Q. 1. Textbook activity question
1. Will the direction of the gravitational force change as we go inside the earth?
Ans The direction of earth’s gravitational force is towards the center of earth so it will not change.
2. Is there a gravitational force between two objects kept on a table or between you and your friend sitting next to you? If yes, why don’t the two move towards each other?
Ans Yes, there is a ‘gravitational force between two objects’ that are kept on a table. They do not ‘move’ towards each other because the intensity of the forces is very weak.
3. An iron ball of mass 3 kg is released from a height of 125 m and falls freely to the ground. Assuming that the value of g is 10 m/s^{2}, calculate.
i ] time taken by the ball to reach the ground
ii ] velocity of the ball on reaching the ground
iii ] the height of the ball at half the time it takes to reach the ground.
Ans : Given:
m = 3 kg, distance travelled by the ball s = 125 m, initial velocity of the ball = u = o acceleration = a = g = 10 m/s^{2}.
(i) Newton’s second equation of motion gives
s = u t + 1/2 a t
125 = 0 t +1/2 × 10 × t^{2} = 5 t^{2}
t^{2} = 125/5 = 25, t = 5 s The ball takes 5 seconds to reach the ground.
(ii) According to Newton’s first equation of motion final velocity = v = u + a t
= 0 + 10 × 5
= 50 m/s
The velocity of the ball on reaching the ground is 50 m/s
(iii) Half time = t = 5/2 = 2.5 s
Ball’s height at this time = s
According to Newton’s second equation
s = u t + 1/2 a t^{2}
s = 0 +1/2 10 × (2.5)^{2} = 31.25 m.
Thus the height of the ball at half time
= 12531.25 = 93.75 m
4. What are Newton’s laws of motion?
Ans Three fundamental laws of Newton’s. The first states that a body continues in a state of rest or uniform motion in a straight line unless it is acted on by an external force. The second states that the rate of change of momentum of a moving body is proportional to the force acting to produce the change. The third states that if one body exerts a force on another, there is an equal and opposite force (or reaction) exerted by the second body on the first.
5. What would happen if the value of G was twice as large?
Ans The gravitational force between any two particles would become double, also the value of g would become double.
6. What would happen if there were no gravity?
Ans If our planet were to lose gravity merely for five seconds, as we know that, it would mean the end of life on Earth. Gravity pushes each other against artefacts. The more massive an object is, the greater its gravitational attraction. Humans and other objects will become weightless without gravity.
If we have no gravity force, the atmosphere would disappear into space, the moon would collide with the earth, the earth would stop rotating, we would all feel weightless, the earth would collide with the sun, and as a consequence. We would all perish.
7. A tennis ball is thrown up and reaches a height of 4.05 m before coming down. What was its initial velocity? How much total time will it take to come down? Assume g = 10 m/s^{2}.
Ans For the upward motion of the ball, the final velocity of the ball = v = 0 Distance travelled by the ball = 4.05 m
acceleration a = – g = 10 m/s^{2}
Using Newton’s third equation of motion v^{2} = u^{2} + 2 a s
0 = u^{2} + 2 (10) × 4.05
∴ u^{2} = 81
u = 9 m/s The initial velocity of the ball is 9 m/s
Now let us consider the downward motion of the ball. Suppose the ball takes t seconds to come down. Now the initial velocity of the ball is zero, u = 0. Distance travelled by the ball on reaching the ground = 4.05 m. As the velocity and acceleration are in the same direction,
a = g = 10 m/s
According to Newton’s second equation of motion s = u t + 1/2a t^{2}
4.05 = 0 +1/2 10 t^{2}
t^{2} = 4.05/5 = 0.81, t = 0.9 s
The ball will take 0.9 s to reach the ground. It will take the same time to go up. Thus, the total time taken = 2 × 0.9 = 1.8 s
8. Starting from rest, what will be Mahendra’s velocity after one second if he is falling down due to the gravitational force of the earth? Ans: Given:
u = 0, F = 733 N, Mahendra’s mass = m = 75 kg time t = 1 s
Mahendra’s acceleration
a = F / m = 733 / 75 m/s^{2}
According to Newton’s first equation of motion, v = u + at
Mahendra’s velocity after 1 second
v = 0 + 9.77 × 1 m/s
v = 9.77 m/s
This is 1.83 × 10^{9} times Mahendra’s velocity in example 2, on page 6.
9. If the area ESF is equal to area ASB, what will you infer about EF?
Ans EF and AB are distances covered by planet in the same time. Explanation: According to Kepler’s second law, if area of ESF = area of ASB, then the planet covers distances EF and AB in the same time.
10. What types of forces are you familiar with?
Ans A force is referred as a push or pull of an object that is caused due to its interaction with another object.
There are many types of forces that act on humans in everyday life. A few of them are – Frictional force, gravitational force, Magnetic force, contact forces like push and pull, electromagnetic force, spring force, resistance force, weak and strong interaction forces, etc.
11. Calculate the gravitational force due to the earth on Mahendra in the earlier example.
Ans Mass of the earth = m_{1} = 6 × 1024 kg Radius of the earth = R = 6.4 × 10^{6} m Mahendra’s mass = m_{2} = 75 kg
G = 6.67 × 10^{11} Nm^{2} /kg^{2}
Using the force law, the gravitational force on Mahendra due to earth is given by
This force is 1.83 × 10^{9} times larger than the gravitational force between Mahendra and Virat. F = G m1 m2
F = 6.67 ×10^{−11} × 75 × 6 × 10^{24}/ (6.4 × 10^{6} )^{2}
N = 733 N
12. Assuming the acceleration in Example 2 above remains constant, how long will Mahendra take to move 1 cm towards Virat?
Ans:
= 1935 s = 32 minutes 15 seconds.
13. In the above example, assuming that the bench on which Mahendra is sitting is frictionless, starting with zero velocity, what will be Mahendra’s velocity of motion towards Virat after 1 s ? Will this velocity change with time and how?
Ans Given: Force on Mahendra = F = 4.002 × 10^{7} N, Mahendra’s mass = m = 75 kg
According to Newton’s second law, the acceleration produced by the force on Mahendra = m = 75 kg.
a = F/m = 4.002 × 10^{7} = 5.34 × 10^{9} m/s^{2}.
Using Newton’s first equation, we can calculate Mahendra’s velocity after 1s, Newton’s first equation of motion is
v = u + at;
As Mahendra is sitting on the bench, his initial velocity is zero (u = 0) Assuming the bench to be frictionless,
v = 0 + 5.34 × 10^{9} × 1 m/s
= 5.34 × 10^{9} m/s
Mahendra’s velocity after 1 s will be 5.34 × 10^{9} m/s .
14. What do you know about the gravitational force?
Ans The gravitational force is a force that attracts any two objects with mass.
We call the gravitational force attractive because it always tries to pull masses together, it never pushes them apart. In fact, every object, including you, is pulling on every other object in the entire universe!
15. Suppose you are standing on a tall ladder. If your distance from the centre of the earth is 2R, what will be your weight?
Ans Weight, W = Gmm/r^{2} = Gmm / (2R)^{2} = GMm / 4R^{2}
= 1/4 (GMm/R^{2} )
= weight on the surface of the earth / 4
16. Show that in SI units, the unit of G is Newton m^{2} kg^{2}. The value of G was first experimentally measured by Henry Cavendish. In SI units its value is 6.673 × 10^{11} N m^{2} kg^{2}.
Ans F = G m1 m2
G = Gm1m2 / r^{2}
G= F.r^{2}/ m1m2
G (SI unit) = N. ^{m2}/ kg.kg = Nm^{2}/ kg^{2}
17. If a person weighs 750 N on earth, how much would be his weight on the Moon given that moon’s mass is 1/3.7 of that of the earth and its radius is ^{1} of that of the earth ?
Ans Given :
Weight on earth = 750 N,
The weight on the moon is nearly 1/6^{th} of the weight on the earth. We can write the weight on moon as mg_{m} (g_{m} is the accelaration due to gravity on the moon). Thus g_{m} is 1/6^{th} of the g on the earth.
18. What are the effects of a force acting on an object? Ans: Effects of Force –
A force acting on an object causes the object to change its shape or size, to start moving, to stop moving, to accelerate or decelerate. When there’s the interaction between two objects they exert a force on each other, these exerted forces are equal in size but opposite in direction.
19. Will your weight remain constant as you go above the surface of the earth?
Ans i. Weight of a body depends on acceleration due to gravity. W = m × g.
ii. As we go above the surface of the earth height increases and hence acceleration decreases.
iii. As acceleration due to gravity decreases consequently weight decreases.
20. According to Newton’s law of gravitation, every object attracts every other object.
Ans (i) The apple attract the earth with the same force with which the earth attracts the apple.
 According to newtons’s third law , these two forces are equal and opposite in direction.
 For same magnitude of force, the acceleration produced in a body is inversely proportional to its mass.
 As the mass of the earth is very large compared to that of the apple, the acceleration of the earth is too small compared to the acceleration of the apple that it cannot be noticed.
Hence , the apple falls towards the earth while the earth does not move towards the apple.
21. According to Newton’s law of gravitation, earth’s gravitational force is higher on an object of larger mass. Why doesn’t that object fall down with higher velocity as compared to an object with lower mass?
Ans i. The acceleration due to gravity g on an object only depends on mas (M) and radius (R) of the earth g =
GM ./R^{2}
 It does not depend on mass (m) of the object.
 The acceleration produced at a given point is the same for all objects.
 Hence object of larger mass does not fall down with higher velocity as compared to an object with lower mass.
Q.2. Multiple Choice Questions :
1. What will be the weight of a person on earth, who weighs 9N on the moon?
a. 3 N b. 15 N c. 45 N d. 54 N
Ans Option d.
2……………………. waves are called the waves on fabric of spacetime.
a. Light waves b. Electromagnetic c. Gravitational d. Gamma
Ans Option c.
3.
3. By using the second law of Kepler’s planetary motion, what is true for the given diagram?
a. Area ASB = Area PSR. b. V_{1} = V_{2}
c. t_{1} = t_{2 d. V t = V t = (ASB + PSR)}
Ans Option a.
4.Value of G is ………….
a. 6.67 x 10^{11} Nm^{2}kg^{2}
b. 6.67 x 10^{23} Nm^{2} kg^{2}
c. 9.8 x 10^{11} Nm^{2}kg^{2}
d. 9.8 m/s^{2}
Ans Option a.
5. The force of gravitation increases as –
a. mass of object increases
b. the distance increases
c. mass of the object is halved
d. squaring the distance
Ans Option a.
6. While rotating a stone tied to a string; on suddenly leaving the string, what happens to the stone?

 Stone falls freely due to gravity
 Stone moves along tangential force
 Stone is projected upwards.
 Stone continues the circular motion
Ans Option b.
7. Which of the following is observed only during circular motion?

 uniform acceleration
 uniform velocity
 Escape velocity
 Centripetal Force
Ans Option d.
8. CGS unit of Gravitational constant is …………..
a.dyne cm^{2} g^{1} s^{1 }
b.dyne cm^{2} g^{2}
c. dyne m^{2} kg^{2}
d. dyne cm^{2} kg^{2}
Ans Option b.
9. LIGO stands for
a.Laser Interferometry Gravitational wave observatory
b.Light Intermissionary Gravity observatory
c. Laser Intermissionary Gravitational observatory
d. Light Interference Gravitational waves observatory
Ans Option a.
10 Value of g is zero at ……………
a. Extreme height b. Centre of the earth c. Polar zone d. Equatorial zone
Ans Option b.
11. If the distance between the two objects is doubled, the force between them
a. increase by factor 4 b. doubles c. decrease by factor 4 d. become hay
Ans Option c.
12. Value of g is more at …………..
a. Polar region b. equatorial region
c. about 100 km from surface d. a pit 20km deep
Ans Option a.
13. What is the weight of an object with 10 kg mass?
a. 10 N b. 9.8 N c. 98 N d. 0.98 N
Ans Option c.
Q. 3. Find the odd one out :
1. Weight, Centripetal force, Potential energy, Acceleration due to gravity
Ans Centripetal force – This force exists only in circular motion while other quantities exist at any instance of time.
2. Mass, Weight, Acceleration due to gravity, Radial distance
Ans: Mass – It remains constant all around while other quantities vary with altitude or depth on any planet.
3. Speed, Velocity, Displacement, Acceleration
Ans : Speed – It is a scalar quantity (nondirectional) while others are vector quantities (directional).
4. Gravitational Force, Centripetal force, Gravitational constant, Weight of an object
Ans: Gravitational constant – Its value remains same across the universe while remaining vary with space and time.
5.Grams, Newton, Centimeters, Dyne
Ans Newton – It is an S.I. unit while others are C.G.S units.
Q.3.Find corelated terms :
1. Initial velocity during free fall : 0 m/s : acceleration during free fall : ……………
Ans Initial velocity during free fall : 0 m/s : acceleration during free fall : 9.8 m/s^{2}
2. Object in motion : kinetic energy :: Object in stable position : ……………..
Ans Object in motion : kinetic energy :: Object in stable position : Potential energy
3. Force : dynes :: velocity : ……………
Ans Force : dynes :: velocity : cm/s or cm s^{1}
4. Increasing magnitude : Acceleration :: Decreasing magnitude : ……………
Ans Increasing magnitude : Acceleration :: Decreasing magnitude : Deceleration
5. Planetary motion : Kepler :: Gravitation : ……………
Ans Planetary motion : Kepler :: Gravitation : Newton
Q.4. Match the pair :
1
I 
II 
III 
i. Acceleration due to gravity 
m/s^{2} 
Zero at the centre 
ii. Gravitational constant 
Kg 
Measure of inertia 
Nm^{2}/Kg^{2} 
Same in entire universe 

N 
Depends on height 
Ans
I 
II 
III 
i. Acceleration due to gravity 
m/s^{2} 
Depends on height 
ii. Gravitational constant 
Nm^{2}/Kg^{2} 
Same in entire universe 
2
I 
II 
III 
i. Mass 
m/s^{2} 
Zero at the centre 
ii. Weight 
Kg 
Measure of inertia 
Nm^{2}/Kg^{2} 
Same in the entire universe 

N 
Depends on height 
Ans
I 
II 
III 
i. Mass 
Kg 
Measure of inertia 
ii. Weight 
N 
Zero at the centre 
3
I 
II 
III 
i. Centripetal Force 
N 
Beyond gravitational limit 
ii. Potential energy 
a=0 m/s^{2} 
Radial towards centre 
Joule 
Increases with height 

m/s 
Object at rest 
Ans
I 
II 
III 
i. Centripetal Force 
N 
Radial towards centre 
ii. Potential energy 
Joule 
Increases with height 
4
I 
II 
III 
i. Uniform motion 
N 
Beyond gravitational limit 
ii. Escape velocity 
a=0 m/s^{2} 
Radial towards centre 
Joule 
Increases with height 

m/s 
Object at rest 
Ans
I 
II 
III 
i. Uniform motion 
a=0 m/s^{2} 
Object at rest 
ii. Escape velocity 
m/s 
Beyond gravitational limit 
Q. 5.State True or False:
1. Value of g varies with altitude:
Ans Value of g varies with altitude – True
Value of g decreases with increase in altitude and maximum on surface of the Earth.
2. High tide and Low tide is result of gravitational force of Sun.
Ans High tide and Low tide is result of gravitational force of Sun – False.
High tide and Low tide is result of gravitational force of moon.
3. The orbit in which Earth revolves around Sun is perfectly circular.
Ans The orbit in which Earth revolves around Sun is perfectly, circular – False
The orbit in which Earth revolves around Sun is the elliptical.
4. In circular motion, object experiences centripetal force.
Ans In circular motion, object experiences centripetal force – True
In circular motion, object experiences centripetal force acts on the object radially towards the centre of circular path of the motion.
5. Object thrown upwards exhibit positive acceleration.
Ans Object thrown upwards exhibit positive acceleration – False
Object thrown upwards has negative acceleration However, after reaching certain height it falls freely with positive acceleration, vertically downwards
6. True unit of weight is Newton.
Ans True unit of weight is Newton – True
Weight is the force acting on object due to gravity given by weight –E – Mg; m being mass of the object
7. Newton gave the three laws of planetary motion.
Ans Newton gave the three laws of planetary motion – False
Kepler gave the three laws of planetary motion.
8. At a stable position, any object has stored energy called kinetic energy.
Ans At a stable position, any object has stored energy called kinetic energy – False
At a stable position, any object has stored energy called potential energy.
Q. 6. Name the following :
1. Give formula for gravitational potential energy at height h from the surface of earth. Ans: Gravitational Potential energy = GMm/R+h
2. If g = GM/r2 then where will the value of g be high at Goa Beach or on top of Mount Everest?
Ans Goa beach. 3.
Ans If F = Gm1 m2/d^{2 } then F = Gm1m2/ (3d)^{2}
Q.7. Give scientific reasons :
1. Weight of the person on the earth and on the moon is very different.
Ans i. Weight is the force experienced by the object due to gravitational pull downwards.
ii. It is determined by mass of the object and the gravitational acceleration on the body. Greater the mass, more is the gravitational acceleration.
iii. As the gravitational force on earth is larger due to its bigger size compared to the gravitational force on the moon, the weight of person is less on the moon than on earth.
iv. Therefore, weight of the person varies according to the gravitational force acting on the object.
2. True free fall is possible only in vacuum.
Ans i] When we release an object from height, the only force that acts on it is the gravitational force of the earth and the object falls down under its influence.
ii. Whenever an object moves under the influence of the force of gravity alone, it is said to be falling freely.
iii. Thus the released object is in a free fall.
iv. In free fall, the initial velocity of the object is zero and goes on increasing due to the acceleration due to gravity of the earth.
v.During free fall, the frictional force due to air opposes the motion of the object and a buoyant force also acts on the object.
Vi Thus, true free fall is possible only in vacuum.
3. The centre of mass of any object having uniform density is at its centroid.
Ans i. The centre of mass of an object is the point inside or outside the object at which the total mass of the object can be assumed to be concentrated.
ii. The centre of mass of a spherical object having uniform density is at its geometrical centre.
iii. The center of mass of a distribution of mass in space is the unique point where the weighted relative
iv.position of the distributed mass sums to zero, or the point where if a force is applied it moves in the direction of the force without rotating.
vThe distribution of mass is balanced around the center of mass and the average of the weighted positions of the distributed mass that defines its center of gravity.
vi The concept of the ‘center of mass’ in the form of the ‘center of gravity’ of any object.
vii. Therefore, The centre of mass of any object having uniform density is at its centroid.
4. If the value of ‘g’ suddenly becomes twice its value, it will become two times more difficult to pull a heavy object along the floor.
Ans i. Weight of an object depends on the value of ‘g’ as it is the force acting on the mass of that object by gravitation pull.
ii.Weight is directly proportional to value of ‘g’, and this increase as ‘g’ increase.
iii. As the value of ‘g’ is doubled, the weight of the object increases by factor 2, making the object heavy.
iv. Therefore, as weight is doubled, it becomes two times harder to pull the heavy object across the floor.
5. When we drop a feather and a heavy stone at the same time from a height, they do not reach the earth at the same time.
Ans i. In vacuum, the feather and stone indeed reach the earth at the same time.
ii. The feather experiences a buoyant force and a frictional force due tov air and therefore floats and reaches the ground slowly, later than the heavy stone.
iii. The buoyant and frictional forces on the stone are much less than the weight of the stone and does not affect the speed of the stone much.
6. Space travellers as well as objects in the spacecraft appear to be floating.
Ans i. The space craft launched remains at a specific height under the influence of gravitational force of earth.
ii. Through the space craft is at a considerable height from the surface of the earth, the value of ‘g’ is not zero.
iii. In the space station the value of ‘g’ is only 11% less than its value on the surface doesn’t make a good reason for weightlessness.
iv. The main reason for the weightlessness is caused by their being in a state of free fall.
v.As the velocity of free fall does not depend on properties of an object, velocity along the orbit of the spacecraft, travellers and all objects is same in the spacecraft.
vi. The only force acting is gravitational force of earth and thus in free fall state, objects and travellers in the vi.spacecraft appear to be floating.
7. Value of ‘g’ at the centre of the earth is zero.
Ans i. As we go down towards the centre of the earth, the distance decreases and value of ‘g’ is expected to increase as it is inversely proportional to the distance.
i.However, with increase in depth, the mass of the earth which contributes towards the gravitation also significantly decreases.
ii. Due to this, the value of ‘g’ decreased as the depth increase towards the centre.
iii. At centre of the earth, the mass contributing to gravitation becomes zero and therefore, value of ‘g’ becomes zero.
Q.8. Solve Numerical problems. :
1. The radius of planet A is half the radius of planet B. If mass of A is M_{A}, what must be mass of B so that value of g on B is half that of its value on A?
Ans R_{A} = radius of A ; g on A = g_{A;} R_{B} = radius of B g on B = g_{B}
2. The mass and weight of an object on earth are 5 kg and 49N respectively. what will be their values on the moon? Assume that acceleration due to gravity on the moon is 1/6^{th} of that on the earth.
Ans Given :mass = m = 5kg
weight = w = 49N gm = ^{1} ge
6
Mass is universally same and does not change
∴ Mass on moon = 5 kg
gm = 1/6 ge = 1/6 × 9.8 = 1.633
∴ Weight on moon = mg_{m}
= 5×1.633
= 8.17 N
∴ Mass on moon = 5kg Weight on moon = 8.17 N
3. An object takes 5 s to reach the ground from a height of 5m on a planet. What is value of g on the planet?
4.
Ans:Mass = 5 kg
Height = 490 m
g = 9.8 m/s
For a free falling body initial velocity is zero.
According to Neutons second equation of motion
5.
Ans Given : r = 1 m, m_{1} = 75 kg, m_{2} = 80 kg
and G = 6.67 x 10 ^{11} Nm^{2}/kg^{2} According to Newton’s law,
F = Gm1 m2/r^{2}
F = 6.67×10^{−11} ×75×80 N/1^{2}
= 4.002 × 10^{7}N
The gravitational force between Mahendra and Virat is 4.002 × 10^{7} N
6. Let the period of revolution of a plant at a distance R from a star be T prove that if it was at distance 2R, its period of revolution will be √8 T.
Ans In first case : distance d_{1} = R time taken = t_{1} = T On increasing the distance to 2R, Distance = D_{2} = 2R
Time taken = t_{2}
Using Kepler’s third law in both cases,
t_{1}^{2}/d_{1}^{3} = constant … (1) and t_{2}^{2}/d_{2}^{3} = constant … (2)
Therefore :
t_{1}^{2}/d_{1}^{3} = t_{2}^{2}/d_{2}^{3}
Substituting values given:
T^{2}/R^{3 }= t_{2}^{2}/ (2R)^{3}
T^{2}/R^{3 }= t^{2}/ (8R)^{3}
t_{2}^{2} = 8T^{2}
Taking square root on both sides.t2 = √ 8T Hence Proverd
Q.10. Write Short Notes:
1. Explain the Weightlessness in space.
Ans i. Space travelers as well as objects in the spacecraft appear to be floating.
ii. Though the spacecraft is at a height from the surface of the earth, the value of g there is not zero. In the space station the value of g is only 11% less than its value on the surface of the earth.
iii. Thus, the height of a spacecraft is not the reason for their weightlessness. Their weightlessness is caused by their being in the state of free fall.
iv. Though the spacecraft is not falling on the earth because of its velocity along the orbit, the only force acting on it is the gravitational force of the earth and therefore it is in a free fall.
v. As the velocity of free fall does not depend on the properties of an object, the velocity of free fall is the same for the spacecraft, the travelers and the objects in the craft.
vi Thus, if a traveler releases an object from her hand, it will remain stationary with respect to her and will appear to be weightless.
2. Write short note on given figure.
Ans i. The following figure represents Kepler’s second law of planetary motion, the rotation of a planet around the sun.
ii. The law states that – The line joining the planet and the Sun sweeps equal areas in equal intervals of time.
iii. AB and CD are distances covered by the planet in equal time i.e. after equal intervals of time, the positions of the planet starting from A and C are shown by B and D respectively.
iv. The straight lines AS and CS sweep equal area in equal interval of time i.e. area ASB and CSD are equal.
3. Write short note on Mass and Weight
Ans i. Mass is the amount of matter present in the object. The SI unit of mass is kg. Mass is a scalar quantity.
i. Its value is same everywhere. Its value does not change even when we go to another planet.Higher the mass, higher is the inertia.
ii. The weight of an object is defined as the force with which the earth attracts the object.
iii. The force (F) on an object of mass m on the surface of the earth can be written using an equation. Weight, W= F = m.g.
4. Write short note on Earth’s gravitational acceleration
Ans i. The earth exerts gravitational force on objects near it. According to Newton’s second law of motion, a force acting on a body results in its acceleration.
ii. Thus, the gravitational force due to the earth on a body results in its acceleration.
iii. This is called acceleration due to gravity and is denoted by ‘g’.
iv. Acceleration is a vector. As the gravitational force on any object due to the earth is directed towards the centre of the earth, the direction of the acceleration due to gravity is also directed towards the centre of the earth i.e. vertically downwards.
Q.10. Complete the given flow chart / table / diagram
1
Scientists 
Discovery 
Sir Issac Newton 
…………… 
…………… 
Free Fall experiment 
Johannes Kepler 
…………… 
…………… 
Value of ‘G’ 
Ans
Scientists 
Discovery 
Sir Issac Newton 
Universal law of gravitation 
Galileo 
Free Fall experiment 
Johannes Kepler 
Law describing planetary motion 
Henry Cavendish 
Value of ‘G’ 
Q.11. Laws/define/principles :
1. Define the following:
i.Acceleration due to gravity
ii. Free fall
Ans i. The acceleration produced in a body under the influence of the force of gravity alone is called acceleration due to gravity.
ii. The motion of any object under the influence of the force of gravity alone is called as free fall.
2. Define escape velocity.
Ans The minimum velocity with which a body should be projected from the surface of a planet or moon, so that it escapes from the gravitational influence of the planet or moon is called as escape velocity.
Q.12. Distinguish between :
 ] Light Waves and Gravitational Waves
Ans
Light Waves 
Gravitational Waves 

Light waves are type of waves called electromagnetic waves. 
Gravitational waves are type of waves called waves on fabric of spacetime. 

i. 

These are strong waves and hence easy to detect. 
These are weak waves and hence difficult to detect. 

ii. 

iii. 
Light waves are visible / invisible spectra. 
These waves are not visible. 
These waves travel from one place to other through medium opaque. 
These waves created a gravitational field in particular boundary of the object. 

iv. 

2. Mass and Weight
Ans
Mass 
Weight 

Mass is the amount matter present in an object. 

i. 
Weight is the force on an object due to gravitational pull. 

Mass is measured in kilograms as SI unit. 

ii. 
Weight is measured in Newton as SI unit. 

Mass of any object is universally constant and does not change. 
Wight changes with the change in Gravitational acceleration. 

iii. 

Mass is a primary quantity and cannot be formulated for measurement. 
Weight is a secondary quantity and depends on mass and gravitational acceleration, given as W = mg. 

iv. 

Mass is measured by a beam balance. 

v. 
Weight is measured by a spring balance. 

3. Potential Energy and Kinetic Energy
Ans :
Potential Energy 
Kinetic Energy 

Energy possessed by an object due to its position or state is called potential energy. 
Energy possessed by an object due to its motion is called kinetic energy. 

i. 

It depends on the height and gravitational acceleration. 
It depends on mass and velocity height or gravitational acceleration. 

ii. 

iii. 
It does not depend on velocity of the object. 
It depends on the final velocity of the object. 
iv. 

It is given by PI = mgh 
It is given by KE = 1/2 mv^{2} 

v. 
Potential energy can be positive or negative. 
Kinetic energy is always positive. 
Q.13. Give examples :
 Give any four examples of naturally occurring circular motion.
Ans: i. Motion of Earth revolving around the sun.
ii.Motion of satellite revolving around the Earth.
iii. Rotation of blades of fan.
iv. Car touring through a curved road.
v. Electrons revolving around the nucleus of atom.
2. Give examples of Electromagnetic waves.
Ans i. Light rays
i i. Gamma rays
iii. Xrays
iv. Ultraviolet rays
v. Microwaves
vi. Infrared waves
vii. Radio waves
Q.15. Give explanation using the given statements :
1. Activity :
a. Tie a stone to one end of a string. Take the other end in your hand and rotate the string so that the stone moves along a circle.
b. Release the string. In this case, the stone will fly off along a straight line when the string is released.
Questions :
i. When the stone is being rotated through string, which motion is the stone in ? ii. After releasing, in what direction does the stone gets thrown away?
iii . What is the major force acting on the stone during rotation ?
Ans i. Circular motion.
ii. In the direction tangential to the circular path.
Iii Centripetal force.
2. Read the statements given below. Identify and write the concept upon which the given statement is based.
i. Gravitational force is a universal force.
ii. The force which acts on planet and is responsible for its circular motion
iii. The earth attracts every object near it towards itself which is the force.
iv Shashi drops a feather and a heavy stone at the same time from a height, they reach the earth at the same time. Where is this experiment performed?
v. Rashi throws an object vertically upward, and the object comes to rest and stays there, what is the distance of an object from earth.
2. In f = Gm1 m2/ d^{2} , the value of which variable is 6.673 × 10^{11} NM ^{2 }kg ^{2 }Ans i. Gravitational force acts not only between two objects on earth but also between any two objects in the universe.
ii.Centripetal force
iii. Gravitational force
iv. The experiment is performed in vacuum.
v. The object is at infinite distance from earth, here object will come to rest as it will be free from the gravity of earth.
Vii. G – Universal Gravitational constant.
3. The value of g also changes if we go inside the earth. The value of R i.e distance decreases and one would think that the value of g should increase as per the formula. However, the part of the earth which contributes towards the gravitational force felt by the object also decreases. As a combined result of change in r and M, the value of g decreases as we go deep inside the earth.
i. What is the expected change in the value of ‘g’ as the distance decrease?
ii. What will be the value of ‘g’ at the center of the earth?
Ans i. The value of g is expected to increase as the distance decreases.
ii.The value of the g at the centre of the Earth will be zero.
Q.15. Complete the table / chart :
1 Complete the table :
Terms 
Units of Measurement 
Universal gravitational constant 
……………… 
Weight 
……………… 
……………… 
Kg 
Velocity 
……………… 
Acceleration due to gravity 
……………… 
……………… 
s 
Ans:
Terms 
Units of Measurement 
Universal gravitational constant 
Nm^{2}/kg^{2} or Nm^{2}•kg^{2} 
Weight 
N or kgm/s^{2} 
Mass 
Kg 
Velocity 
m/s 
Acceleration due to gravity 
m/s^{2} 
Time 
s 
Q.17. Explain with the help of examples 3
1] a.
b.
a.
Explain the given diagram.
Ans i.This diagram shows circular motion and Centripetal force.
i. Tie a stone to one end of a string. Take the other end in your hand and rotate the string so that the stone moves along a circle ii . As long as we are holding the string, we are pulling the stone towards us i.e. towards the centre of the circle and are applying a force towards it.
iii. The force stops acting if we release the string. In this case, the stone will fly off along a straight line which is the tangent to the circle at the position of the stone when the string is released, because that is the direction of its velocity at that instant of time.
iv. A force acts on any object moving along a circle and it is directed towards the centre of the circle. This is called the Centripetal force.
v.`Centripetal’ means centre seeking, i.e. the object tries to go towards the centre of the circle because of this force.
Vi The Centripetal force is the reason for the circular motion of any object.
Q.18. Solve Numerical problems :
1] The mass of the earth is 6 × 10^{24} kg. The distance between the earth and Sun is 1.5 × 10^{11}m. If gravitational force between them is 3.5 × 10^{22}N, what is the mass of the sun? (G = 6.7 × 10^{11} NM^{2} kg^{2})
Ans Given :Mass of earth = Me = 6 × 10^{24}kg Distance between = d = 1.5 × 10^{11} m Gravitational Force = F = 3.5 × 10^{22} N
G = 6.7 × 10^{11} Nm^{2} Kg^{2}
Mass of sun = M_{s}
We know; Gravitation force is given by F = G MeMs/d^{2} ; M_{s }=F.d^{2}/ Gm_{e}F= G MeMs
M_{s}= 3.5 X10^{22}X1.5 X1.5 X10^{22}/ 6.7 X 10^{11}X 6X 10^{24}
M_{s} = 3.5 × 1.5 × 1.5× 10^{(22} ^{+} ^{22} ^{+} ^{11} ^{−} ^{24)}/6.7 × 6
M_{s} = 7.87 × 10^{(31) }/ 40.2 ∴ M_{s} = 0.1958 ×10^{31}
∴ M_{s} = 1.96 ×10^{30} kg
∴ Mass of the sun = 1.96 ×10^{30} kg
2. A stone thrown vertically upwards with initial velocity ‘u’ reaches a height ‘h’ before coming down. Show that time taken to go up is same as time taken to come down.
Ans.
3. An object thrown vertically upwards reaches height of 500m. What was its initial velocity? How long will it take to come back to the earth? Assume g = 10m/s
Ans Given : Height = s = 500 m
g = 10 m/s^{2}
since, object is thrown upwards against gravity. acceleration = – g = − 10 m/s^{2} and final velocity = v = O m/s
∴ t = 10s
Same time is required for the object to reach the ground so total time taken will be T = 2×t = 2×10 = 20s
∴ Initial velocity (upwards) 100 m/s Total time taken by object = 20s
4. What would be the value of ‘g’ on the surface of the earth if its mass was twice and its radius half of what it is now ?
Ans g_{1} = GM/ r^{2} When mass is double and its radius half then we get g_{2} _{}
g_{2} = 8 X 9.8 = 78.4 m/s^{2}
5. The masses of earth and moon are 6 × 10^{24} kg and 7.4 × 10^{22} kg respectively. The distance between them is 3.84 × 10^{5} km. Calculate gravitational force of attraction between the two? (G = 6.7 × 10^{11} NM^{2} kg^{2}).
Ans Given :Mass of earth = Me = 6 × 10^{24}kg Mass of moon = M_{M} = 7.4 × 10^{22}kg
Distance between them = d = 3.84 × 10^{5} km
= 3.84 × 10^{8} m
6 . A ball falls off a table and reaches ground in 1 s. assuming g = 10m/s^{2}, calculate its speed on reaching the ground and the height of the table.
Ans Given :Time taken = t = 1s
= g = 10 m/s^{2} It’s a free fall motion ∴ height = s = 1/2 gt^{2 }s = 1/2 ×10× (t)^{2 }; s = 5 m
using third kinematical equation v^{2} = u^{2} + 2as
u = 0 and a = g in free fall
∴ v^{2} = 2gs
∴ v^{2} = 2 × 10 × 5
∴ v^{2} = 100
∴ v = √100 ∴ v = 10 m/s
∴ Height of table = 5m
Speed of reaching to the ground = 10m/s
Q.18. Write laws, theories and explain. :
 Identify the laws shown in the figure and state three respective laws.
Ans Figure explains Kepler’s laws. They are stated as follows :
i. Kepler’s First Law :
The orbit of planet is an ellipse, with the sun at one of the foci.
ii. Kepler’s Second Law :
The line joining the planet and the sun sweeps equal areas in equal intervals of time. According to this law, Area ASB and PSR are equal, covered by planet in equal time intervals.
iii.Keepler’s Third Law :
The square of the period of revolution around the sun is directly proportional to the cube of the mean distance of a planet from the sun If ‘R’ is the mean distance and T is the period of revolution then, T^{2} α R^{3} i.e. T^{2} / R^{3} = constant
Newton formulated the Gravitational Force equation in which the force is inversely proportional to the distance between the object. This inverse square dependence in his law helped by Kepler’s third law.
2. The value of ‘g’ changes with the height ? Explain ?
Ans i. As we go above the earth’s surface, the value of r i.e. the distance increases and the value of g decreases.
i. However, the decrease is rather small for heights which are small in comparison to the earth’s radius.
ii. For example, remember that the radius of the earth is 6400 km. If an airplane is flying at a height 10 km above the surface of the earth, its distance from the earth’s surface changes from 6400 km to 6410 km and the change in the value of g due to it is negligible.
iii. On the other hand, when we consider an artificial satellite orbiting the earth, we have to take into account
iv. the change in the value of g due to the large change in the distance of the satellite from the centre of the earth.
v. Some typical heights and the values of g at these heights can be measured by the illustrations.
vi. Thus, with the change in height the value of acceleration due to gravity changes.
Q.19. Complete the paragraph.
 Complete the paragraph:
(escape, motion, gravity, directly, infinite, zero, rest, inversely )
An object going vertically upwards from the surface of the earth, having an initial velocity equal to the …………… velocity, escapes the gravitational force of the earth. The force of gravity, being…………….. proportional to the square of the distance, becomes ………….. only at distance from the earth. This means that for the object to be free from the of earth, it has to reach infinite distance from the earth. i.e. the object will come to at infinite distance and will stay there.
Ans An object going vertically upwards from the surface of the earth, having an initial velocity equal to the escape velocity, escapes the gravitational force of the earth. The force of gravity, being inversely proportional to the square of the distance, becomes zero only at infinite distance from the earth. This means that for the object to be free from the gravity of the earth, it has to reach infinite distance from the earth. i.e. the object will come to rest at infinite distance and will stay there.
2. Complete the paragraph:
(negative, decreases, potential, relative, zero, large, increases, small)
The energy stored in an object because of its position or state is called energy. This energy is ……………. and as we go to greater heights from the surface of the earth. We had assumed that the potential energy of an object of mass m, at a height h from the ground is mgh and on the ground it is ………….. When h is compared to the radius R of the earth, we can assume g to be constant and can use the above formula (mgh). But for large values of h, the value of g decreases with increase in h. For an object at infinite distance from the earth, the value of g is zero and earth’s gravitational force does not act on the object. So it is more appropriate to assume the value of potential energy to be zero there. Thus, for smaller distances, i.e. heights, the potential energy is less than zero, i.e. it is ……………
Ans The energy stored in an object because of its position or state is called potential energy. This energy is relative and increases as we go to greater heights from the surface of the earth. We had assumed that thepotential energy of an object of mass m, at a height h from the ground is mgh and on the ground it is zero When h is small compared to the radius R of the earth, we can assume g to be constant and can use the above formula (mgh). But for large values of h, the value of g decreases with increase in h. For an object at infinite distance from the earth, the value of g is zero and earth’s gravitational force does not act on the object. So it is more appropriate to assume the value of potential energy to be zero there. Thus, for smaller distances, i.e. heights, the potential energy is less than zero, i.e. it is negative.
3. Complete the paragraph:
(centroid , uniform, circular, mass, spherical , assumed, fixed, point)
The centre of mass of an object is the …………… inside or outside the object at which the total of theobject can be ………….. to be concentrated. The centre of mass of a ………….. object having density is at its geometrical centre. The centre of mass of any object having uniform density is at its ……………. .
Ans i. point
ii. mass
iii assumed iv spherical
v. uniform vi.centroid
4. Complete the paragraph:
(Newton, directly, cube, inversely, attracts, universal, repulse, square.)
All the considerations including Kepler’s laws led to formulate a theory with respect to gravitation. It was the theory of …………… gravity. According to this theory, every object in the Universe every other object with a definite force. This force is proportional to the product of the masses of the two objects and is …………… proportional to the of the distance between them.
Ans All the considerations including Kepler’s laws led Newton to formulate a theory with respect to gravitation. It was the theory of universal gravity. According to this theory, every object in the Universe attracts every other object with a definite force. This force is directly proportional to the product of the masses of the two objects and is inversely proportional to the square of the distance between the
Q.20 Write answers based on given diagram/figure:
1. Answer the following based on figures.
Questions :

 What is observed in the given figure and who formulated it?
 What is the first law of planetary motion ?
 Applying second law of planetary motion, what can we say about the motion of the planet ?
Ans i. The figure shows the orbit of a planet moving around the Sun and Kepler has formulated it.
ii. The first law of planetary motion is ‘The orbit of a planet is an ellipse with the Sun at one of the foci’.
iii. Applying second law of planetary motion, we can say that – The straight lines AS and CS sweep equal area in equal interval of time i.e. area ASB and CSD are equal.
2. Observe the diagram and answer the question.

 Name of the force and the scientist who discovered it ?
 Book written by this scientist.
 Define the famous law given by this scientist.
Ans i . Gravitational force discovered by Sir Issac Newton
ii Book written by Sir Issac Newton is “Principia”.
iii. Newton’s universal law of gravitation states that, ‘Every object in the universe attracts every other object with a definite force which is directly proportional to the product of the masses of the two objects and inversely proportional to the square of the distance between them.
Q.21. Answer the following :
1. Variation in value of ‘g’ with change along the surface of the earth. Discuss.
Ans i. The shape of the earth is not exactly spherical and so the distance of a point on the surface of the earth from its centre differs somewhat from place to place.
 Due to its rotation, the earth bulges at the equator and is flatter at the poles.
 Its radius is largest at the equator and smallest at the poles.
 The value of g is thus highest (9.832 m/s2) at the poles and decreases slowly with decreasing latitude. It is lowest (9.78 m/s2) at the equator.
2 .Explain the change in Gravitational Acceleration due to change along the surface of the earth. Ans: i. Gravitation Acceleration g is given by formula = GM/R^{2} which shows it is inversely proportional to the distance of surface from centre of the earth.
ii.The radius of earth i.e. the distance of the surface from centre of the earth varies along the poles and equator.
iii.This is because earth is not exactly spherical but flattened at poles and slightly bulged at the equatorial region.
iv. As the distance is smaller at poles, ‘g’ is maximum while it is greater at equator and thus ‘g’ is lowest at equator on the surface of the earth.
2. What is the difference between mass and weight of an object. Will the mass and weight of the object on earth be same as their values on Mars ? Why ?
Ans
Mass 
Weight 

Weight is the force on an object due to gravitational pull. 

i. 
Mass is the amount matter present in an object. 

Mass of any object is universally constant and does not change. 
Wight changes with the change in Gravitational acceleration. 

ii. 

iii. 
Mass is measured by a beam balance. 
Weight is measured by a spring balance. 
iv. 
It’s a scalar quantity. 
It’s a vector quantity. 

 On Mars, the mass will remain same but weight will vary from that on the earth.
 This is because the gravitational accelerating of Mars is only 3.711 m/s^{2} and hence will be different resulting in change of weight. The ‘g’ on earth is 9.8m/s^{2}. Hence the weight on Mars will be lesser or almost 1/3^{rd}.
3. Give the three kinematical equations modified during a free fall.
Ans : During a free fall, the initial velocity u=0 as the object starts falling from the rest. Acceleration on the object is due to gravity and thus a=g. Thus, the three equations modify as.
i.First kinematical equation. It shows relations between velocities, acceleration and time, give as v=u+atSince u=o and a=g, it becomes V = gt.
ii.Second kinematical equation. It shows relation between displacement, initial velocity, acceleration and time, given as S = ut + 1/2 at^{2}. Since u = o and a = g, it becomes s = 1/2 gt^{2}, where ‘s’ represents height from
which object falls.
iii. Third kinematical equation. It shows relation between velocities, acceleration and displacement, given as v^{2}= u^{2} + 2 as. Since u = o and a = g, it becomes V^{2} = 2gs.
3. Explain the variation of values of gravitational acceleration.
Ans Variation in value of ‘g’ is observed as follows :
i. Change along with surface of the earth.
a. Due to rotational motion, earth bulges at equator and is flattened at the poles. Its radius is largest at equator and smallest at the poles.
b. Thus, value of ‘g’ is highest at poles and lowest at the equator as value of ‘g’ is inversely proportional to the radius.
ii. Change with height.
a. As the distance from the centre increase, the value of a decreases.
b. However, the decrease in value of ‘g’ is negligible as the distance increased is considerable with earth’s radius.
iii. Change with depth.
a. As the distance decrease the value of ‘g’ is expected to increase and thus ‘g’ should increase along with depth.
b. But in this case, ‘g’ decrease with depth because the mars of the earth which contributes towards gravitation decreases.
c. Therefore, value of gravitation acceleration varies along different conditions.
Q.22. Answer the following in detail :
1. Observe the given figure and answer the following questions.
The orbit of a planet is an ellipse with the Sun at one of its foci. The Sun’s position is indicated as ‘O’. X_{1}Y_{1} and X_{2}Y_{2} are the distances covered by the planet in equal time, X_{1}O and X_{2}O lines sweep equal area in equal intervals of time. Hence, areas X_{1}OY_{1} and X_{2}OY_{2} are equal.
i.Which laws do we understand from the description given above?
ii. At what point amongst X_{1} and X_{2} velocity of the planet be more? Why?
iii. If shaded area X_{3}OY_{3} is twice the are X_{1}OY_{1} then what will be the relation between time taken by the planet to move X_{1} to Y_{1}, say t_{1} and time taken by the planet to move from X_{3} to Y_{3}, say t_{2}?
IV. Kepler’s third law is expressed mathematically as ^{1}m/r^{n} = Constant. What are the values of m and n in the give expression
Ans i. Kepler’s laws of planetary motion are understood from the given description.
ii. Given : Area of X_{1}OY_{1} = Area of X_{2}OY_{2}. But are X_{1}Y_{1} is greater than are X_{2}Y_{2}. This means planet has to cover larger distance while moving from X_{1} to Y_{1} than from X_{2} to Y_{2} in the same time. This requires velocity of planet to be more while moving from X_{1} to Y_{1}, than from X_{2} to Y_{2}. Hence, velocity of the planet is more at X_{1} than at X_{2}.
iii. As area X_{3}OY_{3} = 2 x area X_{1}OY_{1}, the time taken to go from X_{3} to Y_{3} is equal to twice the time taken to go from X_{1} to Y_{1}.
∴ t_{2} = 2t_{1}
iv. In the given expression, m = 2 and n = 3.
..
For a planet revolving around sun; Let m be the mass of planet which takes time T for one revolution moving with velocity v and r be the radius of the circular path.
Centripetal force will be F = ………….. (1)
Speed =……………..
Thus, in one revolution,
Distance covered = ……………….(Perimeter of the orbit)
Time required = T (……………… )
∴ …………………….= 2πr/T
Thus, this is expression of centripetal force independent of time taken but depends on radius of the path.
3. Waves are created on the surface of water when we drop a stone into it. Similarly you must have seen the waves generated on a string when both its ends are held in hand and it is shaken. Light is also a type of wave called the electromagnetic wave. Gamma rays, Xrays, ultraviolet rays, infrared rays, microwave and radio waves are all different types of electromagnetic waves. Astronomical objects emit these waves and we receive them using our instruments. All our knowledge about the universe has been obtained through these waves. Gravitational waves are a very different type of waves. They have been called the waves on the fabric of spacetime. Einsteine predicted their existence in 1916. These waves are very weak and it is very difficult to detect them. Scientists have constructed extremely sensitive instruments to detect the gravitational waves emitted by astronomical sources. Among these, LIGO (Laser Interferometric Gravitational Wave Observatory) is the prominent one. Exactly after hundred years of their prediction, scientists detected these waves coming from an astronomical source. Indian scientists have contributed significantly in this discovery. This discovery has opened a new path to obtain information about the Universe.
I. Which type of wave a light ray is ? (1 marks) II.What type of rays are Gravitational waves called ? (1 marks) III. Why are the gravitational waves difficult to detect ?(1 marks) IV.Give any four examples of electromagnetic waves. (2 marks)
Ans I. Light ray is a type of wave called the electromagnetic wave.
II Gravitational waves are a very different type of waves called as the waves on the fabric of spacetime.
III As the Gravitational waves are very weak and it is very difficult to detect them.
IV Gamma rays, Xrays, ultraviolet rays, infrared rays, microwave and radio waves are all different types of electromagnetic waves. [any four]