Q: Question 1How does the force of gravitation between two objects change when the distance between them is reduced to half?

We know,F = G m1m2d2\dfrac{\text{m}_1\text{m}_2}{\text{d}^2}d2m1m2Where,F = force of attractionG = constant of proportionalitym1 = mass of first objectm2 = mass of second objectd = distance between the two objects.When distance is reduced to half then, d' = d2\dfrac{\text{d}}{2}2dSo, substituting we get,F' = G m1m2(d2)2\dfrac{\text{m}_1\text{m}_2}{\Big(\dfrac{\text{d}}{2}\Big)^2}(2d)2m1m2 = G m1m2d24\dfrac{\text{m}_1\text{m}_2}{\dfrac{\text{d}^2}{4}}4d2m1m2 = 4G m1m2d2\dfrac{\text{m}_1\text{m}_2}{\text{d}^2}d2m1m2 = 4FHence, when the distance between the objects is reduced to half, the gravitational force increases four times.

Q: Question 6What 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?

(i) We know,F = G m1m2d2\dfrac{\text{m}_1\text{m}_2}{\text{d}^2}d2m1m2Where,F = force of attractionG = constant of proportionalitym1 = mass of first objectm2 = mass of second objectd = distance between the two objects.When mass of one object is doubled.m' = 2m1So, substituting we get,F' = G 2m1m2d2\dfrac{\text{2m}_1\text{m}_2}{\text{d}^2}d22m1m2 = 2G m1m2d2\dfrac{\text{m}_1\text{m}_2}{\text{d}^2}d2m1m2 = 2FHence, when the mass of one object is doubled, the gravitational force increases two times.(ii) When distance is doubled then, d' = 2rSo, substituting we get,F' = G m1m2(2d)2\dfrac{\text{m}_1\text{m}_2}{(2d)^2}(2d)2m1m2 = G m1m24d2\dfrac{\text{m}_1\text{m}_2}{4d^2}4d2m1m2 = F4\dfrac{\text{F}}{4}4FHence, when the distance between the objects is doubled, the gravitational force becomes one fourth of its original force.Now, if it's tripledd' = 3rSo, substituting we get,F' = G m1m2(3d)2\dfrac{\text{m}_1\text{m}_2}{(3d)^2}(3d)2m1m2 =G m1m29d2\dfrac{\text{m}_1\text{m}_2}{9d^2}9d2m1m2 = F9\dfrac{\text{F}}{9}9FHence, when the distance between the objects is tripled, the gravitational force becomes one ninth of its original force.(iii) If masses of both the objects are doubled, thenm1' = 2m1andm2' = 2m2So, substituting we get,F' = G 2m12m2d2\dfrac{\text{2m}_1\text{2m}_2}{\text{d}^2}d22m12m2 = 4G m1m2d2\dfrac{\text{m}_1\text{m}_2}{\text{d}^2}d2m1m2 = 4FHence, when the masses of both the objects are doubled, the gravitational force becomes four times the original force.

Question 1

Question 1How does the force of gravitation between two objects change when the distance between them is reduced to half?

Accepted Answer

We know,F = G m1m2d2\dfrac{	ext{m}_1	ext{m}_2}{	ext{d}^2}d2m1​m2​​Where,F = force of attractionG = constant of proportionalitym1 = mass of first objectm2 = mass of second objectd = distance between the two objects.When distance is reduced to half then, d' = d2\dfrac{	ext{d}}{2}2d​So, substituting we get,F' = G m1m2(d2)2\dfrac{	ext{m}_1	ext{m}_2}{\Big(\dfrac{	ext{d}}{2}\Big)^2}(2d​)2m1​m2​​ = G m1m2d24\dfrac{	ext{m}_1	ext{m}_2}{\dfrac{	ext{d}^2}{4}}4d2​m1​m2​​ = 4G m1m2d2\dfrac{	ext{m}_1	ext{m}_2}{	ext{d}^2}d2m1​m2​​ = 4FHence, when the distance between the objects is reduced to half, the gravitational force increases four times.

Question 2

Question 2Gravitational force acts on all objects in proportion to their masses. Why then, a heavy object does not fall faster than a light object?

Accepted Answer

All objects fall on ground with constant acceleration, called acceleration due to gravity (in the absence of air resistances). This acceleration remains constant and is independent of an object's mass. Hence, heavy objects do not fall faster than light objects.

Question 3

Question 3What 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.)

Accepted Answer

Mass of the body (m) = 1 kgMass of the earth (M) = 6 × 1024 kgRadius of the Earth (R) = 6.4 × 106 mG = 6.67 x 10-11 Nm2kg-2According to the formula for gravitational force we know,F = G MmR2\dfrac{\text{M}\text{m}}{\text{R}^2}R2Mm​Substituting we get,F=6.67×10−11×6×1024×1(6.4×106)2=6.67×10136.4×6.4×1012=9.77 N≈9.8 N\text{F} = \dfrac{6.67 \times 10^{-11}\times 6 \times 10^{24} \times 1}{(6.4 \times 10^6)^2} \[1em] = \dfrac{6.67 \times 10^{13}}{6.4 \times 6.4 \times10^{12}} \[1em] = 9.77 \text{ N} \approx 9.8 \text{ N}F=(6.4×106)26.67×10−11×6×1024×1​=6.4×6.4×10126.67×1013​=9.77 N≈9.8 NHence, magnitude of the gravitational force = 9.8 N

Question 4

Question 4The 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?

Accepted Answer

According to universal law of gravitation, two objects with masses attract each other with equal gravitational force, but in opposite directions.This force is given by: F = G MmR2\dfrac{	ext{M}	ext{m}}{	ext{R}^2}R2Mm​Hence, the earth attracts the moon with the same force with which the moon attracts the earth.

Question 5

Question 5If the moon attracts the earth, why does the earth not move towards the moon?

Accepted Answer

According to the universal law of gravitation, two objects with masses attract each other with equal gravitational force, but in opposite directions. Hence, moon and earth both attract each other with the same force. Therefore, the earth does not move towards the moon.

Question 6

Question 6What 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?

Accepted Answer

(i) We know,F = G m1m2d2\dfrac{	ext{m}_1	ext{m}_2}{	ext{d}^2}d2m1​m2​​Where,F = force of attractionG = constant of proportionalitym1 = mass of first objectm2 = mass of second objectd = distance between the two objects.When mass of one object is doubled.m' = 2m1So, substituting we get,F' = G 2m1m2d2\dfrac{	ext{2m}_1	ext{m}_2}{	ext{d}^2}d22m1​m2​​ = 2G m1m2d2\dfrac{	ext{m}_1	ext{m}_2}{	ext{d}^2}d2m1​m2​​ = 2FHence, when the mass of one object is doubled, the gravitational force increases two times.(ii) When distance is doubled then, d' = 2rSo, substituting we get,F' = G m1m2(2d)2\dfrac{	ext{m}_1	ext{m}_2}{(2d)^2}(2d)2m1​m2​​ = G m1m24d2\dfrac{	ext{m}_1	ext{m}_2}{4d^2}4d2m1​m2​​ = F4\dfrac{	ext{F}}{4}4F​Hence, when the distance between the objects is doubled, the gravitational force becomes one fourth of its original force.Now, if it's tripledd' = 3rSo, substituting we get,F' = G m1m2(3d)2\dfrac{	ext{m}_1	ext{m}_2}{(3d)^2}(3d)2m1​m2​​ =G m1m29d2\dfrac{	ext{m}_1	ext{m}_2}{9d^2}9d2m1​m2​​ = F9\dfrac{	ext{F}}{9}9F​Hence, when the distance between the objects is tripled, the gravitational force becomes one ninth of its original force.(iii) If masses of both the objects are doubled, thenm1' = 2m1andm2' = 2m2So, substituting we get,F' = G 2m12m2d2\dfrac{	ext{2m}_1	ext{2m}_2}{	ext{d}^2}d22m1​2m2​​ = 4G m1m2d2\dfrac{	ext{m}_1	ext{m}_2}{	ext{d}^2}d2m1​m2​​ = 4FHence, when the masses of both the objects are doubled, the gravitational force becomes four times the original force.

Question 7

Question 7What is the importance of universal law of gravitation?

Accepted Answer

The universal law of gravitation successfully explained several phenomena that were believed to be unconnected:The force that binds us to the earthThe motion of the moon around the earthThe motion of planets around the sunThe tides due to the moon and the sun.

Question 8

Question 8What is the acceleration of free fall?

Accepted Answer

The acceleration of free fall is denoted by g and its value on the surface of the earth is 9.8 ms-2.

Question 9

Question 9What do we call the gravitational force between the earth and an object?

Accepted Answer

Gravitational force between the earth and an object is known as the weight of the object.

Question 10

Question 10Amit 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.]

Accepted Answer

The weight of a body on the earth's surface (W) = mg (where m = mass of the body and g = acceleration due to gravity)Given, value of g is larger at poles when compared to the equator. So, for the same mass of gold, weight will be less at the equator as compared to the poles. Therefore, Amit's friend will not agree with the weight of gold bought.

Question 11

Question 11Why will a sheet of paper fall slower than one that is crumpled into a ball?

Accepted Answer

A sheet of paper has a larger surface area in comparison to a crumpled paper ball. The larger surface area will suffer greater air resistance. Hence, a sheet of paper falls slower than the crumpled ball.

Question 12

Question 12Gravitational force on the surface of the moon is only 16\dfrac{1}{6}61​ 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?

Accepted Answer

Given,Acceleration due to earth's gravity g = 9.8 ms-2Object's mass, m = 10 kgAcceleration due to moon gravity = gmWeight on the earth = WeWeight on the moon = WmWeight = mass x gravitygm = 16\dfrac{1}{6}61​ ge (given)So, Wm = m gm = m x 16\dfrac{1}{6}61​ x geWm = 10 x 16\dfrac{1}{6}61​ x 9.8 = 16.34 NWe = m x ge = 10 x 9.8We = 98 NHence, weight on moon and earth = 16.34 and 98 N respectively.

Question 13

Question 13A 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.

Accepted Answer

Given,Initial velocity (u) = 49 ms-1Final velocity v at maximum height = 0Acceleration due to earth gravity g = -9.8 ms-2 (negative as ball is thrown up).Let H be the maximum height to which the ball rises.By third equation of motion,2gH = v2 - u2Substituting we get,2 × (-9.8) × H = 0 - (49)2-19.6 H = -2401H = −2401−19.6\dfrac{-2401}{-19.6}−19.6−2401​ = 122.5 mHence, the maximum height to which it rises = 122.5 m(ii) Total time T = Time to ascend (Ta) + Time to descend (Td)According to the first equation of motion,v = u + gtSubstituting we get,0 = 49 + (-9.8) x TaTa = 499.8\dfrac{49}{9.8}9.849​ = 5 sAlso, Td = 5 s∴ T = Ta + Td = 5 + 5 = 10 sHence, total time it takes to return to the surface of the earth = 10 s

Question 14

Question 14A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.

Accepted Answer

Given,Initial velocity (u) = 0Tower height = total distance = 19.6 mg = 9.8 ms-2Final velocity (v) = ?According to the third equation of motion,v2 = 2gs + u2Substituting we get,v2 = (2 x 9.8 x 19.6) + 0v = 384.16\sqrt{384.16}384.16​v = 19.6 ms-1Hence, final velocity = 19.6 ms-1

Question 15

Question 15A stone is thrown vertically upward with an initial velocity of 40 ms-1. Taking g = 10 ms-2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone?

Accepted Answer

Given,Initial velocity (u) = 40 ms-1Final velocity (v) = 0g = 10 ms-2Max height = ?According to the third equation of motion,v2 = u2 - 2gsNote: [negative g as the object goes up]Substituting we get,0 = (40)2 - 2 x 10 x s⇒ 1600 = 20s⇒ s = 160020\dfrac{1600}{20}201600​ = 80 mTotal Distance = sa + sd = 80 + 80 = 160 mwhere, sa and sd are distance going up and coming down, respectively.Displacement = 0 [as the stone comes to the same point from where it was thrown ].Hence, total distance covered = 160 m and displacement = 0, as the first point is the same as the last point

Question 16

Question 16Calculate 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.

Accepted Answer

Given,Mass of the sun (m) = 2 × 1030 kgMass of the earth (M) = 6 × 1024 kgG = 6.67 x 10-11 Nm2kg-2Average distance R = 1.5 × 1011 mAccording to the formula for gravitational force we know,F = G MmR2\dfrac{	ext{M}	ext{m}}{	ext{R}^2}R2Mm​Substituting we get,F = 6.67×10−11×6×1024×2×1030(1.5×1011)2\dfrac{6.67 	imes 10^{-11}	imes 6 	imes 10^{24} 	imes 2 	imes 10^{30} }{(1.5 	imes 10^{11})^2}(1.5×1011)26.67×10−11×6×1024×2×1030​ = 3.56 x 1022 NHence, magnitude of the gravitational force = 3.56 x 1022 N

Question 17

Question 17A 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 ms-1. Calculate when and where the two stones will meet.

Accepted Answer

Given,(a) In the case, when the stone falls from the top of the tower,Initial velocity u = 0Distance travelled = xTime taken = tAccording to the second equation of motion,S = ut + 12\dfrac{1}{2}21​gt2⇒ S = 0 x t + 12\dfrac{1}{2}21​gt2⇒ S = 12\dfrac{1}{2}21​gt2   [Eq 1](b) When another stone is projected vertically upwards,Initial velocity u = 25 ms-1Distance travelled = (100 - x)Time taken = tUsing S = ut + 12\dfrac{1}{2}21​gt2we get,S' = 25t - 12\dfrac{1}{2}21​gt2   [Eq 2]From equations (1) and (2)S + S' = 12\dfrac{1}{2}21​gt2 + 25t - 12\dfrac{1}{2}21​gt2100 = 12\dfrac{1}{2}21​gt2 + 25t - 12\dfrac{1}{2}21​gt2100 = 25tt = 10025\dfrac{100}{25}25100​ = 4 sec.Hence, after 4 secs, the two stones will meet.From (1)S = 12\dfrac{1}{2}21​ x 10 x 42 = 5 x 16 = 80m.Hence, after 4 sec, 2 stones meet a distance of 80 m from the top.

Question 18

Question 18A 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 4s.

Accepted Answer

Given,Ball returns to thrower in 6s. So, timeup + timedown = 6 s hence, timeup = 3 s.g = 9.8 ms-2(a) Final velocity (v) = 0From first equation of motionv = u – gtup⇒ u = v + gtup⇒ u = 0 + (9.8 x 3)⇒ u = 29.4 ms-1Hence, the velocity with which it was thrown up = 29.4 ms-1.(b) According to the second equation of motion,S = ut - 12\dfrac{1}{2}21​gt2Substituting we get,S = (29.4 x 3) - (12\dfrac{1}{2}21​ x 9.8 x 3 x 3)= 88.2 - 44.1 = 44.1 mHence, maximum height stone reaches is 44.1 m(c) In 3 sec, it reaches the maximum height.Distance travelled in another 1 sec = s'According to the second equation of motion,S = ut + 12\dfrac{1}{2}21​gt2g is +ve as ball is going downSubstituting we get,S = 0 + (12\dfrac{1}{2}21​ x 9.8 x 1 x 1)= 4.9 mHence, in 4 sec the ball will be 4.9 m from the top or 39.2 m (i.e., 44.1 - 4.9) from the bottom.

Question 19

Question 19In what direction does the buoyant force on an object immersed in a liquid act?

Accepted Answer

The buoyant force acts vertically upwards on the object that is immersed in a liquid.

Question 20

Question 20Why a block of plastic when released under water come up to the surface of water?

Accepted Answer

When plastic is immersed in water, two forces act on it:Gravitational force or its weight in vertically downward directionBuoyant force in vertically upward direction.As the density of plastic is less than that of water, the buoyant force is greater than its weight. Hence, the plastic comes up to the surface.

Gravitation

Exercises

Intext Questions 1

Intext Questions 2

Intext Questions 3

Intext Questions 4

Intext Questions 5