Class 10 Force Science Book Solution | Numerical Solution

1. Choose the correct option for the following questions:

1. What is the relation between the distance between two objects (d) and the gravitational force (F) produced between them:

Ans: $$F=\frac{1}{d^2}$$

2. What is the change in the gravitational force between two objects when their mass is doubled?

Ans: the force becomes four-time

3. If the gravitational force between two objects on Earth is 60N . what is the gravitational force hetween those two objects on the moon?

Ans: 60N

4. ‘Which one of the following statements is correct‘?

Ans: both i and ii.

5. At which Of the following places do you weigh the most

Ans: kechanakabal of Jhapa

6. The radius of the Earth is 63.71 km and the weight of an object on the Earth is 800 N. What is the weight of the object at a height of 63.71 km from the surface of the earth?

Ans: 200N

7. If the mass and the radius of a celestial body are two times the mass and the radius of the earth respectively, what is the value of acceleration due to the gravity of that body?

Ans: 4.9 m/s²

8. what will be the weight of a man on the moon? if his weight on earth is 750 N? (The acceleration due to the gravity of the moon = 1.63 m/s²)

Ans: 124.62 N

9. The mass of planet B is twice the mass of planet A but its radius is half of the radius of planet A. Similarly, the mass of planet C is half of the mass of planet A, but its radius is twice the radius of planet A. If the weight of an object in planets A. B and C is w₁,w₂,w₃ respectively, whichof the following order is correct?

Ans: w₂>w₁>w₃

10. Which one of the following conclusions is correct while observing a freely falling object every second?

Ans: acceleration increases uniformly

2. Differentiate between:

(a) Gravitational constant G and acceleration due to gravity

Ans: The difference between gravitational force and acceleration due to gravity are:

2. Mass and weight
Ans: The difference between mass and weight is described below.

3. Give reasons:

(a) Acceleration due to gravity is not the same in all parts of the earth.

Ans: Every part of the earth is of different elevation. As we know acceleration due to gravity is inversely proportional to the square of the distance. So, acceleration due to gravity is not the same in all parts of the earth.

(b) Jumping from a significant height may cause more injury.

Ans: As we know the acceleration due to gravity goes on increasing as the height decreases. Jumping from a great height increases the risk of severe injury due to higher impact force and lack of control over landing due to more weight.

(c) The mass of Jupiter is about 319 times the mass of the Earth, but its acceleration due to gravity is only about 2.6 times the acceleration due to the gravity of the Earth.

Ans: The mass of Jupiter is about 319 times the mass of the Earth but its radius is more compare to the Earth. As acceleration due to gravity is directly proportional to the mass and inversely proportional to the square of the radius. So, its acceleration due to gravity is only about 2.6 times the acceleration due to the gravity of the Earth.

(d) Among the objects dropped from the same height in the polar region and the equatorial region of the earth, the object dropped in the polar region falls faster.

Ans: The acceleration due to gravity on the polar region is more than in the equatorial region. Due to which the object dropped in the polar region from the same height falls faster.

(e) Out of two paper sheets, one is folded to form a ball. If the paper ball and the sheet of paper are dropped simultaneously in the air, the folded paper will fall faster.

Ans: The sheet of paper has more area compared to the the paper ball. Due to more area it faces more air resistance. So, the folded paper will fall faster.

(f) When a marble and a feather are dropped simultaneously in a vacuum, they reach the ground together (at the same time).

Ans: Inside the vacuum, the air resistance is completely zero. So both of them fall with constant acceleration due to gravity(freefall).

(g) As you climb Mount Everest, the weight of the goods that you carry decreases.

Ans: The weight of a body is directly proportional to the acceleration due to gravity. And, acceleration due to gravity decreases with an increase in height. So, the weight of the goods that we carry decreases as we climb Mount Everest.

(h) It is difficult to lift a big stone on the surface of the earth, but it is easy to lift a smaller one.

Ans: The big stone has more mass than the small stone. By the relation, w=mg; the object having more mass has more weight as ‘g’ is constant on the earth. So, It is difficult to lift a big stone on the surface of the earth, but it is easy to lift a smaller one.

(i) Mass of an object remains constant but its weight varies from place to place.

Ans: The acceleration due to gravity changes with changes in elevation. As weight of a body depends upon mass and acceleration due to gravity, being the mass constant, the weight varies from place to place.

(j) One will have an eerie feeling when he/she moves down while playing a Rote Ping.

Ans: The eerie feeling experienced when moving downward during a Rote Ping (a spinning amusement ride) is due to the combination of centrifugal force pushing outward and the sensation of falling. As the ride spins and tilts, it can create a disorienting sensation, making it feel like you’re falling or defying gravity, which some people find eerie or thrilling.

4. Answer the following questions:

(a) What is gravity?

Ans: The force with which a body is pulled towards the centre of the Earth is called gravity.

(b) State Newton’s universal law of gravitation.

Ans: Newton’s universal law of gravitation states that,” the force of attraction between any two bodies in the universe is directly proportional to the product of their masses and inversely proportional to the square of their centres.”

(c) Write the nature of gravitational force.

Ans: The nature of gravitational force are:

1. It increases with an increase in mass of the body.
2. It decreases with an increase in distance between the centres of the body.

(d) Define gravitational constant (G).

Ans: Universal gravitational constant(G) is defined as the force of attraction between two bodies having unit mass(1kg) each separated by unit distance(1m).

(e) Under what conditions is the value of gravitational force equal to the gravitational constant (F=G)?

Ans: When two bodies of masses 1 kg each are separated by 1m distance, then the value of gravitational force equal to the gravitational constant (F=G).

For instance, we have Newton’s law of gravitation as, $$F=\frac{GMm}{R^2}…………(i)$$

If M=1kg, m=1kg and R=1m, then eqn(i) becomes, $$F=\frac{G*1*1}{1^2}………….(ii)$$

or, $$F=G………………….(iii)$$

(f) Write two effects of gravitational force.

Ans: The two effects of gravitational force are:

1. Weight: Gravitational force gives objects weight, pulling them towards massive bodies.
2. Orbits: Gravitational force maintains stable orbits of celestial objects around each other.

(g) Mathematically present the difference in the gravitational force between two objects when the mass of each is made double and the distance between them is made one-fourth their initial distance.

Ans: we have, Newton’s universal law of gravitation as $$F=\frac{Gm_1m_2}{d^2}…………..(i)$$

If the distance between them is made by one-fourth keeping their masses doubled, then d’=1/4d , m₁‘=2m₁ & m₂‘=2m₂so eqn. (i) becomes, $$F’=\frac{G*2m_1*2m_2}{(d/4)^2}$$

or, $$F’=4\frac{Gm_1m_2}{d^2/16}$$

or, $$F’=64\frac{Gm_1m_2}{d^2}$$

or, $$F’=64F…………(ii)$$

Hence, the gravitation force will be increased by 64 times.

(h) What is gravitational force?

Ans: The force of attraction between any two bodies in the universe is called gravitational force.

(i) Define acceleration due to gravity.

Ans: Acceleration due to gravity, often denoted as “g,” is the rate at which objects fall toward the Earth’s surface under the influence of gravity, typically approximately 9.81 meters per second squared (m/s²) on the surface of the Earth.

(j) What is free fall? Give two examples of it.

Ans: Free fall is the motion of an object under the sole influence of gravity, with no other forces acting upon it. Two examples of free fall are:

1.A skydiver jumping out of an airplane and falling toward the Earth. During this descent, the only significant force acting on the skydiver is gravity, causing them to accelerate toward the ground until they deploy their parachute.

2.Dropping a ball from a certain height. As the ball falls, it is in free fall, subject only to the force of gravity until it makes contact with the ground or another surface.

(k) Under what conditions is an object said to be in free fall?

Ans: An object is said to be in free fall when it is falling under the influence of gravity alone, with no other significant forces acting upon it.

(l) Write the conclusions of the feather and coin experiment.

Ans: The conclusion that can be drawn from this experiment is that the acceleration due to gravity is independent of the mass of the falling body.

(m) What is weightlessness?

Ans: Weightlessness may be defined as the state or condition of a body of zero weight or feeling itself to be weightless.

(n) Mention any four effects of gravitational force.

Ans: Any four effects of gravitational force are:

1. Weight: Gravitational force gives objects weight, pulling them towards massive bodies.
2. Orbits: Gravitational force maintains stable orbits of celestial objects around each other.
3. Tides: Gravitational force from the Moon and Sun causes the rise and fall of sea levels.
4. Acceleration: Gravitational force makes objects fall toward Earth’s surface.

(o) Prove that acceleration due to the gravity of the Earth is inversely proportional to the square of its radius.

Ans:Let us consider ‘M’ be the mass of the earth, ‘R’ be the radius of the Earth and ‘m’ be the mass of the body. we have the force of gravitation as, $$F=\frac{GMm}{R^2}……………(i)$$ Also, acceleration due to gravity(g) is given by $$g=\frac{F}{m}……………(ii)$$ or, $$g=\frac{GMm}{R^2*m}…………(iii)$$ or, $$g=\frac{GM}{R^2}……………(iv)$$

As ‘G’ & ‘M’ are constant, $${g}\propto\frac{1}{R^2}$$

(p) Mention the factors that influence acceleration due to gravity.

Ans: The factor that influence acceleration due to gravity are:

1. mass of the bodies
2. radius of the body.

(q) The acceleration due to the gravity in the Earth surface is 9.8 m/s2. What does this mean?

Ans: The acceleration due to the gravity in the Earth surface is 9.8 m/s2, it means that a body falls 9.8 metre every second when it is thrown from certain height above the earth surface.

(r) Mass of the Moon is about 1/81 times the mass of the Earth and its radius is about 37/10 times the radius of the Earth. If the earth is squeezed to the size of the moon, what will be the effect on its acceleration due to gravity? Explain with the help of mathematical calculation.

Ans: To determine the effect on the acceleration due to gravity if the Earth were squeezed to the size of the Moon, we can use the formula for gravitational acceleration:

$$g = \frac{G \cdot M}{R^2}$$

Where:

• (g) is the gravitational acceleration
• (G) is the gravitational constant $$6.67 \times 10^{-11} \, \text{m}^3/\text{kg} \, \text{s}^2$$
• (M) is the mass of the celestial body
• (R) is the radius of the celestial body

Let’s first calculate the ratio of the gravitational acceleration of the Moon to that of the Earth.

For the Moon:
$$M_{\text{Moon}} = \frac{1}{81} \times M_{\text{Earth}}$$
$$R_{\text{Moon}} = \frac{37}{10} \times R_{\text{Earth}}$$

Now, let’s calculate the ratio of the gravitational acceleration of the Moon to that of the Earth:

$$\frac{g_{\text{Moon}}}{g_{\text{Earth}}} = \frac{\frac{G \cdot M_{\text{Moon}}}{R_{\text{Moon}}^2}}{\frac{G \cdot M_{\text{Earth}}}{R_{\text{Earth}}^2}}$$

Now, we can plug in the values:

$$\frac{g_{\text{Moon}}}{g_{\text{Earth}}}= \frac{\frac{G \cdot \frac{1}{81} \cdot M_{\text{Earth}}}{(\frac{37}{10} \cdot R_{\text{Earth}})^2}}{\frac{G \cdot M_{\text{Earth}}}{R_{\text{Earth}}^2}}$$

Now, simplify:

$$\frac{g_{\text{Moon}}}{g_{\text{Earth}}}= \frac{\frac{1}{81} \cdot \frac{1}{(\frac{37}{10})^2}}{1}$$

$$\frac{g_{\text{Moon}}}{g_{\text{Earth}}}= \frac{1}{81} \cdot \frac{100}{37^2} = \frac{100}{81} \cdot \frac{1}{37^2} = \frac{100}{81} \cdot \frac{1}{1369}$$

$$\frac{g_{\text{Moon}}}{g_{\text{Earth}}}= \frac{100}{110949}$$

So, if the Earth were squeezed to the size of the Moon, the acceleration due to gravity on the Earth would be approximately 1109.49 times stronger than the current acceleration due to gravity on the Moon.

(s) The acceleration due to gravity of an object of mass 1 kg in outer space is 2m/s². What is the acceleration due to gravity of another object of mass 10 kg at the same point? Justify with arguments.

Ans: The acceleration due to gravity of another object of mass 10 kg at the same point is also 2m/s² because the acceleration due to gravity is independent of mass of an object.

(t) A man first measures the mass and weight of an object in the mountain and then in the Terai. Compare the data that he obtains.

Ans: The mass measured by a man in the mountain and in the Terai is same but the weight of the man will be more in the Terai than in the mountain because acceleration due to gravity is more in the terai than in the mountain.

(u) A student suggests a trick for gaining profit in a business. He suggests buying oranges from the mountain selling them to Terai at the cost price. If a beam balance is used during this transaction, explain, based on scientific fact, whether his trick goes wrong or right.

Ans: In the Terai the weight of same quantity of oranges will be more than in the mountain due to more acceleration due to gravity. If a beam balance is used, the suggested trick goes right as more weight costs more for same quantity of oranges.

(v) How is it possible to have a safe landing while jumping from a flying aeroplane using a parachute? Is it possible to have a safe landing on the moon in the same way? Explain with reasons.

Ans: Using a parachute while jumping from a flying airplane allows for a safe landing on Earth because of the presence of Earth’s atmosphere and gravitational pull. Here’s how it works:

Parachute on Earth: When a person jumps from a flying airplane, they initially fall freely under gravity (free fall). At a certain altitude, they deploy a parachute, which is designed to create air resistance and slow down their descent. The atmosphere provides the necessary resistance, which, along with gravity, enables a controlled descent. The parachute’s design and material properties further ensure a gentle landing, as it allows for a gradual decrease in speed.
However, on the Moon, a parachute landing is not feasible in the same way for the following reasons:

Lack of Atmosphere: The Moon has virtually no atmosphere, so there is no air to provide the necessary resistance for a parachute to slow down a descent. Parachutes rely on air molecules to create drag, and without an atmosphere, this is not possible.

Weaker Gravity: While the Moon has less gravity than Earth, it’s not strong enough to significantly slow a descent without air resistance. The difference in gravitational pull alone wouldn’t be sufficient to ensure a safe landing.

(w) The acceleration of an object moving on the earth is inversely proportional to the mass of the object, but for an object falling towards the surface of the earth, the acceleration does not depend on the mass of the object, why?

Ans: The acceleration of an object moving on the earth is inversely proportional to the mass of the object, as stated in Newton’s second law of motion. However, for an object falling towards the surface of the earth, the acceleration does not depend on the mass of the object. This is because, in free fall, objects move under the sole influence of gravity and do not encounter air resistance. The acceleration of a free-falling object (on earth) is 9.8 m/s² which is the same for all free-falling objects regardless of their mass.

5 . Solve the following mathematical problems:

a. The mass of two objects A and B are 20 kg and 40 kg respectively. If the distance between their centers is 5m, calculate the gravitational force produced between them.

Given :
Mass of objects A (m₁) = 10 kg
Mass of objects B (m₂) = 20 kg
Distance between them (d) = 1.5 x 10¹¹ m
Gravitational constant (G) = 6.67X10^–11 Nm²/kg²
The force of gravitation (F) =?
According to the formula,

$$F\;=G\;\;\frac{\;m_1m_2}{R^2}\;$$

$$\;\;\;=\;\frac{\;6.67\times10^{-11}\times20\times40}{5^2}$$

$$\;\;\;=\;2.134\times10^{-9}$$

The gravitational force produced between them is 2.134 x 10^-9 N .

C. Mass of the sun and Jupiter are 2 x 10 ³⁰ kg and 1.9 x 10²⁷ kg respectively. If the distance between the sun and Jupiter is 1.8 x 10⁸ Km. calculate the gravitational force between Sun and Jupiter.

Given :
Mass of Sun (m₁) = 2 x 10 ³⁰ kg
Mass of earth (m₂) = 1.9 x 10²⁷ kg
Distance between them (d) = 1.8 x 10⁸ Km = 1.8 x 10¹¹ m
Gravitational constant (G) = 6.67X10^–11 Nm²/kg²
The force of gravitation (F) =?

According to the formula,

$$F\;=G\;\;\frac{\;m_1m_2}{R^2}\;$$

$$\;\;\;=\;\frac{6.67\;\times\;10^{-11}\times2\times10^{30}\times1.9\times10^{27}}{{(1.8\times10^{11})}^2}$$

$$=7.82\times10^{24\;}N$$

The gravitational force between Sun and Jupiter is 7.82 X 10²⁴ N.

(d) Gravitational force produced between the Earth and Moon is 2.01 x 10²⁰ N. If the distance between these two masses is 3.84 x 10⁵ km and the mass of the earth is 5.972 x 10²⁴ kg. calculate the mass of the moon.

Given :
Mass of Earth (m₁) = 5.972 x 10²⁴ kg
Mass of Moon (m₂) = ?
Distance between them (d) = 3.84 x 10⁵ km = 3.84 x 10⁸ m
Gravitational constant (G) = 6.67 X 10^–11 Nm²/kg²
The force of gravitation (F) =2.01 x 10²⁰ N

According to the formula,

$$F\;=G\;\;\frac{\;m_1m_2}{R^2}\;$$

$$\;\;\;2.01\;\times\;10^{20}=\;\frac{6.67\;\times\;10^{-11}\times5.972\;x\;10^{24}\times\;m_2}{{(3.84\;\times\;10^8)}^2}$$

$$m_2=7.34\;\times10\;22\;kg$$

The mass of moon is 7.34 X 10²² Kg.

(e) Gravitational force produced between the Earth and the Sun is 3.54 x 10²² N. If the masses of the Earth and sun are 5.972 X 10²⁴ kg and 2 x 10³⁰ kg respectively. what is the distance between them?

Given :
Mass of Earth (m₁) =5.972 X 10²⁴ kg
Mass of Sun (m₂) = 2 x 10³⁰ kg
Distance between them (d) = ?
Gravitational constant (G) = 6.67 X 10^–11 Nm²/kg²
The force of gravitation (F) =3.54 x 10²² N

According to the formula,

$$F\;=G\;\;\frac{\;m_1m_2}{R^2}\;$$

$$3.54\;\times\;10^{22}=\;\frac{6.67\;\times\;10^{-11}\times5.972\times10^{24}\;\times2\times\;10^{30}}{d^2}$$

$$d_2=1.5\times10^{11}\;m$$

The distance between Earth and the Sun is 1.5 x 10¹¹ m.

(f) The mass of the moon is 7.342 x 10²² kg. If the average distance between the earth and the moon is 384400 km. calculate the gravitational force exerted by the moon on every kilogram of water on the surface of the earth.

Given :
Mass of Moon (m₁) = 7.342 x 10²² kg
Mass of water (m₂) = 1 kg
Distance between them (d) = 384400 km = 3.844 x 10⁸ m
Gravitational constant (G) = 6.67 X 10^–11 Nm²/kg²
The force of gravitation (F) =3.54 x 10²² N

According to the formula,

$$F\;=G\;\;\frac{\;m_1m_2}{R^2}\;$$

$$=\;\frac{6.67\;\times\;10^{-11}\times7.342\times10^{22}\;\times1}{{(3.844\;\times\;10^8)}^2}$$

$$\;=\;3.314\times10^{-5\;}\;N$$

The gravitational force extered by moon on the 1 kg of water on earth is 3.314 x 10 ^-5 N.

(g) If the mass of the moon is 7.342 x 10²² kg and its radius is 1737 km. calculate its acceleration due to gravity.

Given,

Mass of the moon (M) = 7.342 x 10²² kg

Radius of the moon (R)=1737 km =1.737 x 10⁶ m

Gravitational constant (G) = 6.67 x 10^-11 Nm²kg^-2

Acceleration due to gravity (g) =?

According to formula,

$$g\;=\;\frac{GM}{R^2}$$

$$=\;\frac{6.67\;\times\;10^{-11}\times7.342\times10^{22}}{{(1.737\times10^6)}^2}$$

$$= 1.63 m/s2$$

The acceleration duw to gravity is 1.63 m/s² .

(h) Mass of the Earth is 5.972 x 10²⁴ kg and the diameter of the moon is 3474 km. If the earth is compressed to the size of the moon, how many times will be the change in acceleration due to the gravity of the earth so formed than that of the real Earth?

Given,

Mass of the Earth (M) = 5.972 x 10²⁴ kg

Radius of the moon (R)=1737 km =1.737 x 10⁶ m

Gravitational constant (G) = 6.67 x 10^-11 Nm²kg^-2

Acceleration due to gravity (g) =?

According to formula,

$$g\;=\;\frac{GM}{R^2}$$

$$=\;\frac{6.67\;\times\;10^{-11}\times5.972\times10^{24}}{{(1.737\times10^6)}^2}$$

$$\;=\;132.02\;m/s^2$$

As we know gravity of the earth is 9.8 m/s².

So, the acceleration due to gravity increases by 132.02/9.8 = 13.47 times.

(i) If the mass of Mars is 6.4 x 10²³ kg and its radius is 3389 km. calculate its acceleration due to gravity. What is the weight of an object of mass 200 kg on the surface of Mars?

Mass of the Mars (M) = 6.4 x 10²³ kg

Radius of the moon (R)=3389 km =3.389 x 10⁶ m

Gravitational constant (G) = 6.67 x 10^-11 Nm²kg^-2

Acceleration due to gravity (g) =?

According to formula,

$$g\;=\;\frac{GM}{R^2}$$

$$=\frac{6.67\times10^{-11}\times6.4\times10^{23}}{{(3.389\times10^6)}^2}$$

$$=3.75\;m/s^2$$

As we know , Weight of an object(W) = mg

or, w = 200 x 3.75

or , w = 750 N.

The acceleration due to gravity and weight of an object on Mars is 3.75m/s² and 750 N respectively.

(j) The acceleration due to the gravity of the earth is 9.8 m/s². If the mass of Jupiter is 319 times the mass of the Earth and its radius is 11 times the radius of the Earth, calculate the acceleration of gravity of Jupiter. What is the weight ofan object of mass 100 kg on Jupiter?

Solution:

Mass of the Jupiter (M) = 319 times mass of earth

Radius of the jupiter (R)=11 times thje radius of the earth

Gravitational constant (G) = 6.67 x 10^-11 Nm²kg^-2

Acceleration due to gravity of earth (g₁) =9.8 m/s²

Acceleration due to gravity of jupiter (g) =?

According to formula,

$$g\;=\;\frac{GM}{R^2}$$

$$g=\frac{G\;(314\;M_e)}{{(11R_e)}^2}$$

$$g=\frac{314\;G\;M_e}{{(11)}^2\;\times R_e^2}$$

$$g=\frac{314\;\times9.8}{{(11)}^2\;}$$

$$g=25.83\;m/s^2$$

if the mass n the Jupiter is 100 kg then its weight will be W= mg

or, W = 100 x 25.83

or, W = 2583 N

The acceleration due to gravity and weight of 100 kg is 25.83 m/s² and 2583N respectively.

(l) Mass of the earth is 5.972 x 10²⁴ kg and its radius is 6371 km. If the height of Mt. Everest is 8848.86 m from the sea level, calculate the weight of an object of mass 10 kg at the peak of Mt. Everest.

Given:

Mass of the Mars (M) = 5.972 x 10²⁴ kg

Radius of the moon (R)=6371 km =6.371 x 10⁶ m

Gravitational constant (G) = 6.67 x 10^-11 Nm²kg^-2

Acceleration due to gravity (g) =?

According to formula,

$$g\;=\;\frac{GM}{R^2}$$

$$=\frac{6.67\times10^{-11}\times5.972\times10^{24}}{{(6.371\times10^6)}^2}$$

$$=\;9.81$$

But at the height of mount Everest, the acceleration due to gravity will be :

$$g^1\;=g(1\;-\frac{2h}R)$$

$$g^1\;=9.81(1\;-\frac{17696}{6400000})$$

$$g^1\;=9.78\;m/s^2$$

Here the weight of 10 kg mass on top of mount everest is W = mg

or, W = 10 x 9.78

or, W = 97.8 N

Here the weight of 10 kg mass on top of mount everest is 97.8 N.

(m) The acceleration due to gravity of the Mars is 3.75m/s². How much mass can a weight-lifter lift on Mars who can lift 100 kg mass on the Earth?

Solution:

Acceleration due to gravity on Mars = 3.75m/s²

He can lift up to 100 kg on Mars then the weight of object = M x g

or , W = 100 x 9.81

or, W = 981 N

then again the mass on Mars will be W/g

or m = 981 / 3.75

or, m =261.6 Kg

He can uplift up to 261.6 kg on Mars.

(n) If a stone is dropped from a height of 15 m. how long will it take to reach the ground? Calculate the velocity of the stone when it hits the ground.

Solution:

Height (H) = 15 m

As we know ,

H = u² + 1/2 gt²

or, 15 = 0 +1/2 x 9.81 x t²

or, t = 1.75 sec

Also to calculate the velocity of stone to hit the ground,

V² = U² + 2gH

or, V² = 0 + 2 x 9.81 x15

or V = 17.15 m/s

The time to reach the ground and the velocity hitting the ground is 1.75 sec and 17.15 m/s respectively.

(0) If a cricket ball is thrown vertically upwards into the sky with a velocity of 15 m/s. to what maximum height will the ball reach?

Solution :

Initial velocity (U) = 15 m/s

Final velocity at maximum height (V) = 0 m/s

As we know,

V² = U² – 2 gH

or, 0 = 15² – 2 x 9.81 xH

or, H = 11. 47 m

The maximum height the ball will reach is 11.47 m.

Additional Problems:

A.Tick the best alternatives:

1.What is the acceleration due to gravity on the surface of the moon?

Ans: (b) 1.6m/s²

2. What would be the force of attraction between any two bodies when masses of two bodies is made double and distance between their centers is made halved?

Ans: (c) 16 times                                     

(Hint:   $$F=\frac{GMm}{R^2}$$  ; M=2M, m=2m, R=R/2)

3. Which one is the correct relation?

Ans: (b) $$gα\frac{1}{r^2}$$

4. Under what condition does our weight become zero?

Ans:(a) freefall

5. What is the weight of the body on the earth, if its weight is 5N in moon?

Ans:(c) 30N

6. What is the value of G on the surface of the moon?

Ans:(a) 6.67*10^-11 Nm²/kg²

B. Answer the following questions.

1. State Newton’s universal law of gravitation. Derive the mathematical expression for it.

Ans: Newton’s universal law of gravitation states that,” the force of attraction between any two bodies in the universe is directly proportional to the product of their masses and inversely proportional to the square of their centres.”

Derivation of Newton’s universal law of gravitation:

Let us consider, two bodies A and B having masses “m₁” and “m₂” respectively are separated by “d” distance apart. If the force of attraction between them is “F”.

Then, according to the Newton’s universal law of gravitation, $$Fα{m1}*{m2}………(i)$$(when d is kept constant), and, $$Fα\frac{1}{d^2}………………(ii)$$ (when m₁ and m₂ are kept constant)

Combining relations (i) and (ii), we get, $$Fα\frac{m1m2}{d^2}……….(iii)$$

or, $$F=G\frac{m1m2}{d^2}………(iV)$$ [Where “G” is called the universal gravitational constant whose value is 6.67*10^-11 Nm²/kg²]

Equation (iv) is the required equation.

2. How does the gravitation force of attraction between two objects having fixed mass vary if the distance between them is tripled?

Ans: we have, Newton’s universal law of gravitation as $$F=\frac{GMm}{R^2}………..(i)$$

If the distance between them is tripled keeping masses fixed, then R=3R, so eqn. (i) becomes, $$F’=\frac{GMm}{(3R)^2}$$

or, $$F’=\frac{GMm}{9R^2}$$

or, $$F’=\frac{1}{9}\frac{GMm}{R^2}$$

or, $$F’=\frac{1}{9}F…………(ii)$$

Hence, the gravitation force will be reduced by 9 times.

3. Define the universal gravitational constant. why is Newton’s law of gravitation called universal law?

Ans: Universal gravitational constant(G) is defined as the force of attraction between two bodies having unit mass(1kg) each separated by unit distance(1m). Its numerical value is 6.67*10^-11 Nm²/kg²

Newton’s law of gravitation is called universal law because it describes the fundamental force of gravity that operates between all objects with mass in the universe.

4. Under What condition do the values of gravitational force and gravitational constant become the same/equal?

Ans: The gravitational force and gravitational constant become the same or equal when the mass of each body is 1 kg and the distance between their centres is 1m. For instance, we have Newton’s law of gravitation as, $$F=\frac{GMm}{R^2}…………(i)$$

If M=1kg, m=1kg and R=1m, then eqn(i) becomes, $$F=\frac{G*1*1}{1^2}………….(ii)$$

or, $$F=G………………….(iii)$$

Hence, when mass of each body is 1 kg and distance between their centres is 1m, then the values of gravitational force and gravitational constant become same/equal.

5. The earth’s orbit is oval in shape. Explain how the magnitude of the gravitational force between the Earth and the sun changes as the Earth moves from position A to B as shown in the figure.

Ans: Since the earth’s orbit is oval in shape, as seen in the figure; the distance between the sun and position ‘A’ is more than the distance between the sun and position ‘B’. From Newton’s law of gravitation, $$F=\frac{GMm}{R^2}……….(i)$$ This implies the gravitational force between two bodies is inversely proportional to the square of the distance between their centres. i.e.$$Fα\frac{1}{R^2}………………(ii)$$. It means that when ‘R’ decreases ‘F’ increases and vice-versa.

So, The magnitude of the gravitational force between the sun and the earth at position ‘B’ is more than the force between the sun and earth at position ‘A’.

6. The acceleration due to the gravity of the moon is 1.67m/s^-2 , What does it mean?

Ans: The acceleration due to the gravity of the moon is 1.67m/s-2, it means that the body falls 1.67 metres in 1 second when dropped from a certain height on the surface of the moon.

7. Prove that the acceleration due to gravity is independent of the mass of the falling body.

Ans: Let us consider ‘M’ be the mass of the earth and ‘m’ be the mass of the falling body. we have the force of gravitation as, $$F=\frac{GMm}{R^2}……………(i)$$ Also, acceleration due to gravity(g) is given by $$g=\frac{F}{m}……………(ii)$$ or, $$g=\frac{GMm}{R^2*m}…………(iii)$$ or, $$g=\frac{GM}{R^2}……………(iv)$$ Here, the mass of the falling body(m) cancels on both the numerator and denominator.

Hence, eqn(iv) proved that the acceleration due to gravity is independent of the mass of the falling body.

8. Two heavenly bodies having mass ‘m1’ and ‘m2’ are separated by a distance ‘d’. What happens:

i) if the masses of both the bodies are doubled keeping the distance between them constant?

ii) if the distance between them is made half keeping their masses constant?

iii)if the distance between them is reduced by one-third keeping their masses constant?

Ans: (i) the masses of both the bodies are doubled keeping the distance between them constant:

we have, Newton’s universal law of gravitation as $$F=\frac{Gm1m2}{d^2}…………(i)$$

If the masses of both bodies are made doubled keeping the distance between them constant, then m1=2m1 & m2=2m2, so eqn. (i) becomes, $$F’=\frac{G*2m1*2m2}{(d)^2}$$

or, $$F’=4\frac{Gm1m2}{d^2}$$

or, $$F’=4\frac{Gm1m2}{d^2}$$

or, $$F’=4F…………(ii)$$

Hence, the gravitation force will be increased by 4 times.

(ii) the distance between them is made half keeping their masses constant:

we have, Newton’s universal law of gravitation as $$F=\frac{Gm1m2}{d^2}……………..(i)$$

If the distance between them is made halfed keeping masses constant, then d’=d/2. so eqn. (i) becomes, $$F’=\frac{Gm1m2}{(d/2)^2}$$

or, $$F’=\frac{Gm1m2}{d^2/4}$$

or, $$F’=4\frac{Gm1m2}{d^2}$$

or, $$F’=4F…………(ii)$$

Hence, the gravitation force will be increased by 4 times.

(iii) the distance between them is reduced by one-third keeping their masses constant:

we have, Newton’s universal law of gravitation as $$F=\frac{Gm1m2}{d^2}…………..(i)$$

If the distance between them is reduced by one-third keeping their masses constant, then d’=1/3d so eqn. (i) becomes, $$F’=\frac{Gm1m2}{(1/3d)^2}$$

or, $$F’=\frac{Gm1m2}{d^2/9}$$

or, $$F’=9\frac{Gm1m2}{d^2}$$

or, $$F’=9F…………(ii)$$

Hence, the gravitation force will be increased by 9 times.

9. Explain the coin and feather experiment. What conclusion can be drawn from the experiment?

Ans: Take about a 1m long glass tube as shown in the figure.

Put a coin and feather at one end and close it with a cap. Then turn it upside down. It will be observed that the coin drops faster than the feather. Then, take the air out of the tube and repeat the same experiment. You will see that the coin and the feather falling simultaneously. It happens so because the acceleration due to gravity is the same in all objects irrespective of their masses.

The conclusion that can be drawn from this experiment is that the acceleration due to gravity is independent of the mass of the falling body.

10. under what conditions do a feather and an iron ball fall at the same rate? explain with reason.
Ans: In the vacuum, the feather and an iron ball fall at the same rate because the acceleration produced on them is equal despite their unequal masses.

11. What happens to the weight of an object when it is taken from Earth to the moon? Why?
Ans: The weight of an object will decrease when it is taken from the earth to the moon’s surface as the acceleration due to the gravity of the moon is less than the earth’s surface.
weight= mass * acceleration due to gravity.

12. What is freefall? In which conditions is a body said to be in freefall?
Ans: Freefall is defined as the fall of an object under the effect of gravity only without any resistance.
The conditions in which the body is said to be in freefall are given below.
a. no air resistance
b. falling with constant acceleration due to gravity
c. when there is no effect of any force on the falling body.

13. What is weightlessness? Give any two clear examples.
Ans: Weightlessness is the condition in which the effective weight of a body becomes zero due to the zero reaction force of the earth or other heavenly bodies.
Ans: falling from the top, astronauts on the spaceship.

14. Why does a person feel weightlessness during freefall?
Ans: A person feels weightlessness during freefall because they are falling under gravity’s influence without any supporting force counteracting it, creating a sensation of floating as all objects inside the falling body, including the person, are accelerating at the same rate.

C. Differentiate between

1. Gravitational force and Acceleration due to gravity

3. Difference between the acceleration due to gravity (g) and gravitational constant (G)
Ans: The difference between the acceleration due to gravity and the gravitational constant is given below.

4. Freefall and Weightlessness

5. Weightlessness in space and Weightlessness in earth

Ans: The difference between Weightlessness in space and Weightlessness in earth are :

D. Explain with reason

1. The value of acceleration due to gravity varies from place to place on the surface of the earth.
Ans: The acceleration due to gravity is inversely proportional to the Radius of the earth, the earth is a flat surface and its radius varies from place to place so the value of acceleration due to gravity varies from place to place on the surface of the earth.

2. The weight of the body becomes greater at the pole than at the equator.
Ans: We know that the weight of the body is the product of the mass of the body and acceleration due to gravity and later one changes from place to place as the radius in the pole region is less than the equator region so the acceleration due to gravity is greater at pole region. Hence, The weight of the body becomes greater at the pole than at the equator.

3. The mass of Jupiter is 319 times more than that of the Earth but the acceleration due to gravity of Jupiter is only 2.5 times greater than that of Earth.
Ans: We know that the acceleration due to gravity is inversely proportional to the radius of the surface. Mass of the jupiter is 319 times more than that of the Earth but its acceleration due to gravity is only 2.5 times more than the Earth’s because the radius of Jupiter is about 11 times greater than the Earth.

4. A paratrooper jumping from a flying aeroplane lands safely.
Ans: When parachutists jump and open the parachute, the parachute increases the air resistance due to which net acceleration due to gravity decreases, thus a paratrooper jumping from a flying aeroplane lands safely.

5. A feather and a stone having equal mass fall at different rates in the air.
Ans: Due to their area there will be a difference in acceleration due to the gravity value on each which leads to different air resistance hence a feather and a sone having equal mass fall at different rates in the air.

E. Solve the following numerical problems.

1. Calculate the gravitational force in between two objects of mass 25kg and 20kg if the distance between them is 5m.

Given;

Mass of first object ( m₁) = 25 Kg

Mass of second object ( m₂) = 20 Kg

Distance between them (d) = 5 m

Gravitational constant (G) = 6.67X10^–11 Nm²/kg²
The force of gravitation (F) =?
According to the formula,

$$F\;=G\;\;\frac{\;m_1m_2}{R^2}\;$$

$$=\;\frac{6.67X10^{-11}\;\times25\times20}{5^2}$$

$$=\;\frac{3.335\;\times\;10^{-8}\;}{25}$$

$$=\;1.33\;\times10^{-9\;}N$$

Hence, the force of gravitation between them is 1.33 X 10^-9 N.

2. Mass of the sun is 2×10³⁰ kg and that of the earth is 6 x 10²⁴kg and the distance between them is 1.5 x 10¹¹ m. What is the gravitational force produced between them?

Given :
Mass of Sun (m₁) = 10 kg
Mass of earth (m₂) = 20 kg
Distance between them (d) = 1.5 x 10¹¹ m
Gravitational constant (G) = 6.67X10^–11 Nm²/kg²
The force of gravitation (F) =?
According to the formula,

$$F\;=G\;\;\frac{\;m_1m_2}{R^2}\;$$

$$=\frac{\;6.67\;\times\;10^{-11}\;\times2\times10^{30}\times6\times10^{24}\;}{{(1.5\times10^{11})}^2}$$

$$=\frac{\;8\;\times10^{44}}{2.25\times10^{22}}$$

$$=3.6\times10^{22}$$

Hence, the force of gravitation between them is 3.6 X 10²⁰ N.

3. Mass and the radius of the moon are 7.2×10²² kg an 1.7×10⁶ km respectively. Calculate the acceleration due to gravity of the moon and the weight of a man of mass 70kg on the moon.

Given,

Mass of the moon (M) = 7.2 x 10²² kg

Radius of the moon (R)=1.7 x 10⁶m

Gravitational constant (G) = 6.67 x 10^-11 Nm²kg^-2

Acceleration due to gravity (g) =?

According to formula,

$$g\;=\;\frac{GM}{R^2}$$

$$=\;\frac{6.67\times10^{-11}\times7.2\times10^{22}}{{(1.7\times10^6)}^2}$$

So, acceleration due to gravity (g) is 1.67m/s².

Again, mass of a man (m) = 70kg

Then, W=mg= 70 x 1.67=116.9N

Therefore, acceleration due to gravity on the surface of the moon is 1.67m/s² and the weight of the man on the moon is 116.9N.

4. Calculate the weight of an object of mass 100 kg on the surface of a planet whose mass and diameter are 4.8 x 1024 kg and 1200 km respectively.

Given,

Mass of the Object (M) = 4.8 x 10²⁴ kg

Radius of the moon (R)=12000/2 km = 6000 x 10³ = 6 x 10⁶ m

Gravitational constant (G) = 6.67 x 10^-11 Nm²kg^-2

Acceleration due to gravity (g) =?

According to formula,

$$g\;=\;\frac{GM}{R^2}$$

$$=\frac{6.67\times10^{-11}\times4.8\times10^{24}}{{(6\times10^6)}^2}$$

$$=\frac{3.2\times10^{14}}{3.6\times10^{13}}$$

$$=8.89 m/s2$$

So, acceleration due to gravity (g) is 8.89m/s².

Again, mass of a man (m) = 100kg

Then, W=mg= 100 x 8.89=889 N

Therefore, the weight of an object of mass 100 kg will be 116.9N.

5. The mass of the earth is 6 x 10²⁴ kg and its radius is 6400 km. What is the mass of an man weighing 977N?

Given,

Mass of the earth (M) = 6 x 10²⁴ kg

Radius of the earth (R)=6400 km = 6.4 x 10⁶ m

Gravitational constant (G) = 6.67 x 10^-11 Nm²kg^-2

Acceleration due to gravity (g) =?

According to formula,

$$g\;=\;\frac{GM}{R^2}$$

$$=\frac{6.67\times10^{-11}\times6\times10^{24}}{{(6.4\times10^6)}^2}$$

$$=\frac{4\times10^{14}}{4.096\times10^{13}}$$

$$=9.766\;m/s^2$$

So, acceleration due to gravity (g) is 9.77m/s².

Again, mass of a man (m) = ?

Then, W=mg

m = w/g

= 977/9.77

=100 kg

Therefore, the mass of a man is 100 kg.

6. If an object at a distance 6.4 x 10⁶ m from the centre of the earth weighs 10N, what will be its weight when it is shifted to the distance of 1.28 x10⁷ m from the centre of the earth?

Given ;

Distance from center of earth to object (R₁) = 6.4 x 10⁶ m

Distance from center of earth to object (R₂) = 1.28 x10⁷ m

Weight measured from distance R₁ = 10 N

Weight measured from distance R₂ =?

Then we know;

$$\frac{w_2}{w_1}=\;\frac{R_1^2}{R_{2^2}}$$

$$\frac{w_2}{10}=\;\frac{{(6.4\times10^6)}^2}{{(1.28\times10^7)}^2}$$

$$W_2=\;2.5\;N$$

Therefore, the weight measured from distance R₂ is 2.5 N.

Example 7

Calculate the force of gravitation between two bodies having masses of 10kg and 20kg which are kept 200cm apart.

Given :
Mass of first body (m₁) = 10 kg
Mass of second body (m₂) = 20 kg
Distance between them (d) = 200 cm = 2m
Gravitational constant (G) = 6.67X10^–11 Nm²/kg²
The force of gravitation (F) =?
According to the formula,

$$F\;=G\;\;\frac{\;m_1m_2}{R^2}\;$$

$$F\;=\frac{6.67\times10^{-11}\times10\times20}{2^2}$$

$$F\;=\frac{6.67\times10^{-9}\times2}4$$

$$=3.335\times10^{-9}\;N$$
Hence, the force of gravitation between them is 3.335 x 10^-9 N

Example 8
The mass of the earth is 6 x 10²⁴ kg and its radius is 6380 km. Calculate the gravitational force produced between the earth and an object of mass 200kg kept on the surface of the earth.
Given :

$$Mass\;of\;the\;earth(M)=6\;\times\;10^{24\;}kg$$
Mass of the object (m) = 200 kg
Radius of the earth (R)= d = 6380 km = 6380 x 10³m = 6.38 x 10⁶m
Force of attraction ( F) =?
According to the formula,
$$F=\frac{GMm}{R^2}$$

$$F=\frac{6.67\times10^{-11}\times6\times10^{24}\times200}{{(6.38\times10^6)}^2}$$

$$=\frac{6.67\times6\times200\times10^{24-11}}{6.38\times6.38\times10^{6+6}}$$

$$=\frac{8004\times10^{13}}{40.70\times10^{12}}$$

$$=196.6\;\times10\;N$$

$$=1966\;N$$

So, the gravitational force produced between them is 1966N.

Example 9
The mass of the earth is 6 x 10²⁴ kg and its radius is 6380 km. How much force of gravity of earth acts on 1 kg object?
Given,
Mass of the earth (M) = 6 x 10²⁴ kg
Radius of earth (R) = 6380 km = 6.38 x 10⁶m
Mass of the object (m) =1 kg
Force of gravity (F) = W =?
According to the formula,
$$Gravity\;(F)\;=\;\frac{GMm}{R^2}\;$$
$$=\frac{6.67\times10^{-11}\times6\times10^{24}\times1}{{(6.38\times10^6)}^2}$$
$$=\frac{40.02\times10^{13}}{40.70\times10^{12}}$$
$$=9.8\;N$$
Hence, the force of gravity of the earth on 1 kg object is 9.8N.

Example 10
The radius of the moon is 1.7X10⁶ m and its mass is 7.2 x 10²² kg. How much force of gravity of moon acts on 1kg mass on its surface?
Given,
Mass of the moon (M) = 7.2 x 107 kg
Radius of the moon (R) = 1.7X10⁶m
Mass of the object (m) = 1 kg
Force of gravity (F) = W =?

According to the formula,

$$Gravity\;(F)\;=\;\frac{GMm}{R^2}\;$$

$$=\frac{6.67\times10^{-11}\times7.2\times10^{22}\times1}{{(1.7\times10^6)}^2}$$

$$=\frac{48.02\times10^{11}}{2.89\times10^{12}}$$

$$=1.67\;N$$

The weight of 1 kg is 9.8 N on the earth whereas the weight is 1.67 N on the surface of the moon.

Example 11

The mass of the moon is 7.2 x10²² kg and its radius is 1.7 x10⁶ m then find its acceleration due to gravity. Also, calculate the weight of a man having 70kg mass on the moon.

Given,

Mass of the moon (M) = 7.2 x 10²² kg

Radius of the moon (R)=1.7 x 10⁶m

Gravitational constant (G) = 6.67 x 10^-11 Nm²kg^-2

Acceleration due to gravity (g) =?

According to formula,

$$g\;=\;\frac{GM}{R^2}$$

$$=\;\frac{6.67\times10^{-11}\times7.2\times10^{22}}{{(1.7\times10^6)}^2}$$

So, acceleration due to gravity (g) is 1.67m/s².

Again, mass of a man (m) = 70kg

Then, W=mg= 70 x 1.67=116.9N

Therefore, acceleration due to gravity on the surface of the moon is 1.67m/s² and the weight of the man on the moon is 116.9N.

Example 12

The mass of the Jupiter is 1.9 x10²⁷ kg and its radius is 71x 10 ⁶m. Calculate its acceleration due to gravity. Also, find the weight of a man of 70kg on the Jupiter.

Given,

Mass of the Jupiter (M) = 1.9 x 10²⁷ kg

Radius of the Jupiter (R) = 71* 10⁶ m

Gravitational Constant (G) = 6.67 x 10^-11 Nm²kg^-2

Acceleration due to gravity (g) =?

According to the formula,

$$g=\;\frac{GM}{R^2}$$

$$=\frac{6.67\times10^{-11}\times1.9\;\times10^{27}}{{(71\times10^6)}^2}$$

$$=25\;m/s^2$$

Again, Mass of man (m) = 70kg

According to formula, W = mg = 70 25m/s’= 1750N

Therefore, acceleration due to gravity on the surface of the Jupiter is 25m/s²
and weight of the man on the Jupiter is 1750N.

Example 13

How much mass can be lifted by weightlifter on the surface of the moon if he can lift 100kg on the earth?[g_e=9.8m/s² and g_m=1.67m/s²]

Given, on the earth, mass lifted = 100kg

Weight on earth (w,) = mxg.=100 x 9.8=980N

Thus, that man can lift 980N.

Again, On the moon (m)=? W=980N, g_m=1.67m/s²

According to formula, W=m.g,, or m=W/ g_m = 980/1.67 =586.8kg

Therefore, that man can lift 586.8kg on the moon.