law of Orbits):
Every planet revolves around the Sun in an elliptical orbit, with the Sun located at one of the two foci of the ellipse. This means that planetary orbits are not perfect circles but rather stretched circles.
2. Second Law (Law of Areas):
An imaginary line connecting a planet to the Sun sweeps out equal areas in equal intervals of time. This law implies that a planet moves faster when it is closer to the Sun and slower when it is farther away. The speed of the planet changes as its distance from the Sun changes, to cover the same amount of area in the same amount of time.
Mathematically, the areal speed \( \frac{dA}{dt} \) is constant.
3. Third Law (Law of Periods):
The square of the orbital period (\( T \)) of any planet is directly proportional to the cube of the semi-major axis (\( r \), or average distance) of its elliptical orbit.
This can be written as \( T^2 \propto r^3 \), or \( T^2 = Kr^3 \), where \( K \) is a constant that is the same for all planets orbiting the Sun. This law explains that planets farther from the Sun have longer orbital periods.
In simple words: Kepler's laws tell us how planets move around the Sun. The first law says planets move in stretched circles (ellipses), not perfect circles. The second law says a planet moves faster when it's closer to the Sun and slower when it's far away. The third law says that planets further from the Sun take much longer to complete one orbit.

🎯 Exam Tip: When explaining Kepler's Laws, clearly state each law and include its key characteristic (ellipse for first, equal areas/variable speed for second, and \( T^2 \propto r^3 \) for third).

 

Question 10. Explain weightlessness in space.
Answer: Weightlessness is the sensation of having no weight, which occurs when there is no force of support acting on an object. We normally feel our weight due to the normal force exerted by the ground or a supporting surface, which balances the gravitational pull. When this supporting force is absent or equal to the gravitational force causing free fall, we experience weightlessness. For example, if a person jumps freely downwards, they feel weightless during the fall because there is no ground pushing back.
In artificial satellites, objects and astronauts experience a state of apparent weightlessness. This happens because the satellite and everything inside it are continuously falling around the Earth in orbit. They are all in a state of free fall. The gravitational force from Earth provides the necessary centripetal force to keep the satellite in its orbit.
Consider an object of mass \( m \) inside a satellite. If its weight is \( W \) and the reaction force from the satellite's floor is \( R \), then for the object to move in orbit with orbital velocity \( v_0 \) and radius \( r \), the net force providing centripetal acceleration is \( W - R \).
So, \( \frac{mv_0^2}{r} = W - R \)
However, the weight \( W \) of the object inside the satellite is exactly the gravitational force that provides the centripetal force for its orbital motion: \( W = \frac{mv_0^2}{r} \).
Substituting this into the equation:
\( \frac{mv_0^2}{r} = \frac{mv_0^2}{r} - R \)
This implies that \( R = 0 \).
Since the reaction force \( R \) is what we perceive as weight, a zero reaction force means the object feels weightless. Therefore, in an orbiting artificial satellite, objects appear to float because they are constantly falling around the Earth alongside the satellite, rather than being pushed against a surface.
In simple words: Weightlessness means you feel no push from the ground or a surface. In a spaceship orbiting Earth, everything inside is constantly falling around the Earth. Because of this continuous fall, nothing pushes against you, so you feel like you have no weight.

🎯 Exam Tip: The key concept for weightlessness in orbit is "continuous free fall." Emphasize that gravity is still present; it's the lack of a supporting force that creates the sensation of weightlessness.

 

Question 1. Derive the formulae for kinetic energy and binding energy of a satellite.
Answer:
Energy of a Satellite:
A satellite revolving in its orbit possesses both kinetic energy due to its orbital motion and potential energy due to its position in the gravitational field. The total energy of the satellite is the sum of these two energies.
1. Kinetic Energy (\( E_k \)):
A satellite of mass \( m \) moving with orbital velocity \( v_0 \) in an orbit of radius \( r \) has kinetic energy:
\( E_k = \frac{1}{2} mv_0^2 \)
For a stable circular orbit, the centripetal force \( \frac{mv_0^2}{r} \) is provided by the gravitational force \( \frac{GMm}{r^2} \).
\( \frac{mv_0^2}{r} = \frac{GMm}{r^2} \)
From this, \( v_0^2 = \frac{GM}{r} \).
Substituting \( v_0^2 \) into the kinetic energy formula:
\( E_k = \frac{1}{2} m \left(\frac{GM}{r}\right) \)
\( E_k = \frac{GMm}{2r} \)
2. Potential Energy (\( U \)):
The gravitational potential energy of a satellite of mass \( m \) at a distance \( r \) from the center of a planet of mass \( M \) is:
\( U = -\frac{GMm}{r} \)
The negative sign indicates that the satellite is bound to the planet; external energy is needed to free it.
3. Total Energy (\( E_t \)):
The total mechanical energy of the satellite is the sum of its kinetic and potential energies:
\( E_t = E_k + U \)
\( E_t = \frac{GMm}{2r} + \left(-\frac{GMm}{r}\right) \)
\( E_t = \frac{GMm}{2r} - \frac{2GMm}{2r} \)
\( E_t = -\frac{GMm}{2r} \)
The total energy is negative, which again signifies that the satellite is gravitationally bound to the planet and requires external energy to escape its orbit.
4. Binding Energy (\( E_b \)):
Binding energy is the minimum amount of energy required to remove a satellite from its orbit and move it to infinity, where its total energy would be zero.
\( E_b = (\text{Total energy at infinity}) - (\text{Total energy in orbit}) \)
\( E_b = 0 - \left(-\frac{GMm}{2r}\right) \)
\( E_b = \frac{GMm}{2r} \)
Thus, the binding energy is equal to the positive value of the total energy of the satellite.
In simple words: A satellite has energy from its movement (kinetic energy) and its position (potential energy). We can use physics rules to find out exactly how much of each it has. Adding these gives the total energy. The binding energy is how much energy we would need to give the satellite to make it fly away from the planet forever.

🎯 Exam Tip: Pay close attention to the signs in energy calculations; potential energy and total energy in a bound orbit are negative, while kinetic energy and binding energy are positive.

 

Question 12. Write Newton's gravitation law. Derive it in vector form an explain that it follows action-reaction law.
Answer:
Newton's Law of Universal Gravitation:
Newton's law of universal gravitation states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This force acts along the line joining the two particles.
Mathematically, the magnitude of the gravitational force \( F \) between two objects with masses \( M_1 \) and \( M_2 \), separated by a distance \( r \), is given by:
\( F = \frac{GM_1M_2}{r^2} \)
where \( G \) is the universal gravitational constant.
Derivation in Vector Form:
Let's consider two particles, \( M_1 \) and \( M_2 \), with position vectors \( \vec{r_1} \) and \( \vec{r_2} \) respectively.
The position vector of \( M_2 \) relative to \( M_1 \) is \( \vec{r_{21}} = \vec{r_2} - \vec{r_1} \).
The unit vector in the direction from \( M_1 \) to \( M_2 \) is \( \hat{r}_{21} = \frac{\vec{r_{21}}}{|\vec{r_{21}}|} = \frac{\vec{r_2} - \vec{r_1}}{|\vec{r_2} - \vec{r_1}|} \).
The gravitational force exerted on \( M_2 \) by \( M_1 \) (\( \vec{F_{21}} \)) is directed towards \( M_1 \). Therefore, the force vector is opposite to the direction of \( \hat{r}_{21} \).
\( \vec{F_{21}} = -\frac{GM_1M_2}{r_{21}^2} \hat{r}_{21} \)
Similarly, the position vector of \( M_1 \) relative to \( M_2 \) is \( \vec{r_{12}} = \vec{r_1} - \vec{r_2} \).
The unit vector in the direction from \( M_2 \) to \( M_1 \) is \( \hat{r}_{12} = \frac{\vec{r_{12}}}{|\vec{r_{12}}|} = \frac{\vec{r_1} - \vec{r_2}}{|\vec{r_1} - \vec{r_2}|} \).
The gravitational force exerted on \( M_1 \) by \( M_2 \) (\( \vec{F_{12}} \)) is directed towards \( M_2 \). So, the force vector is opposite to the direction of \( \hat{r}_{12} \).
\( \vec{F_{12}} = -\frac{GM_1M_2}{r_{12}^2} \hat{r}_{12} \)
Since \( \vec{r_{12}} = -\vec{r_{21}} \), it follows that \( \hat{r}_{12} = -\hat{r}_{21} \) and \( r_{12} = r_{21} \).
Therefore, we can write:
\( \vec{F_{12}} = -\frac{GM_1M_2}{r_{21}^2} (-\hat{r}_{21}) \)
\( \vec{F_{12}} = \frac{GM_1M_2}{r_{21}^2} \hat{r}_{21} \)
Comparing with the expression for \( \vec{F_{21}} \):
\( \vec{F_{12}} = - \left(-\frac{GM_1M_2}{r_{21}^2} \hat{r}_{21}\right) \)
\( \vec{F_{12}} = -\vec{F_{21}} \)
Follows Action-Reaction Law (Newton's Third Law):
The vector form \( \vec{F_{12}} = -\vec{F_{21}} \) clearly shows that the gravitational force between two particles is an action-reaction pair. This means that the force exerted by \( M_1 \) on \( M_2 \) is equal in magnitude and opposite in direction to the force exerted by \( M_2 \) on \( M_1 \). This consistent with Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. Gravitational forces always act in pairs, pulling the objects towards each other.
In simple words: Newton's gravity law says everything pulls on everything else, stronger for heavier objects and weaker if they are far apart. In vector form, it shows that the pull from object 1 on object 2 is exactly the same strength but in the opposite direction as the pull from object 2 on object 1. This means gravity follows Newton's Third Law of Motion, like a push and a push back.

🎯 Exam Tip: When deriving the vector form, remember to correctly use unit vectors and show how \( \vec{F_{12}} = -\vec{F_{21}} \) to demonstrate Newton's Third Law. The negative sign is crucial for indicating opposite directions.

 

Question 1. Calculate the gravitational force between two metal spheres of masses 50 kg and 100 kg respectively and the separation between their centres is 50 cm.
Answer: To calculate the gravitational force, we use Newton's Law of Universal Gravitation.
The formula for gravitational force (\( F \)) is:
\( F = \frac{GM_1M_2}{r^2} \)
Given values:
Mass of the first sphere, \( M_1 = 50 \text{ kg} \)
Mass of the second sphere, \( M_2 = 100 \text{ kg} \)
Distance between their centers, \( r = 50 \text{ cm} = 0.50 \text{ m} \)
Universal Gravitational Constant, \( G = 6.67 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \)
Now, substitute these values into the formula:
\( F = \frac{(6.67 \times 10^{-11}) \times (50) \times (100)}{(0.50)^2} \)
\( F = \frac{6.67 \times 10^{-11} \times 5000}{0.25} \)
\( F = 6.67 \times 10^{-11} \times 20000 \)
\( F = 133400 \times 10^{-11} \)
\( F = 1.334 \times 10^{-6} \text{ N} \)
The gravitational force between the two spheres is \( 1.334 \times 10^{-6} \text{ N} \). This force is very small because gravitational forces are only significant for very large masses like planets.
In simple words: We used the gravity formula to find the pull between two metal balls. One ball is 50 kg, the other is 100 kg, and they are 50 cm apart. After putting the numbers into the formula, we found the pull is very tiny, about \( 1.334 \times 10^{-6} \) Newtons.

🎯 Exam Tip: Ensure all units are consistent (e.g., meters for distance) before plugging values into the formula. Remember to square the distance in the denominator.

 

Question 3. A body is thrown with a velocity of 10 km/s from the Earth's surface, which is only a little less from the value of escape velocity. Calculate the height up to which the body will be reached.
Answer: We can calculate the maximum height reached by the projectile using the formula:
\( h = \frac{v^2 R}{2gR - v^2} \)
Given values:
Initial velocity of the body, \( v = 10 \text{ km/s} = 10 \times 10^3 \text{ m/s} \)
Acceleration due to gravity, \( g = 10 \text{ m/s}^2 \)
Radius of the Earth, \( R = 6.4 \times 10^6 \text{ m} \)
Now, substitute these values into the formula:
\( h = \frac{(10 \times 10^3)^2 \times (6.4 \times 10^6)}{2 \times (10) \times (6.4 \times 10^6) - (10 \times 10^3)^2} \)
\( h = \frac{(10^4)^2 \times (6.4 \times 10^6)}{128 \times 10^6 - (10^4)^2} \)
\( h = \frac{10^8 \times 6.4 \times 10^6}{12.8 \times 10^7 - 10^8} \)
\( h = \frac{6.4 \times 10^{14}}{1.28 \times 10^8 - 1 \times 10^8} \)
\( h = \frac{6.4 \times 10^{14}}{0.28 \times 10^8} \)
\( h = \frac{6.4}{0.28} \times 10^6 \text{ m} \)
\( h \approx 22.857 \times 10^6 \text{ m} \)
\( h = 2.28 \times 10^7 \text{ m} \)
Converting to kilometers:
\( h = 2.28 \times 10^4 \text{ km} \)
The body will reach a height of approximately 22,800 kilometers above the Earth's surface. This demonstrates that even a small difference from escape velocity can lead to a very high altitude.
In simple words: We threw an object from Earth at almost escape speed. We used a special formula with its speed, Earth's radius, and gravity to find out how high it would go. It reached a height of about 22,800 kilometers before falling back.

🎯 Exam Tip: Be careful with powers of 10 and unit conversions (km/s to m/s, km to m) when solving problems involving very large or very small numbers.

 

Question 4. On the surface of the Earth a body's weight is 72 N. How much would be the gravitational force at a height of half the radius of the Earth?
Answer: We are given the weight of a body on the Earth's surface and asked to find its weight at a specific height.
Given:
Weight of the body on the surface, \( W = 72 \text{ N} \)
Height above the surface, \( h = \frac{R}{2} \) (half the Earth's radius)
The acceleration due to gravity (\( g' \)) at a height \( h \) is related to the acceleration due to gravity on the surface (\( g \)) by the formula:
\( g' = g \left(\frac{R}{R+h}\right)^2 \)
Substitute \( h = \frac{R}{2} \) into the formula:
\( g' = g \left(\frac{R}{R + \frac{R}{2}}\right)^2 \)
\( g' = g \left(\frac{R}{\frac{3R}{2}}\right)^2 \)
\( g' = g \left(\frac{2}{3}\right)^2 \)
\( g' = g \left(\frac{4}{9}\right) = \frac{4g}{9} \)
Now, the new weight (\( W' \)) at this height will be \( W' = mg' \).
Since \( W = mg \) (weight on the surface):
\( W' = m \left(\frac{4g}{9}\right) \)
\( W' = \frac{4}{9} (mg) \)
\( W' = \frac{4}{9} W \)
Substitute the given weight \( W = 72 \text{ N} \):
\( W' = \frac{4}{9} \times 72 \text{ N} \)
\( W' = 4 \times 8 \text{ N} \)
\( W' = 32 \text{ N} \)
So, the gravitational force (weight) at a height of half the Earth's radius would be 32 N. This shows how gravity weakens as you move farther from the Earth's center.
In simple words: An object weighs 72 N on Earth. If we take it to a height that is half of Earth's radius, its weight will become less. We used a formula that shows how gravity changes with height, and found its new weight would be 32 N.

🎯 Exam Tip: Remember that weight is directly proportional to the local acceleration due to gravity, and gravity decreases with the square of the distance from the Earth's center.

 

Question 5. Escape velocity on the surface of the Earth is 11.2 km/s. If any object is thrown with double the escape velocity, what would be its speed at maximum distance? Neglect the presence of Sun and other celestial bodies.
Answer: We can solve this problem using the principle of conservation of energy.
The total energy of the object on the Earth's surface (\( E_1 \)) must be equal to its total energy at the maximum distance (\( E_2 \)).
Total Energy \( E = \text{Kinetic Energy} (K) + \text{Potential Energy} (U) \)
\( K_1 + U_1 = K_2 + U_2 \)
Given:
Escape velocity on the surface, \( v_e = 11.2 \text{ km/s} \)
Initial velocity of the object, \( v_1 = 2v_e \)
We know the escape velocity formula: \( v_e = \sqrt{\frac{2GM}{R}} \), where \( G \) is the gravitational constant, \( M \) is Earth's mass, and \( R \) is Earth's radius.
From this, \( v_e^2 = \frac{2GM}{R} \implies \frac{GM}{R} = \frac{v_e^2}{2} \).
On the Earth's surface (initial state):
\( K_1 = \frac{1}{2} m v_1^2 = \frac{1}{2} m (2v_e)^2 = \frac{1}{2} m (4v_e^2) = 2mv_e^2 \)
\( U_1 = -\frac{GMm}{R} = -m \left(\frac{GM}{R}\right) = -m \left(\frac{v_e^2}{2}\right) = -\frac{1}{2} mv_e^2 \)
Since the object is thrown with a velocity greater than the escape velocity, it will escape Earth's gravitational field. Therefore, the maximum distance it reaches is infinity (\( r_{\text{max}} = \infty \)).
At infinity (final state):
\( U_2 = -\frac{GMm}{\infty} = 0 \)
Let the speed of the object at infinity be \( v_2 \).
\( K_2 = \frac{1}{2} m v_2^2 \)
Now, apply conservation of energy:
\( K_1 + U_1 = K_2 + U_2 \)
\( 2mv_e^2 - \frac{1}{2} mv_e^2 = \frac{1}{2} m v_2^2 + 0 \)
\( \frac{3}{2} mv_e^2 = \frac{1}{2} m v_2^2 \)
Divide both sides by \( \frac{1}{2} m \):
\( 3v_e^2 = v_2^2 \)
\( v_2 = \sqrt{3} v_e \)
Substitute the given value of escape velocity \( v_e = 11.2 \text{ km/s} \):
\( v_2 = \sqrt{3} \times 11.2 \text{ km/s} \)
\( v_2 \approx 1.732 \times 11.2 \text{ km/s} \)
\( v_2 \approx 19.3984 \text{ km/s} \)
Rounding to one decimal place, \( v_2 \approx 19.4 \text{ km/s} \).
So, the speed of the object at maximum distance (infinity) would be approximately 19.4 km/s. This residual velocity is what the object retains after overcoming Earth's gravity.
In simple words: An object is thrown from Earth at double the speed needed to escape gravity. We used energy rules to figure out its speed once it has fully escaped Earth's pull and is infinitely far away. Its final speed will be about 19.4 km/s.

🎯 Exam Tip: For problems involving objects escaping a gravitational field, always use the conservation of energy and remember that at "maximum distance" (infinity), the potential energy becomes zero.

 

Question 6. Three bodies of same masses are kept at the top of an equilateral triangle with side 'a'. What speed would the three bodies be moved in a circle so that the triangle moves in the circular orbit and there should be no change in the side of the triangle?
Answer:

The gravitational force between any two masses is given by \( F = \frac{GM^2}{a^2} \). The equilateral triangle has internal angles of 60°. When we consider the forces acting on a mass at one vertex (say, A) from the other two masses (B and C), the resultant force will point towards the center O of the triangle. This resultant force acts as the centripetal force, keeping the masses in a circular orbit without changing the triangle's shape.
From the right-angled triangle formed by one vertex, the center, and the midpoint of the opposite side, the distance from a vertex to the center (r) can be found.
In right-angled triangle ABD (where D is the midpoint of BC):
\( AD^2 + BD^2 = AB^2 \)
\( AD^2 + (\frac{a}{2})^2 = a^2 \)
\( AD^2 = a^2 - \frac{a^2}{4} = \frac{3a^2}{4} \)

\( \implies AD = \frac{a\sqrt{3}}{2} \)
By the geometrical property of a median in an equilateral triangle, the distance from a vertex to the centroid (O) is \( r = AO = \frac{2}{3} AD \).

\( \implies r = \frac{2}{3} \times \frac{a\sqrt{3}}{2} = \frac{a}{\sqrt{3}} \)
The centripetal force required for mass M to move in a circle of radius r is \( \frac{Mv^2}{r} \).
The gravitational forces on mass A from B and C are each \( F = \frac{GM^2}{a^2} \). The angle between these two forces is 60°. The resultant force (F_net) on mass A towards the center O is found using vector addition. \( F_{net} = 2F \cos(30^\circ) = 2 \times \frac{GM^2}{a^2} \times \frac{\sqrt{3}}{2} = \frac{\sqrt{3}GM^2}{a^2} \)
This resultant force provides the necessary centripetal force:
\( \frac{Mv^2}{r} = \frac{\sqrt{3}GM^2}{a^2} \)
Substitute \( r = \frac{a}{\sqrt{3}} \):
\( \frac{Mv^2}{(a/\sqrt{3})} = \frac{\sqrt{3}GM^2}{a^2} \)
\( \frac{\sqrt{3}Mv^2}{a} = \frac{\sqrt{3}GM^2}{a^2} \)
Divide by \( \sqrt{3}M/a \):
\( v^2 = \frac{GM}{a} \)

\( \implies v = \sqrt{\frac{GM}{a}} \)
This means the speed needed for the masses to move in a circle and maintain the triangle's shape is \( v = \sqrt{\frac{GM}{a}} \).
In simple words: For the three masses in an equilateral triangle to spin without changing their shape, the pull between them (gravity) must match the force needed to keep them moving in a circle. We find the distance from each mass to the center, then calculate the total gravitational pull towards that center. This pull equals the centripetal force, which helps us find the required speed.

🎯 Exam Tip: Remember that for an equilateral triangle, the centroid is also the circumcenter, and the distance from any vertex to the center is \( \frac{a}{\sqrt{3}} \). The resultant gravitational force always points towards the center.

 

Question 7. The potential energy of 3 kg body is -54 J. Calculate its escape velocity.
Answer:
Given:
Potential energy on the Earth's surface \( U_i = -54 \, J \)
Mass of the body \( m = 3 \, kg \)
If the body escapes, its potential energy at infinity \( U_f = 0 \).
The energy needed for escape is the difference between the final and initial potential energy:
Escape energy \( = U_f - U_i = 0 - (-54) = 54 \, J \)
This escape energy is provided by the kinetic energy of the body. The formula for kinetic energy is \( \frac{1}{2} m v_e^2 \), where \( v_e \) is the escape velocity.
So, \( \frac{1}{2} m v_e^2 = 54 \)
Substitute \( m = 3 \, kg \):
\( \frac{1}{2} \times 3 \times v_e^2 = 54 \)
\( 1.5 \times v_e^2 = 54 \)
\( v_e^2 = \frac{54}{1.5} \)
\( v_e^2 = 36 \)

\( \implies v_e = \sqrt{36} = 6 \, m/s \)
Thus, the escape velocity for this body is 6 m/s.
In simple words: We are given how much energy is holding the body to the Earth. To escape, it needs that exact amount of energy in motion (kinetic energy). We use the kinetic energy formula with the given mass to find the speed needed to break free.

🎯 Exam Tip: Always remember that potential energy in a gravitational field is typically negative, and escape energy is the positive amount needed to reach zero potential energy at infinity.

 

Question 9. 'Lagrangian Point' is a location in space between the Earth and the Sun; where the net gravitational force on any body due to both the Earth and the Sun is zero. Calculate the distance of this point from Earth. The distance between Earth and Sun is approximately \( 10^8 \, km \), and the mass of the Sun is \( 3.24 \times 10^5 \) times the mass of the Earth.
Answer:
Given:
Distance between Earth and Sun \( d = 10^8 \, km = 10^{11} \, m \)
Mass of Sun \( M_S = 3.24 \times 10^5 M_E \), where \( M_E \) is the mass of Earth.
Let \( x \) be the distance of the Lagrangian point P from Earth.
Then the distance of point P from the Sun is \( (d - x) \).
At the Lagrangian point, the gravitational force from the Earth and the Sun on a small test mass \( m \) are equal and opposite.
Force from Earth on \( m \): \( F_E = \frac{G M_E m}{x^2} \)
Force from Sun on \( m \): \( F_S = \frac{G M_S m}{(d-x)^2} \)
Equating these forces for the net force to be zero:
\( \frac{G M_E m}{x^2} = \frac{G M_S m}{(d-x)^2} \)
Cancel \( Gm \):
\( \frac{M_E}{x^2} = \frac{M_S}{(d-x)^2} \)
Substitute \( M_S = 3.24 \times 10^5 M_E \):
\( \frac{M_E}{x^2} = \frac{3.24 \times 10^5 M_E}{(d-x)^2} \)
Cancel \( M_E \):
\( \frac{1}{x^2} = \frac{3.24 \times 10^5}{(d-x)^2} \)
Take the square root of both sides:
\( \frac{1}{x} = \frac{\sqrt{3.24 \times 10^5}}{d-x} \)
\( \frac{1}{x} = \frac{18 \times 10 \sqrt{10}}{d-x} \)
\( \frac{1}{x} = \frac{180 \sqrt{10}}{d-x} \)
We know \( \sqrt{10} \approx 3.162 \). So, \( 180 \sqrt{10} \approx 180 \times 3.162 = 569.16 \).
\( \frac{1}{x} = \frac{569.16}{d-x} \)
\( d - x = 569.16x \)
\( d = 569.16x + x \)
\( d = (569.16 + 1)x \)
\( d = 570.16x \)
\( x = \frac{d}{570.16} \)
Substitute \( d = 10^{11} \, m \):
\( x = \frac{10^{11}}{570.16} \)
\( x \approx 1.75 \times 10^8 \, m \)
So, the distance of the Lagrangian point from Earth is approximately \( 1.75 \times 10^8 \) meters or 1.75 x 10^5 km. This point is crucial for stable orbits in space missions.
In simple words: A Lagrangian Point is where the push and pull of gravity from the Earth and Sun cancel each other out, so an object stays put. We use the given distances and mass ratios to set up a math problem where the Earth's gravity pull equals the Sun's gravity pull. Solving this helps us find how far this special spot is from the Earth.

🎯 Exam Tip: When dealing with gravitational force cancellation points, ensure you correctly set up the inverse square law for both forces and solve for the unknown distance. Approximations for square roots should be clear.

 

Question 10. Assume that any object is revolving around a big star in a circular orbit of radius R. Its time period is T. If the gravitational force between the object and the star is directly proportional to \( (R^{-5/2}) \) then how is its time period dependent on radius?
Answer:
Given that the gravitational force \( F_g \) is proportional to \( R^{-5/2} \):
\( F_g \propto R^{-5/2} \)
So, \( F_g = k R^{-5/2} \), where \( k \) is a constant.
For an object to revolve in a circular orbit, the gravitational force provides the necessary centripetal force \( F_c \).
\( F_c = \frac{mv^2}{R} \), where \( m \) is the mass of the object and \( v \) is its orbital velocity.
Also, the orbital velocity \( v \) is related to the time period \( T \) by \( v = \frac{2\pi R}{T} \).
So, \( F_c = \frac{m}{R} \left( \frac{2\pi R}{T} \right)^2 = \frac{m}{R} \frac{4\pi^2 R^2}{T^2} = \frac{4\pi^2 m R}{T^2} \)
Equating the gravitational force to the centripetal force:
\( k R^{-5/2} = \frac{4\pi^2 m R}{T^2} \)
Now, we want to find how \( T \) depends on \( R \). Let's rearrange the equation for \( T^2 \):
\( T^2 = \frac{4\pi^2 m R}{k R^{-5/2}} \)
\( T^2 = \frac{4\pi^2 m}{k} R^{1 - (-5/2)} \)
\( T^2 = \frac{4\pi^2 m}{k} R^{1 + 5/2} \)
\( T^2 = \frac{4\pi^2 m}{k} R^{7/2} \)
Since \( \frac{4\pi^2 m}{k} \) is a constant, we can write:
\( T^2 \propto R^{7/2} \)
Taking the square root of both sides:
\( T \propto \sqrt{R^{7/2}} \)

\( \implies T \propto R^{7/4} \)
Therefore, the time period of the object is proportional to \( R^{7/4} \). This shows how changing the orbit's radius greatly impacts the time it takes to complete one revolution.
In simple words: We are told how the pulling force changes with distance. We also know that the pulling force must be equal to the force that keeps the object moving in a circle. By putting these two ideas together and using the formula for speed in a circle, we can figure out how the time it takes to go around (time period) changes as the size of the circle changes.

🎯 Exam Tip: For problems involving proportionality, always start by defining the proportionality constant. Then equate the forces (gravitational and centripetal) and solve for the required variable in terms of the radius.

 

Question 11. Calculate the escape velocity on the Moon. Given Earth's radius is 4 times the radius of Moon and mass of Earth is 80 times of the Moon.
Answer:
Given:
Mass of Earth \( M_E = 80 M_m \) (where \( M_m \) is the mass of Moon)
Radius of Earth \( R_E = 4 R_m \) (where \( R_m \) is the radius of Moon)
Escape velocity on Earth \( v_e = 11.2 \, km/s \)
The formula for escape velocity is \( v_e = \sqrt{\frac{2GM}{R}} \).
For Earth:
\( v_E = \sqrt{\frac{2GM_E}{R_E}} \)
For Moon:
\( v_m = \sqrt{\frac{2GM_m}{R_m}} \)
To find \( v_m \), we can take the ratio of the escape velocities:
\( \frac{v_m}{v_E} = \frac{\sqrt{\frac{2GM_m}{R_m}}}{\sqrt{\frac{2GM_E}{R_E}}} \)
\( \frac{v_m}{v_E} = \sqrt{\frac{2GM_m}{R_m} \times \frac{R_E}{2GM_E}} \)
\( \frac{v_m}{v_E} = \sqrt{\frac{M_m}{R_m} \times \frac{R_E}{M_E}} \)
Now, substitute the given relations \( M_E = 80 M_m \) and \( R_E = 4 R_m \):
\( \frac{v_m}{v_E} = \sqrt{\frac{M_m}{R_m} \times \frac{4R_m}{80M_m}} \)
Cancel \( M_m \) and \( R_m \):
\( \frac{v_m}{v_E} = \sqrt{\frac{4}{80}} \)
\( \frac{v_m}{v_E} = \sqrt{\frac{1}{20}} \)
\( v_m = v_E \sqrt{\frac{1}{20}} = \frac{v_E}{\sqrt{20}} \)
Substitute \( v_E = 11.2 \, km/s \):
\( v_m = \frac{11.2}{\sqrt{20}} \)
Since \( \sqrt{20} \approx 4.47 \):
\( v_m = \frac{11.2}{4.47} \)
\( v_m \approx 2.50 \, km/s \)
The escape velocity on the Moon is much lower than on Earth, which is why objects can be launched from the Moon with less force.
In simple words: We know the escape speed for Earth and how the Moon's size and mass compare to Earth's. Using a formula that links escape speed to mass and radius, we set up a comparison. This helps us calculate the escape speed needed to leave the Moon.

🎯 Exam Tip: When dealing with ratios of physical quantities like escape velocity, always write down the general formula first. Then, create a ratio to cancel out common constants, simplifying calculations and reducing errors.

 

Question 12. A space laboratory whose mass is \( 2 \times 10^3 \, kg \) is transferred from an orbit of radius 2R to an orbit of radius 3R. Calculate the work done. Here, R = 6400 km (Earth's radius).
Answer:
Given:
Mass of space laboratory \( m = 2 \times 10^3 \, kg \)
Initial orbit radius \( r_1 = 2R \)
Final orbit radius \( r_2 = 3R \)
Earth's radius \( R = 6400 \, km = 6.4 \times 10^6 \, m \)
The work done \( W \) to change the orbit of a satellite is equal to the change in its total energy. For an object in orbit, the potential energy at a distance \( r \) from the center of Earth is \( U = -\frac{GMm}{r} \). The kinetic energy for a circular orbit is \( K = \frac{1}{2}mv^2 \), and since \( \frac{mv^2}{r} = \frac{GMm}{r^2} \), we have \( K = \frac{GMm}{2r} \).
Total energy \( E = U + K = -\frac{GMm}{r} + \frac{GMm}{2r} = -\frac{GMm}{2r} \).
Initial total energy \( E_1 = -\frac{GMm}{2r_1} = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R} \)
Final total energy \( E_2 = -\frac{GMm}{2r_2} = -\frac{GMm}{2(3R)} = -\frac{GMm}{6R} \)
Work done \( W = E_2 - E_1 = -\frac{GMm}{6R} - \left(-\frac{GMm}{4R}\right) \)
\( W = -\frac{GMm}{6R} + \frac{GMm}{4R} = GMm \left( \frac{1}{4R} - \frac{1}{6R} \right) \)
\( W = GMm \left( \frac{3 - 2}{12R} \right) = \frac{GMm}{12R} \)
We know that \( GM = gR^2 \), where \( g \) is the acceleration due to gravity on Earth's surface (\( g \approx 9.8 \, m/s^2 \)).
So, \( W = \frac{gR^2 m}{12R} = \frac{gRm}{12} \)
Substitute the given values:
\( g = 9.8 \, m/s^2 \)
\( R = 6.4 \times 10^6 \, m \)
\( m = 2 \times 10^3 \, kg \)
\( W = \frac{9.8 \times (6.4 \times 10^6) \times (2 \times 10^3)}{12} \)
\( W = \frac{9.8 \times 6.4 \times 2 \times 10^9}{12} \)
\( W = \frac{125.44 \times 10^9}{12} \)
\( W \approx 10.45 \times 10^9 \, J \)
\( W \approx 1.045 \times 10^{10} \, J \)
The work done to transfer the laboratory to a higher orbit is approximately \( 1.045 \times 10^{10} \) Joules. This positive work means energy must be supplied to move the satellite to a larger, higher energy orbit.
In simple words: We need to find how much energy is needed to move a lab from one orbit to another. We calculate the total energy of the lab in both orbits, which includes its speed and position energy. The difference between these two total energies tells us the work that must be done.

🎯 Exam Tip: For orbital transfer problems, always use the total mechanical energy formula for a satellite (\( E = -\frac{GMm}{2r} \)) and remember that work done is the change in total energy (\( W = E_{final} - E_{initial} \)).

 

Question 13. If radius of the Earth is 6400 km. Then what would be the linear velocity of any object at the equator?
Answer:
Given:
Radius of Earth \( R = 6400 \, km = 6.4 \times 10^6 \, m \)
Time period of Earth's rotation \( T = 24 \, hr \)
First, convert the time period to seconds:
\( T = 24 \, hr \times 60 \, min/hr \times 60 \, s/min = 86400 \, s \)
The linear velocity \( v \) of an object at the equator due to Earth's rotation is given by the formula:
\( v = R\omega \)
Where \( \omega \) is the angular velocity, which is \( \omega = \frac{2\pi}{T} \).
So, \( v = R \frac{2\pi}{T} \)
Substitute the values:
\( v = (6.4 \times 10^6 \, m) \times \frac{2 \times 3.14}{86400 \, s} \)
\( v = \frac{12.8 \times 3.14 \times 10^6}{86400} \, m/s \)
\( v = \frac{40.192 \times 10^6}{86400} \, m/s \)
\( v \approx 465.18 \, m/s \)
To convert to km/h:
\( v = 465.18 \, m/s \times \frac{3600 \, s}{1 \, hr} \times \frac{1 \, km}{1000 \, m} \)
\( v \approx 1674.65 \, km/hr \)
If converting to miles/h (using 1 mile = 1.609 km):
\( v = \frac{1674.65 \, km/hr}{1.609 \, km/mile} \approx 1040.7 \, mile/hr \)
The linear velocity of an object at the equator is approximately 1675 km/h, which is quite fast and contributes to the apparent flattening of Earth at the poles.
In simple words: The Earth spins around once every 24 hours. We know the size of the Earth. To find how fast a point on the equator is moving, we use a simple formula that relates the distance from the center (Earth's radius) to the time it takes to spin around once.

🎯 Exam Tip: Always ensure consistent units throughout your calculations. Convert all quantities to SI units (meters, seconds) before calculating, then convert back if a specific unit is requested for the final answer.

 

Question 14. How much would be the increase in the potential energy if a body of mass m is taken to a height R equal to the radius of the Earth?
Answer:
Given:
Mass of body \( m \)
Height \( h = R \) (where \( R \) is the radius of Earth)
The gravitational potential energy of a body of mass \( m \) at the Earth's surface (distance \( R \) from the center) is \( U_1 = -\frac{GMm}{R} \).
When the body is taken to a height \( h = R \) above the Earth's surface, its distance from the Earth's center becomes \( r = R + h = R + R = 2R \).
The potential energy at this new height is \( U_2 = -\frac{GMm}{2R} \).
The increase in potential energy (\( \Delta U \)) is the difference between the final and initial potential energy:
\( \Delta U = U_2 - U_1 = -\frac{GMm}{2R} - \left(-\frac{GMm}{R}\right) \)
\( \Delta U = -\frac{GMm}{2R} + \frac{GMm}{R} \)
To combine these, find a common denominator:
\( \Delta U = \frac{-GMm + 2GMm}{2R} \)
\( \Delta U = \frac{GMm}{2R} \)
We also know that \( GM = gR^2 \), where \( g \) is the acceleration due to gravity on the Earth's surface.
Substitute \( GM = gR^2 \) into the expression for \( \Delta U \):
\( \Delta U = \frac{gR^2 m}{2R} \)

\( \implies \Delta U = \frac{mgR}{2} \)
The increase in potential energy is \( \frac{mgR}{2} \). This energy is needed to lift the mass against gravity. This shows that potential energy increases as an object moves farther from the Earth's surface.
In simple words: When you lift something up, it gains potential energy. We need to find how much more energy a body has when lifted to a height equal to Earth's radius. We calculate its energy at the start and at the end, then find the difference, using formulas that involve Earth's mass and radius.

🎯 Exam Tip: Always define your initial and final positions clearly when calculating changes in potential energy. Remember that gravitational potential energy is negative and increases (becomes less negative) as the distance from the mass increases.

 

Question 15. A satellite is revolving at a distance x from the centre of the Earth. How much would be the increase in its speed if the radius of the circular orbit reduces by 1%?
Answer:
The orbital speed \( v_0 \) of a satellite revolving at a distance \( x \) from the center of the Earth is given by:
\( v_0 = \sqrt{\frac{GM}{x}} \)
We can also write this as: \( v_0 = (GM)^{1/2} x^{-1/2} \)
Here, \( GM \) is a constant.
To find the change in speed when \( x \) changes, we can use differentiation. Let's take the derivative of \( v_0 \) with respect to \( x \):
\( \frac{dv_0}{dx} = (GM)^{1/2} \times (-\frac{1}{2}) x^{-3/2} = -\frac{1}{2} (GM)^{1/2} x^{-3/2} \)
We can also write \( \frac{dv_0}{dx} = -\frac{1}{2} \frac{\sqrt{GM}}{x^{3/2}} = -\frac{1}{2} \frac{\sqrt{GM}}{x \sqrt{x}} = -\frac{1}{2} \frac{v_0}{x} \)
So, \( dv_0 = -\frac{1}{2} v_0 \frac{dx}{x} \)
We are interested in the percentage increase in speed. So, \( \frac{dv_0}{v_0} \times 100 \).
\( \frac{dv_0}{v_0} = -\frac{1}{2} \frac{dx}{x} \)
Multiply by 100 for percentage change:
\( \frac{\Delta v_0}{v_0} \times 100 = -\frac{1}{2} \left( \frac{\Delta x}{x} \times 100 \right) \)
Given that the radius of the circular orbit reduces by 1%. This means \( \frac{\Delta x}{x} \times 100 = -1\% \). The negative sign indicates a reduction.
So, \( \frac{\Delta v_0}{v_0} \times 100 = -\frac{1}{2} (-1\%) \)
\( \frac{\Delta v_0}{v_0} \times 100 = 0.5\% \)
The percentage increase in orbital velocity is 0.5%. This demonstrates that a smaller orbit requires a slightly faster orbital speed.
In simple words: When a satellite moves closer to Earth, its orbit becomes smaller. To stay in this smaller orbit, the satellite needs to speed up. We use a formula that connects speed to the orbit's size. By calculating how speed changes for a small change in orbit size, we find that a 1% smaller orbit makes the satellite move 0.5% faster.

🎯 Exam Tip: When dealing with percentage changes for quantities involving powers or roots, using proportional reasoning from derivatives (\( \frac{\Delta y}{y} = n \frac{\Delta x}{x} \) for \( y \propto x^n \)) can simplify calculations significantly.

 

Question. How much would be the radius of the Sun so that it becomes a Black Hole? Take the mass of the Sun constant ( \( 10^{30} \, kg \) ) and the maximum limit of velocity of projection should be equal to speed of the light because as given by Einstein the velocity of any object cannot be more than the speed of high. "Hint: Black Hole are those celestial bodies from which no object can escape because their gravitational acceleration is very much.]"
Answer:
Given:
Mass of the Sun \( M = 10^{30} \, kg \)
Gravitational constant \( G = 6.67 \times 10^{-11} \, Nm^2/kg^2 \)
Maximum limit of velocity of projection (speed of light) \( v_{max} = c = 3 \times 10^8 \, m/s \)
For a celestial body to become a black hole, the escape velocity from its surface must be equal to or greater than the speed of light.
The formula for escape velocity is \( v_e = \sqrt{\frac{2GM}{R}} \).
We need to find the radius \( R_{min} \) at which \( v_e = c \).
So, \( c = \sqrt{\frac{2GM}{R_{min}}} \)
Square both sides:
\( c^2 = \frac{2GM}{R_{min}} \)
Rearrange to solve for \( R_{min} \):
\( R_{min} = \frac{2GM}{c^2} \)
Substitute the given values:
\( R_{min} = \frac{2 \times (6.67 \times 10^{-11}) \times (10^{30})}{(3 \times 10^8)^2} \)
\( R_{min} = \frac{13.34 \times 10^{19}}{9 \times 10^{16}} \)
\( R_{min} = \frac{13.34}{9} \times 10^{19-16} \)
\( R_{min} = 1.482 \times 10^3 \, m \)
So, if the Sun were to shrink to a radius of approximately 1.482 kilometers, it would become a black hole. This incredibly small size for such a massive object highlights the extreme density required for a black hole.
In simple words: A black hole is formed when something is so heavy and dense that nothing, not even light, can escape its pull. We use the Sun's mass and the speed of light in a special formula to figure out how small the Sun would have to get to become a black hole.

🎯 Exam Tip: Remember that the critical radius for an object to become a black hole (Schwarzschild radius) is directly proportional to its mass. Always equate the escape velocity to the speed of light to solve this type of problem.

 

Question 17. If an object is placed in a tunnel running through the Earth's centre, it does simple harmonic motion. Calculate the time period for this. If radius of Earth is \( 6.4 \times 10^6 \, m \) and mass is \( 6 \times 10^{24} \, kg \).
Answer:
When an object is placed in a tunnel running through the Earth's center, it undergoes simple harmonic motion because the gravitational force inside the Earth is proportional to the displacement from the center. The time period \( T \) for such motion is given by:
\( T = 2\pi \sqrt{\frac{R}{g}} \)
Where \( R \) is the radius of the Earth and \( g \) is the acceleration due to gravity at the Earth's surface.
Given:
Radius of Earth \( R = 6.4 \times 10^6 \, m \)
Mass of Earth \( M = 6 \times 10^{24} \, kg \)
We need the value of \( g \). We know \( g = \frac{GM}{R^2} \).
\( G = 6.67 \times 10^{-11} \, Nm^2/kg^2 \)
\( g = \frac{(6.67 \times 10^{-11}) \times (6 \times 10^{24})}{(6.4 \times 10^6)^2} \)
\( g = \frac{40.02 \times 10^{13}}{40.96 \times 10^{12}} \)
\( g \approx 0.977 \times 10 = 9.77 \, m/s^2 \)
Now, substitute \( R \) and \( g \) into the time period formula:
\( T = 2\pi \sqrt{\frac{6.4 \times 10^6}{9.77}} \)
\( T = 2\pi \sqrt{0.655 \times 10^6} \)
\( T = 2\pi \times \sqrt{655000} \)
\( T = 2\pi \times 809.3 \)
\( T \approx 2 \times 3.14159 \times 809.3 \)
\( T \approx 5085 \, s \)
Convert seconds to minutes:
\( T = \frac{5085}{60} \, minutes \)
\( T \approx 84.75 \, minutes \)
So, the time period for an object oscillating through the Earth's center would be approximately 84.75 minutes. Interestingly, this is roughly the orbital period of a low-Earth-orbit satellite.
In simple words: If you drop something through a tunnel straight through the Earth, it will swing back and forth like a pendulum. We can calculate how long one full swing takes using the Earth's size and the strength of its gravity.

🎯 Exam Tip: For objects in tunnels through Earth, remember the simple harmonic motion formula \( T = 2\pi \sqrt{\frac{R}{g}} \). Ensure you use the correct values for Earth's radius and surface gravity.

 

Question 18. What would be the change in escape velocity, if the radius of the Earth reduces 4% and mass remains constant?
Answer:
The escape velocity \( v_e \) is given by the formula:
\( v_e = \sqrt{\frac{2GM}{R}} \)
We can write this as \( v_e = (2GM)^{1/2} R^{-1/2} \).
Since \( G \) and \( M \) (mass of Earth) are constant, we can see that \( v_e \) is proportional to \( R^{-1/2} \).
This means \( v_e \propto R^{-1/2} \).
For small changes, the percentage change in \( v_e \) is related to the percentage change in \( R \) by:
\( \frac{\Delta v_e}{v_e} \times 100 = (-\frac{1}{2}) \left( \frac{\Delta R}{R} \times 100 \right) \)
Given that the radius of the Earth reduces by 4%, so \( \frac{\Delta R}{R} \times 100 = -4\% \). (The negative sign indicates reduction).
Substitute this value into the equation:
\( \frac{\Delta v_e}{v_e} \times 100 = (-\frac{1}{2}) (-4\%) \)
\( \frac{\Delta v_e}{v_e} \times 100 = +2\% \)
So, if the radius of the Earth reduces by 4% while its mass remains constant, the escape velocity would increase by 2%. This implies that if Earth became smaller but kept the same mass, it would be harder to escape its gravity.
In simple words: We know the formula for the speed needed to escape Earth's gravity. If Earth shrinks but keeps its total weight, the escape speed will change. Using a special rule for how changes in size affect the escape speed, we find that if the Earth's size goes down by 4%, the escape speed will go up by 2%.

🎯 Exam Tip: When a quantity is proportional to a power of another (e.g., \( y \propto x^n \)), the percentage change in \( y \) is approximately \( n \) times the percentage change in \( x \). Pay attention to the signs for increase (+) or decrease (-).