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Revision Notes for Class 9 Science Chapter 10 Gravitation
Class 9 Science students should refer to the following concepts and notes for Chapter 10 Gravitation in Class 9. These exam notes for Class 9 Science will be very useful for upcoming class tests and examinations and help you to score good marks
Chapter 10 Gravitation Notes Class 9 Science
GRAVITATION
NEWTON’S UNIVERSAL LAW OF GRAVITATION
Every object in the Universe attracts every other object with a force which is
(i) directly proportional to the product of their masses, and
(ii) inversely proportional to the square of the distance between their centres. The direction of
the force is along the line joining the centres of two objects.
F ∝ m1m2 and F∝ 1/ r2 ⇒ F∝ m1m2 / r2
F= G.m1 m2 / r2 where G is a constant proportionality and is called universal gravitational constant.
UNITS AND VALUE OF GRAVITATIONAL CONSTANT
F=Gm1m2/r2
G.m1m2=Fr2 ⇒ G= F.r2 /m1m2
Unit of G is Nm2/kg2.
If m1 = m2 = 1 kg, r = 1 then we have G = F
Hence, Universal gravitational constant G is numerically equal to the gravitational force of attraction between two bodies, each of unit mass kept at unit distance from each other.
Value of G = 6.67 x 10–11 Nm2/kg2
KEPLER’S LAWS OF PLANETARY MOTION
KEPLER’S FIRST LAW
Every planet revolves around the Sun in an elliptical orbit, with the sun situated at any one of the foci of the ellipse.
KEPLER’S SECOND LAW
In the elliptical orbit of the planet, the line joining the centre of the planet to the centre of the Sun sweeps equal intervals of time.
KEPLER’S THIRD LAW
The square of time period of revolution of a planet around the Sun is directly proportional to the cube of the semi-major axis of the elliptical orbit.
r3 ∝ T2 ⇒ r3/T2=constant
where r = radius of orbit = mean distance of planet from the Sun (inm), T = the time period of revolution of planet around the Sun (in second)
IMPORTANCE OF THE UNIVERSAL LAW OF GRAVITATION
The universal law of gravitation successfully explained several phenomena which were believed to be unconnected:
(i) the force that binds us to the earth;
(ii) the motion of the moon around the earth;
(iii) the motion of planets around the Sun; and
(iv) the tides due to the moon and the Sun.
INTEXT QUESTIONS
1. State the universal law of gravitation
Ans. The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. For two objects of masses m1 and m2 and the distance between them r, the force (F) of attraction acting between them is given by the universal law of gravitation as:
F= Gm1m2 / r2
Where, G is the universal gravitation constant given by: G=6.67*10-11Nm2 / kg2
2. Write the formula to find the magnitude of the gravitational force between the earth and an object on the surface of the earth.
Ans. Let ME be the mass of the Earth and m be the mass of an object on its surface. If R is the radius of the Earth, then according to the universal law of gravitation, the gravitational force (F) acting between the Earth and the object is given by the relation.
F Gm1m2/r2
FREE FALL
When an object falls from any height under the influence of gravitational force only, it is known as free fall. In the case of free fall no change of direction takes place but the magnitude of velocity changes because of acceleration. This acceleration acts because of the force of gravitation and is denoted by ‘g’. This is called acceleration due to gravity.
EXPRESSION FOR ACCELERATION DUE TO GRAVITATION ‘G’.
Let mass of the object put under free fall = m.
And acceleration due to gravity = g.
Therefore, according to Newton’s Second Law of Motion which states that Force is the product of mass and acceleration,
F = m x g -----------------(i)
Now, according to Universal Law of gravitation,
F G.M.m/d2……………. (ii)
Thus, from above two expressions, we get
Where, g is acceleration due to gravity,
G is the Universal Gravitational Constant.
M is the mass of earth.
And d is the distance between object and centre of earth.
WHEN OBJECT IS NEAR THE SURFACE OF EARTH
When an object is near the surface of earth, the distance between object and centre of the earth will be equal to the radius of earth because the distance of object is negligible in comparison of the radius of earth.
Let the radius of earth is equal to R.
Therefore, after substituting ‘R’ at the place of ‘d’ we get,
g = GM/R2 .................(iv)
Since, earth is not a perfect sphere rather it has oblique shape. Therefore, radius at the equator is greater than at the poles.
Since, value of ‘g’ is reciprocal of the square of radius of earth, thus, the value of ‘g’ will be greater at the poles and less at the equator.
And the value of ‘g’ will decrease with increase of distance of object from earth.
Calculation of value of ‘g’
We know that
The accepted value of G is 6.673 x 10–11Nm2/kg2.
The mass of earth, M = 6 x 1024 kg
The radius of earth, R = 6.4 x 106 m
Therefore, by using expression, g = GM/R2 , the value of ‘g’ can be calculated.
Therefore, after substituting the value of G, M and R in the expression for ‘g’ we get.
g = 6.673´10-11 ´* 6 *´1024 / (6.4 *´106 )2
⇒ g = 9.8m / s2
Motion of an object under the influence of gravitational force of earth
The expression for ‘g’ is written as
g = GM / R2
Since, the value of ‘g’ does not depend upon the mass or distance of an object, therefore, all objects fall over the earth with the same rate.
The equations for motion are as follows:
v = u + at ……….. (i)
s = ut + 1/2 at2 ………. (ii)
v2 = u2 + 2as ……… (iii)
Therefore, the equations of motion are also applied to calculate the velocity, distance, etc by replacing ‘a’ by ‘g’. After substituting ‘g’ at the place of ‘a’ we get above equations as follows:
v = u + gt ……….. (iv)
s = ut + 1/2 gt2 ………. (v)
v2 = u2 + 2gs ……… (vi)
In the calculation; initial velocity (u), final velocity (v), time taken (t), or distance covered (s), the value of ‘g’ is taken as positive in the case of object moving towards earth and taken as negative in the case of object is thrown in opposite direction of earth.
MASS
Mass is the measurement of inertia and inertia is the property of any object which opposes the change in state of the object. It is inertia because of which an object in rest has tendency to remain in rest and an object in motion has tendency to remain in motion.
Inertia depends upon the mass of an object. Object having greater mass has greater inertia and vice versa. Mass of an object remains constant everywhere, i.e. mass will remain same whether that object is at the moon, at the earth or anywhere in the universe.
WEIGHT:
Earth attracts every object towards it. We know that force is the product of mass and acceleration due to gravity.
This means, F = m x g -----------------------(i)
The force by which earth attracts an object towards it is called the weight of the object, which is the product of mass (m) of the object and acceleration due to gravity (g). Weight is denoted by ‘W’.
Therefore, by substituting in the expression ‘F = mg’ we get, W = m x g ------------------------------(ii)
Since weight is the force which is acting vertically downwards, therefore, weight has both magnitude and direction and hence it is a vector quantity.
Since the value of ‘g’ is always constant at a given place, Therefore, expression ‘W = m x g’ can be written as follows:
W α m --------------------(iii)
This means weight of any object is directly proportional to its mass, i.e. weight will increase with the increase of mass and decrease with decrease in mass.
This is the cause that weight of any object is the measure of its mass.
The unit of weight
Since, weight of an object is equal to the force by which an object is attracted towards earth, therefore, unit of weight is same as the unit of force.
Therefore, Unit of weight is ‘newton (N)’.
WEIGHT OF AN OBJECT ON THE SURFACE OF MOON
Let ME be the mass of the Earth and m be an object on the surface of the Earth. Let RE be the radius of the Earth. According to the universal law of gravitation, weight WE of the object on the surface of the Earth is given by,
NUMERICAL
1. The gravitational force between two objects is F. How will this force change when (i) distance between them is reduced to half (ii) the mass of each object is quadrupled?
2. A sphere of mass 40kg is attracted by a second sphere of mass 15kg when their centres are 20 cm apart, with a force of 0.1 milligram weight. Calculate the value of gravitational constant.
3. A body of mass 1 kg is placed at a distance of 2m from another body of mass 10kg. At what distance from the body of 1 kg, another body of mass 5 kg be placed so that the net force of gravitation acting on the body of mass 1 kg is zero?
4. A geostationary satellite is orbiting the earth at a height 5 R above the surface of earth, where R is the radius of earth. Find the time period of another satellite at a height of 2 R from the surface of earth.
5. The distance of planet Jupiter from the sun 5.2 times that of Earth. Find the period of revolution of Jupiter around sun.
6. If the distance of Earth from the Sun were half the present value, how many days will make one year?
7. Two satellites of a planet have periods 32 days and 256 days. If the radius of orbit of former is R, find the orbital radius of the latter.
8. The mass of Earth is 6 x 1024 kg and that of moon is 7.4 x 1022 kg. If the distance between the Earth and the Moon is 3.84 x 105 km, calculate the force exerted by Earth on the Moon. Given G = 6.7 x 10–11 Nm2/kg2.
9. If the distance between two masses is increased by a factor of 4, by what factor would the mass of one of them have to be altered to maintain the same gravitational force?
10. Two bodies A and B having masses 2kg and 4kg respectively are separated by 2m. Where should a body of mass 1kg be placed so that the gravitational force on this body due to bodies A and B is zero?
11. The mass of Sun is 2 x 1030 kg and mass of Earth is 6 x 1024 kg. If the distance between the centres of Sun and Earth is 1.5 x 108 km, calculate the force of gravitation between them.
12. Two electrons each of mass 9.1 x 10–31 kg are at a distance of 10 A . Calculate the gravitational force of attraction between them. Given 1A = 10-10 m
13. The gravitational force between force two objects is 100 N. How should the distance between these objects be changed so that force between them becomes 50 N?
14. Calculate the force of gravitation between two objects of masses 80kg and 1200 kg kept at a distance of 10 m from each other. Given G = 6.67 x 10–11 Nm2/kg2.
15. Calculate the force of attraction between the Earth and the Sun, given that the mass of Earth is 6 x 1024 kg and that of sun is 2 x 1030 kg. The average distance between thte two is 1.5 x 1011m.
16. A sphere of mass 25kg attracts another sphere of mass 24kg with a force of 0.1milligram weight. If distance between the centres of two spheres is 20cm, what is the value of G?
17. If distance between two masses is quadrupled, what will be the new force of attraction between them? Given the initial gravitational pull is 9.8N.
18. An electron of mass 9.1 x 10–31kg is at a distance of 10 A from a proton of mass 1.67 x 10–27kg. Calculate the gravitational force of attraction between them. Given 1A = 10-10 m
19. Two bodies A and B having masses 20kg and 40kg are separted by 10m. At what distance from body A should another body C of mass 15kg be placed so that net gravitational force on C is zero?
20. Calculate the gravitational force on a body of mass 1kg lying on the surface of earth. Given mass of earth is 6 x 1024 kg and radius of earth is 6400km.
INTEXT QUESTIONS
1. What do you mean by free fall?
Ans. Gravity of the Earth attracts every object towards its centre. When an object is released from a height, it falls towards the surface of the Earth under the influence of gravitational force. The motion of the object is said to have free fall.
2. What do you mean by acceleration due to gravity?
Ans. When an object falls towards the ground from a height, then its velocity changes during the fall. This changing velocity produces acceleration in the object. This acceleration is known as acceleration due to gravity (g). Its value is given by 9.8 m/s2.
INTEXT QUESTIONS
1. What are the differences between the mass of an object and its weight?
2. Why is the weight of an object on the moon 1th/6 its weight on the earth?
Ans. Let ME be the mass of the Earth and m be an object on the surface of the Earth. Let RE be the radius of the Earth. According to the universal law of gravitation, weight WE of the object on the surface of the Earth is given by,
NUMERICAL
1. Calculate the force of gravity acting on your friend of mass 60kg. Given mass of earth = 6 x 1024 kg and radius of Earth = 6.4 x 106m.
2. Mass of an object is 10kg. What is its weight on Earth?
3. What is the mass of an object whose weight is 49N?
4. An object weighs 10N when measured on the surface of the earth. What would be its weight when measured on the surface of the Moon?
5. An object is thrown vertically upwards and rises to a height of 10m. Calculate (i) the velocity with which the object was thrown upwards and (ii) the time taken by the object to reach the highest point.
6. A force of 2 kg wt. acts on a body of mass 4.9kg. Calculate its acceleration.
7. A force of 20N acts upon a body weight is 9.8N. What is the mass of the body and how much is its acceleration?
8. A body has a weight of 10 kg on the surface of earth. What will be its mass and weight when taken to the centre of earth?
9. How much would a 70 kg man weigh on moon? What will be his mass on earth and moon? Given g on moon = 1.7 m/s2.
10. The Earth’s gravitational force causes an acceleration of 5 m/s2 in a 1 kg mass somewhere in space. How much will the acceleration of a 3 kg mass be at the same place?
11. A particle is thrown up vertically with a velocity of 50m/s. What will be its velocity at the highest point of the journey? How high would the particle rise? What time would it take to reach the highest point? Take g = 10 m/s2.
12. If a planet existed whose mass was twice that of Earth and whose radius 3 times greater, how much will a 1kg mass weigh on the planet?
13. A boy on cliff 49m 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?
14. A stone drops from the edge of a roof. It passes a window 2m high in 0.1s. How far is the roof above the top of the window?
15. A stone is dropped from the edge of a roof. (a) How long does it take to fall 4.9m ? (b) How fast does it move at the end of that fall? (c) How fast does it move at the end of 7.9m? (d) What is its acceleration after 1s and after 2s?
EXERCISE QUESTIONS
1. How does the force of gravitation between two objects change when the distance between them is reduced to half?
Ans. According to the universal law of gravitation, gravitational force (F) acting between two objects is inversely proportional to the square of the distance (r) between them, i.e.,
F ∝ 1/r2
If distance r becomes r/2, then the gravitational force will be proportional to 1 ( 1/(r/2)2 )=4/r2
Hence, if the distance is reduced to half, then the gravitational force becomes four times larger than the previous value.
2. Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?
Ans. All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). It is constant and does not depend upon the mass of an object. Hence, heavy objects do not fall faster than light objects.
3. 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).
Ans. According to the universal law of gravitation, gravitational force exerted on an object of mass m is given by:
4. The earth and the moon are attracted to each other by gravitational force. Does the earth attract the moon with a force that is greater or smaller or the same as the force with which the moon attracts the earth? Why?
Ans. According to the universal law of gravitation, two objects attract each other with equal force, but in opposite directions. The Earth attracts the moon with an equal force with which the moon attracts the earth.
5. If the moon attracts the earth, why does the earth not move towards the moon?
Ans. The Earth and the moon experience equal gravitational forces from each other. However, the mass of the Earth is much larger than the mass of the moon. Hence, it accelerates at a rate lesser than the acceleration rate of the moon towards the Earth. For this reason, the Earth does not move towards the moon.
6. What happens to the force between two objects, if
(i) the mass of one object is doubled?
(ii) the distance between the objects is doubled and tripled?
(iii) the masses of both objects are doubled?
Ans. According to the universal law of gravitation, the force of gravitation between two objects is given by: F = Gm1m2 / r 2
(i) F is directly proportional to the masses of the objects. If the mass of one object is doubled, then the gravitational force will also get doubled.
(ii) F is inversely proportional to the square of the distances between the objects. If the distance is doubled, then the gravitational force becomes one-fourth of its original value.
Similarly, if the distance is tripled, then the gravitational force becomes one-ninth of its original value.
(iii) F is directly proportional to the product of masses of the objects. If the masses of both the objects are doubled, then the gravitational force becomes four times the original value.
7. What is the importance of universal law of gravitation?
Ans. The universal law of gravitation proves that every object in the universe attracts every other object.
8. What is the acceleration of free fall?
Ans. When objects fall towards the Earth under the effect of gravitational force alone, then they are said to be in free fall. Acceleration of free fall is 9.8 m s−2, which is constant for all objects (irrespective of their masses).
9. What do we call the gravitational force between the Earth and an object?
Ans. Gravitational force between the earth and an object is known as the weight of the object.
10. Amit buys few grams of gold at the poles as per the instruction of one of his friends. He hands over the same when he meets him at the equator. Will the friend agree with the weight of gold bought? If not, why? [Hint: The value of g is greater at the poles than at the equator].
Ans. Weight of a body on the Earth is given by:
W = mg
Where,
m = Mass of the body
g = Acceleration due to gravity
The value of g is greater at poles than at the equator. Therefore, gold at the equator weighs less than at the poles. Hence, Amit’s friend will not agree with the weight of the gold bought.
11. Why will a sheet of paper fall slower than one that is crumpled into a ball?
Ans. When a sheet of paper is crumbled into a ball, then its density increases. Hence, resistance to its motion through the air decreases and it falls faster than the sheet of paper.
Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the Earth. What is the weight in newtons of a 10 kg object on the moon and on the Earth?
Ans. Weight of an object on the moon = 1 ´ Weight of an object on the Earth
Also,
Weight = Mass × Acceleration
Acceleration due to gravity, g = 9.8 m/s2
Therefore, weight of a 10 kg object on the Earth = 10 × 9.8 = 98 N
And, weight of the same object on the moon = 1/6 * 98 = 16.3N
12. A ball is thrown vertically upwards with a velocity of 49 m/s. Calculate
(i) the maximum height to which it rises.
(ii)the total time it takes to return to the surface of the earth.
Ans. According to the equation of motion under gravity:
v2 − u2 = 2 gs
Where,
u = Initial velocity of the ball
v = Final velocity of the ball
s = Height achieved by the ball
g = Acceleration due to gravity
At maximum height, final velocity of the ball is zero, i.e., v = 0
u = 49 m/s
During upward motion, g = − 9.8 m s−2
Let h be the maximum height attained by the ball.
Hence, 0 - 492 = 2 ´ (-9.8) ´ h
⇒ h = 49 * 49 / 2 * 9.8 = 122.5m
Let t be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion:
v = u + gt
We get,
0 = 49 + t * (-9.8) ⇒ 9.8t = 49 ⇒ t = 49/9.8 = 5s
But,
Time of ascent = Time of descent
Therefore, total time taken by the ball to return = 5 + 5 = 10 s
13. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
Ans. According to the equation of motion under gravity:
v2 − u2 = 2 gs
Where,
u = Initial velocity of the stone = 0
v = Final velocity of the stone
s = Height of the stone = 19.6 m
g = Acceleration due to gravity = 9.8 m s−2
∴ v2 − 02 = 2 × 9.8 × 19.6
v2 = 2 × 9.8 × 19.6 = (19.6)2
v = 19.6 m s− 1
Hence, the velocity of the stone just before touching the ground is 19.6 m s−1.
14. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?
Ans. According to the equation of motion under gravity:
v2 − u2 = 2 gs
Where,
u = Initial velocity of the stone = 40 m/s
v = Final velocity of the stone = 0
s = Height of the stone
g = Acceleration due to gravity = −10 m s−2
Let h be the maximum height attained by the stone.
Therefore,
0 - (40)2 = 2 * h´* (-10) ⇒ h = (40 * 40) / 20 = 80m
Therefore, total distance covered by the stone during its upward and downward journey = 80 + 80 = 160 m
Net displacement of the stone during its upward and downward journey = 80 + (−80) = 0
15. Calculate the force of gravitation between the earth and the Sun, given that the mass of the earth = 6 × 1024 kg and of the Sun = 2 × 1030 kg. The average distance between the two is 1.5 × 1011 m.
Ans. According to the universal law of gravitation, the force of attraction between the Earth and the Sun is given by: F = GMSunMEarth / R2
Where,
MSun = Mass of the Sun = 2 × 1030 kg
MEarth = Mass of the Earth = 6 × 1024 kg
R = Average distance between the Earth and the Sun = 1.5 × 1011 m
G = Universal gravitational constant = 6.7 × 10−11 Nm2 kg−2
F = 6.7 *´10-11 * 2 * 1030 * 6 ´1024 / (1.5 ´1011 )2= 3.57 ´1022 N
16. A stone is allowed to fall from the top of a tower 100 m high and at the same time another stone is projected vertically upwards from the ground with a velocity of 25 m/s. Calculate when and where the two stones will meet.
Ans. Let the two stones meet after a time t.
(i) For the stone dropped from the tower: Initial velocity, u = 0
Let the displacement of the stone in time t from the top of the tower be s.
Acceleration due to gravity, g = 9.8 m s−2
From the equation of motion,
s = ut + 1/2 gt2 = 0 * t + 1/2´*9.8 * t
⇒ s = 7.9t2 - - - - - -(1)
(ii) For the stone thrown upwards: Initial velocity, u = 25 m s−1
Let the displacement of the stone from the ground in time t be s'. Acceleration due to gravity, g = −9.8 m s−2
Equation of motion,
s ' = ut + 1/2 gt2 = 25t - 1/2 * 9.8 ´* t2
⇒ s ' = 25t - 4.9t 2 - - - - - -(2)
The combined displacement of both the stones at the meeting point is equal to the height of the tower 100 m.
∴ s + s ' = 100
⇒ 1/2 gt2 + 25t - 1/2 gt2 = 100
⇒ t = 100/25 = 4s
In 4 s, the falling stone has covered a distance given by equation (1) as
s = 1/2 *´10 * 42 = 80m
Therefore, the stones will meet after 4 s at a height (100 − 80) = 20 m from the ground
17. A ball thrown up vertically returns to the thrower after 6 s. Find
(a) the velocity with which it was thrown up,
(b) the maximum height it reaches, and
(c) its position after 4 s.
Ans. (a) Time of ascent is equal to the time of descent. The ball takes a total of 6 s for its upward and downward journey.
Hence, it has taken 3 s to attain the maximum height. Final velocity of the ball at the maximum height, v = 0
Acceleration due to gravity, g = −9.8 m s−2
Equation of motion, v = u + gt will give,
0 = u + (−9.8 × 3)
u = 9.8 × 3 = 29.4 ms− 1
Hence, the ball was thrown upwards with a velocity of 29.4 m s−1.
(b) Let the maximum height attained by the ball be h.
Initial velocity during the upward journey, u = 29.4 m s−1
Final velocity, v = 0
Acceleration due to gravity, g = −9.8 m s−2
From the equation of motion,
s = ut + 1/2 at2
⇒ h = 29.4 ´* 3 + 1/2 ´ (-9.8) *´ 32 = 44.1m
(c) Ball attains the maximum height after 3 s. After attaining this height, it will start falling downwards.
In this case,
Initial velocity, u = 0
Position of the ball after 4 s of the throw is given by the distance travelled by it during its downward journey in 4 s − 3 s = 1 s.
Equation of motion, s = ut + 1/2 gt2 will give
⇒ s = 0 * t + 1/2 * (9.8) *´12 = 4.9m
Total height = 44.1 m
This means that the ball is 39.2 m (44.1 m − 4.9 m) above the ground after 4 seconds.
Gavitation(physics)
Q1. It is seen that a falling apple is attracted towards the earth. Does the apple attract the earth? If so, we do not see the earth moving towards an apple. Why?
Ans. According to the third law of motion, the apple does attract the earth. But according to the second law of motion, for a given force, acceleration is inversely proportional to themass of an object The mass of an apple is negligibly small compared to that of the earth. So, we do not see the earth moving towards the apple.
Q.2. Define Centripetal force.
Ans. When a body moves in a circular path with a certain speed and changes direction at every point. The change in direction involves change in velocity or acceleration. The force that causes this acceleration and keeps the body moving along the circular path is acting towards the centre. is called the centripetal (meaning ‘centre-seeking’) force. In the absence of this force, the body flies off along a straight line. This straight line will be a tangent to the circular path.
Q.3.State UNIVERSAL LAW OF GRAVITATION
Ans. Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The force is along the line joining the centres of two objects.
Q.4. Give the IMP ORTANCE OF THE UNIVERSAL LAW OF GR AVITATION
ANS. The universal law of gravitation successfully explained several phenomena which were believed to be unconnected:
(i) the force that binds us to the earth;
(ii) the motion of the moon around the earth; (iii) the motion of planets around the Sun; (iv) the tides due to the moon and the Sun.
Q.5. Derive the relation between g & G.
Ans. Take a stone.Throw it upwards.
It reaches a certain height and then it starts falling down.
the force exerted by the earth on the body is
F = G(M*m) / d2 ................(1)
Therefore the magnitude of the gravitational force F will be equal to the product of mass and acceleration due to the gravitational force, that is,
F = m g ..............(2)
From Eqs. (1) and (2) we have,
7. WEIGHT OF AN OBJECT ON THE MOON
8. Free Fall : When object falls towards the earth under the effect of gravity alone then object is said to be in free fall.
The acceleration produced in an object when it falls freely under the effect of gravity alone is called acceleration due to gravity (g).
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CBSE Class 9 Science Chapter 10 Gravitation Notes
We hope you liked the above notes for topic Chapter 10 Gravitation which has been designed as per the latest syllabus for Class 9 Science released by CBSE. Students of Class 9 should download and practice the above notes for Class 9 Science regularly. All revision notes have been designed for Science by referring to the most important topics which the students should learn to get better marks in examinations. Studiestoday is the best website for Class 9 students to download all latest study material.
Notes for Science CBSE Class 9 Chapter 10 Gravitation
Our team of expert teachers have referred to the NCERT book for Class 9 Science to design the Science Class 9 notes. If you read the concepts and revision notes for one chapter daily, students will get higher marks in Class 9 exams this year. Daily revision of Science course notes and related study material will help you to have a better understanding of all concepts and also clear all your doubts. You can download all Revision notes for Class 9 Science also from www.studiestoday.com absolutely free of cost in Pdf format. After reading the notes which have been developed as per the latest books also refer to the NCERT solutions for Class 9 Science provided by our teachers
Chapter 10 Gravitation Notes for Science CBSE Class 9
All revision class notes given above for Class 9 Science have been developed as per the latest curriculum and books issued for the current academic year. The students of Class 9 can rest assured that the best teachers have designed the notes of Science so that you are able to revise the entire syllabus if you download and read them carefully. We have also provided a lot of MCQ questions for Class 9 Science in the notes so that you can learn the concepts and also solve questions relating to the topics. All study material for Class 9 Science students have been given on studiestoday.
Chapter 10 Gravitation CBSE Class 9 Science Notes
Regular notes reading helps to build a more comprehensive understanding of Chapter 10 Gravitation concepts. notes play a crucial role in understanding Chapter 10 Gravitation in CBSE Class 9. Students can download all the notes, worksheets, assignments, and practice papers of the same chapter in Class 9 Science in Pdf format. You can print them or read them online on your computer or mobile.
Notes for CBSE Science Class 9 Chapter 10 Gravitation
CBSE Class 9 Science latest books have been used for writing the above notes. If you have exams then you should revise all concepts relating to Chapter 10 Gravitation by taking out a print and keeping them with you. We have also provided a lot of Worksheets for Class 9 Science which you can use to further make yourself stronger in Science
You can download notes for Class 9 Science Chapter 10 Gravitation for latest academic session from StudiesToday.com
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