CBSE Class 9 Physics Gravitation Worksheets Set 01

Very Short Answer Questions and Numerical Problems on Gravitation

Question 1. Who gave the universal law of gravitation?
Answer:
Isaac Newton.
In simple words: Newton stated the law.

Teacher's Note:
1. The law is called universal because it applies to all bodies.

Question 2. What do you mean by gravitation?
Answer:
Gravitation is the force of attraction between any two particles or bodies in the universe.
In simple words: Every mass attracts every other mass.

Teacher's Note:
1. The force is always attractive.

Question 3. State the universal law of gravitation.
Answer:
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 them.
F = G m₁m₂ / r²
In simple words: Force grows with the masses and falls with the square of the distance.

Teacher's Note:
1. G is the universal gravitational constant.

Question 4. Write an expression for the gravitational force between the earth and a body on the earth.
Answer:
F = GMm/R², where M is the mass of the earth, m is the mass of the body, R is the radius of the earth and G is the gravitational constant.
In simple words: Use the law of gravitation with the earth's mass and radius.

Teacher's Note:
1. This force is the weight of the body.

Question 5. The distance between two particles is decreased. Will the gravitational force between them increase or decrease?
Answer:
It increases, because F ∝ 1/r².
In simple words: Closer means stronger.

Teacher's Note:
1. Inverse square law.

Question 6. An iron ball exerts a force F on a rubber ball. How much force will the rubber ball exert on the iron ball?
Answer:
The same force F, in the opposite direction. By Newton's third law the forces between two bodies are equal and opposite.
In simple words: Action and reaction are equal.

Teacher's Note:
1. Gravitational forces always occur in equal and opposite pairs.

Question 7. Define the universal gravitational constant (G).
Answer:
G is the force of attraction between two objects of unit mass (1 kg each) separated by unit distance (1 m).
In simple words: G is the force between two 1 kg masses 1 m apart.

Teacher's Note:
1. Its value is 6.673 × 10⁻¹¹ N m²/kg².

Question 8. Write down the unit of G in SI. What is its value?
Answer:
The SI unit of G is N m² kg⁻².
G = 6.673 × 10⁻¹¹ N m² kg⁻².
In simple words: The unit follows from F = Gm₁m₂/r².

Teacher's Note:
1. G is the same everywhere in the universe.

Question 9. The value of G = 6.673 × 10⁻¹¹ N m² kg⁻² on the surface of the earth. What is the value of G on the surface of the moon?
Answer:
The same, 6.673 × 10⁻¹¹ N m² kg⁻², because G is a universal constant.
In simple words: G does not change from place to place.

Teacher's Note:
1. Do not confuse G with g.

Question 10. The gravitational force between the earth and an object is known as ______.
Answer:
Gravity (the force of gravity).
In simple words: The earth's pull on an object is gravity.

Teacher's Note:
1. It is the weight of the object.

Question 11. The gravitational force between two objects on the earth is 2 N. What will be the gravitational force between these two objects on the surface of the moon?
Answer:
2 N. The force depends only on G, the masses and the distance, none of which change.
In simple words: The force is the same.

Teacher's Note:
1. Only the weight changes on the moon, not the mutual attraction of two objects.

Question 12. Tides in oceans are formed due to gravitational force between ______ and ______.
Answer:
The moon and the earth (with the sun also contributing).
In simple words: The moon's pull raises the sea water.

Teacher's Note:
1. Spring and neap tides depend on the positions of the sun and moon.

Question 13. Name the force which is responsible for the revolution of planets around the sun.
Answer:
The gravitational force of attraction between the sun and the planet; it provides the centripetal force.
In simple words: Gravity keeps planets in orbit.

Teacher's Note:
1. Centripetal force for orbits comes from gravitation.

Question 14. Name the force which is responsible for the revolution of the moon around the earth.
Answer:
The gravitational force between the earth and the moon, which supplies the centripetal force.
In simple words: Gravity keeps the moon in orbit.

Teacher's Note:
1. The earth attracts the moon.

Question 15. How does the gravitational force between two bodies vary with r?
Answer:
It varies inversely as the square of the distance, F ∝ 1/r².
In simple words: Double the distance and the force becomes one-fourth.

Teacher's Note:
1. Inverse square law.

Question 16. Is the acceleration due to gravity on the surface of the moon less or more than the acceleration due to gravity on the surface of the earth?
Answer:
Less. On the moon g is about 1.63 m/s², which is nearly one-sixth of g on the earth.
In simple words: Moon's g is one-sixth of earth's.

Teacher's Note:
1. The moon has a smaller mass and radius.

Question 17. Write the expression for the acceleration due to gravity.
Answer:
g = GM/R², where M is the mass and R the radius of the earth.
In simple words: g depends on the mass and radius of the planet.

Teacher's Note:
1. It does not depend on the mass of the falling body.

Question 18. What is the SI unit of acceleration due to gravity (g)?
Answer:
m s⁻².
In simple words: It is an acceleration.

Teacher's Note:
1. 1 m/s² = 100 cm/s².

Question 19. What is the value of acceleration due to gravity on the surface of the earth?
Answer:
g = 9.8 m s⁻².
In simple words: About 9.8 m/s².

Teacher's Note:
1. It is often rounded to 10 m/s² in calculations.

Question 20. Is acceleration due to gravity a scalar or a vector quantity?
Answer:
A vector quantity, directed towards the centre of the earth.
In simple words: It has direction.

Teacher's Note:
1. Always pointing downward.

Question 21. What is the value of acceleration due to gravity on the surface of the moon?
Answer:
About 1.6 m s⁻² (1.63 m s⁻²).
In simple words: One-sixth of the earth's value.

Teacher's Note:
1. g = GM/R² with the moon's M and R.

Question 22. Acceleration due to gravity on the surface of the moon = ______ × acceleration due to gravity on the surface of the earth.
Answer:
1/6.
In simple words: The moon's g is one-sixth.

Teacher's Note:
1. This is why astronauts seem to float on the moon.

Question 23. Where is the acceleration due to gravity more: at the poles or at the equator?
Answer:
At the poles. The earth is flattened at the poles (smaller radius) and the rotation reduces g at the equator.
In simple words: Poles have the larger g.

Teacher's Note:
1. g at the poles is about 9.83 m/s², at the equator about 9.78 m/s².

Question 24. What is the value of g at the centre of the earth?
Answer:
Zero.
In simple words: There is no net pull at the centre.

Teacher's Note:
1. Weight is therefore zero at the centre.

Question 25. What will be the direction of acceleration due to gravity if the body is thrown vertically upward?
Answer:
Always downward, towards the centre of the earth, whatever the direction of motion.
In simple words: Gravity always acts downward.

Teacher's Note:
1. It makes the upward speed decrease.

Question 26. Define the mass of a body.
Answer:
Mass is the quantity of matter contained in a body.
In simple words: Mass tells how much matter there is.

Teacher's Note:
1. Its SI unit is the kilogram.

Question 27. Why is mass known as inertial mass?
Answer:
Mass is a measure of the inertia of a body, that is, of its resistance to a change in its state of motion, so it is called inertial mass.
In simple words: More mass means more inertia.

Teacher's Note:
1. A heavier body is harder to accelerate.

Question 28. What is the SI unit of mass? Is mass a scalar or a vector?
Answer:
The SI unit of mass is the kilogram. Mass is a scalar quantity.
In simple words: Mass has only magnitude.

Teacher's Note:
1. Weight, in contrast, is a vector.

Question 29. The mass of a body at the equator is 20 kg. What will be the mass of this body at the poles?
Answer:
20 kg. Mass does not change from place to place.
In simple words: Mass is constant.

Teacher's Note:
1. Weight changes slightly with g, mass does not.

Question 30. Define the weight of a body. Write a relation between mass and weight.
Answer:
Weight is the force with which a body is attracted by the earth.
W = mg
In simple words: Weight is mass times g.

Teacher's Note:
1. The SI unit of weight is the newton.

Question 31. What is the gravitational acceleration of a spaceship at a distance equal to two earth radii from the centre of the earth?
Answer:
g' = g (R/2R)² = g/4.
g' = 9.8/4 = 2.45 m/s².
In simple words: Double the distance from the centre reduces g to one-fourth.

Teacher's Note:
1. Distance from the centre matters, not height.

Question 32. A boy on a cliff 49 m high drops a stone. One second later, he throws a second stone after the first. They both hit the ground at the same time. With what speed did he throw the second stone?
Answer:
First stone: 49 = ½ × 9.8 × t², so t = √10 = 3.16 s.
The second stone is in the air for 3.16 − 1 = 2.16 s: 49 = u(2.16) + ½ × 9.8 × (2.16)², so 49 = 2.16u + 22.9.
u = 26.1/2.16 = 12.1 m/s.
In simple words: Use the equation of motion for both stones with the same fall height.

Teacher's Note:
1. The second stone has a shorter time of flight.

Question 33. A stone drops from the edge of a roof. It passes a window 2 m high in 0.1 s. How far is the roof above the top of the window?
Answer:
Let the speed at the top of the window be v: 2 = v(0.1) + ½ × 9.8 × (0.1)², so v = 19.5 m/s.
Height above the window top = v²/2g = (19.5)²/(2 × 9.8) = 19.4 m.
In simple words: Find the speed at the top of the window and then the height it fell.

Teacher's Note:
1. Free fall from rest: v² = 2gh.

Question 34. A particle is dropped from a tower 180 m high. How long does it take to reach the ground? What is the velocity when it touches the ground? Take g = 10 m/s².
Answer:
180 = ½ × 10 × t², so t² = 36 and t = 6 s.
v = gt = 10 × 6 = 60 m/s.
In simple words: Use h = ½gt² and v = gt.

Teacher's Note:
1. Starting from rest.

Question 35. To estimate the height of a bridge over a river, a stone is dropped freely from the bridge. The stone takes 2 s to touch the water surface in the river. Calculate the height of the bridge from the water level. Take g = 9.8 m/s².
Answer:
h = ½gt² = ½ × 9.8 × 2² = 19.6 m.
In simple words: Height = half g times the square of the time.

Teacher's Note:
1. The initial speed is zero.

Question 36. How much would a 70 kg man weigh on the moon? What will be his mass on the earth and on the moon? Given g on the moon = 1.7 m/s².
Answer:
Weight on the moon = 70 × 1.7 = 119 N.
His mass is 70 kg both on the earth and on the moon.
In simple words: Mass stays the same; weight changes with g.

Teacher's Note:
1. W = mg.

Question 37. A body has a weight of 10 kg wt on the earth. What will be its mass and weight when taken to the centre of the earth?
Answer:
Mass = 10 kg, since mass does not change.
At the centre of the earth g = 0, so the weight is zero.
In simple words: Mass remains; weight becomes zero.

Teacher's Note:
1. g is zero at the centre.

Question 38. A force of 2 kg wt acts on a body of mass 4.9 kg. Calculate its acceleration.
Answer:
F = 2 kg wt = 2 × 9.8 = 19.6 N.
a = F/m = 19.6/4.9 = 4 m/s².
In simple words: Convert the force to newtons first.

Teacher's Note:
1. F = ma.

Question 39. A force of 20 N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration?
Answer:
m = W/g = 9.8/9.8 = 1 kg.
a = F/m = 20/1 = 20 m/s².
In simple words: Weight divided by g gives the mass.

Teacher's Note:
1. The 20 N force is the net force.

Question 40. A man weighs 600 N on the earth. What is his mass? Take g = 10 m/s². If he were taken to the moon, his weight would be 100 N. What is his mass on the moon? What is the acceleration due to gravity on the moon?
Answer:
Mass = 600/10 = 60 kg.
Mass on the moon = 60 kg (mass is constant).
g on the moon = 100/60 = 1.67 m/s².
In simple words: Mass is the same; g on the moon follows from weight / mass.

Teacher's Note:
1. W = mg on each body.

Question 41. A car falls off a ledge and drops to the ground in 0.5 s. Let g = 10 m/s². (i) What is its speed on striking the ground? (ii) What is its average speed during the 0.5 s? (iii) How high is the ledge from the ground?
Answer:
(i) v = gt = 10 × 0.5 = 5 m/s.
(ii) Average speed = (0 + 5)/2 = 2.5 m/s.
(iii) Height = average speed × time = 2.5 × 0.5 = 1.25 m.
In simple words: Free fall from rest: v = gt and h = average speed × t.

Teacher's Note:
1. Uniform acceleration gives the average speed as (u + v)/2.

Question 42. An object is thrown vertically upwards and rises to a height of 10 m. Calculate (i) the velocity with which the object was thrown upwards and (ii) the time taken by the object to reach the highest point.
Answer:
(i) v² = u² − 2gh = 0, so u = √(2 × 9.8 × 10) = 14 m/s.
(ii) t = u/g = 14/9.8 = 1.43 s.
In simple words: At the top the speed is zero.

Teacher's Note:
1. Use v = u − gt.

Question 43. The mass of an object is 10 kg. What is its weight on the earth?
Answer:
W = mg = 10 × 9.8 = 98 N.
In simple words: Weight is mass times g.

Teacher's Note:
1. Weight is a force measured in newtons.

Question 44. An object weighs 10 N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon?
Answer:
Weight on the moon = (1/6) × 10 = 1.67 N.
In simple words: The moon's g is one-sixth.

Teacher's Note:
1. Mass is unchanged.

Question 45. Calculate the value of the acceleration due to gravity on the moon. Given: mass of the moon = 7.4 × 10²² kg, radius of the moon = 1740 km.
Answer:
g = GM/R² = 6.67 × 10⁻¹¹ × 7.4 × 10²² / (1.74 × 10⁶)²
g = 4.94 × 10¹² / 3.03 × 10¹² = 1.63 m/s².
In simple words: Use g = GM/R² with the moon's data.

Teacher's Note:
1. R must be in metres.

Question 46. Suppose a planet exists whose mass and radius are both half those of the earth. Calculate the acceleration due to gravity on the surface of this planet.
Answer:
g' = G(M/2)/(R/2)² = 2GM/R² = 2g.
g' = 2 × 9.8 = 19.6 m/s².
In simple words: Half the mass but a quarter of R² gives twice g.

Teacher's Note:
1. g ∝ M/R².

Question 47. A ball is thrown up with a speed of 15 m/s. How high will it go before it begins to fall? Take g = 9.8 m/s².
Answer:
h = u²/(2g) = 15²/(2 × 9.8) = 225/19.6 = 11.48 m.
In simple words: At the highest point v = 0.

Teacher's Note:
1. Use v² = u² − 2gh.

Question 48. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Answer:
F ∝ 1/r², so when the distance is halved the force becomes 4 times (4F).
In simple words: Halving the distance quadruples the force.

Teacher's Note:
1. Inverse square law.

Question 49. Gravitational force acts on all objects in proportion to their masses. Why then does a heavy object not fall faster than a light object?
Answer:
The acceleration is a = F/m. The force on the body is proportional to its mass (F = GMm/R²), so the mass cancels: a = GM/R² = g.
Hence all bodies fall with the same acceleration (in the absence of air resistance), and a heavy object does not fall faster.
In simple words: A bigger pull is shared by a bigger mass, so the acceleration is the same.

Teacher's Note:
1. Galileo's result: all bodies fall equally in vacuum.

Question 50. What is the magnitude of the gravitational force between the earth and a 1 kg object on its surface? (Mass of the earth is 6 × 10²⁴ kg and radius of the earth is 6.4 × 10⁶ m.)
Answer:
F = GMm/R² = 6.67 × 10⁻¹¹ × 6 × 10²⁴ × 1 / (6.4 × 10⁶)²
F = 4.0 × 10¹⁴ / 4.096 × 10¹³ = 9.77 N (about 9.8 N).
In simple words: This is the weight of a 1 kg mass.

Teacher's Note:
1. Equal to mg for m = 1 kg.