Ch.1: Gravitation

Gravitation | Class 10 | Science-1 | Chapter 1 | Maharashtra State Board

Link for YouTube video 👉 https://youtu.be/s3gyw_32L4g

Note: The examples solved here are explained step by step. You can write them as per your convenience.

Learn Gravitation in a simple and easy way! This chapter explains gravitational force, Newton’s law of gravitation, Kepler’s laws, free fall, acceleration due to gravity, escape velocity, weightlessness, and more with solved examples and clear concepts for better understanding and exam preparation.

Questions & Answers

1. Study the entries in the following table and rewrite them putting the connected items in a single row.

I	II	III
Mass	kg	Measure of inertia
Weight	N	Depends on height
Acceleration due to gravity	m/s²	Zero at the centre
Gravitational constant	Nm²/kg²	Same in the entire universe

2. Answer the following questions.

a. What is the difference between mass and weight of an object. Will the mass and weight of an object on the earth be same as their values on Mars? Why?

Difference between mass and weight:

Mass	Weight
Amount of matter in an object	Force with which earth attracts the object
SI unit is kg	SI unit is Newton (N)
Scalar quantity	Vector quantity
Same everywhere	Changes from place to place
Measure of inertia	Depends on gravity

The mass of an object on Mars will remain the same because mass does not depend on gravity.

The weight will change because the acceleration due to gravity on Mars is different from that on Earth.

b. What are (i) free fall (ii) acceleration due to gravity (iii) escape velocity (iv) centripetal force

(i) free fall

Free fall is the motion of an object when only gravitational force acts on it.

(ii) acceleration due to gravity

The acceleration produced in an object due to the gravitational force of the earth is called acceleration due to gravity and is denoted by g.

(iii) escape velocity

The minimum velocity required by an object to escape completely from the gravitational pull of the earth is called escape velocity.

(iv) centripetal force

The force acting on an object moving in a circular path and directed towards the centre of the circle is called centripetal force.

c. Write the three laws given by Kepler. How did they help Newton to arrive at the inverse square law of gravity?

Kepler’s Laws:

1. Kepler’s First Law: The orbit of a planet around the Sun is an ellipse with the Sun at one focus.

2. Kepler’s Second Law: The line joining the planet and the Sun sweeps equal areas in equal intervals of time.

3. Kepler’s Third Law: The square of the period of revolution of a planet is directly proportional to the cube of its mean distance from the Sun.

T² ∝ r³

Using Kepler’s third law and centripetal force relation, Newton concluded that the gravitational force between the Sun and planets is inversely proportional to the square of the distance between them.

F ∝ 1/r²

Thus he proposed the inverse square law of gravitation.

d. A stone thrown vertically upwards with initial velocity u reaches a height ‘h’ before coming down. Show that the time taken to go up is same as the time taken to come down.

For upward motion:

Using first equation of motion:

v = u − gt

At maximum height:

( v = 0 )

So,

0 = u − gt

t = u/g

Time taken to go up:

t₁ = u/g

For downward motion:

Initial velocity = 0

Final velocity after reaching ground = (u)

Using:

u = gt

t = u/g

Time taken to come down:

t₂=u/g

Therefore,

t₁=t₂

Hence, time taken to go up is equal to time taken to come down.

e. 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. Why?

Weight of an object is:

W = mg

If (g) becomes twice, the weight of the object also becomes twice.

The frictional force depends on the normal reaction, which increases with weight.

Therefore, friction becomes greater and it becomes twice as difficult to pull the object along the floor.

3. Explain why the value of g is zero at the centre of the earth.

As we go inside the earth, the gravitational force decreases because the mass contributing to gravity decreases.

At the centre of the earth, gravitational pulls from all directions become equal and opposite.

Hence, the net gravitational force becomes zero.

Therefore, acceleration due to gravity (g = 0) at the centre of the earth.

4. Let the period of revolution of a planet at a distance R from a star be T. Prove that if it was at a distance of 2R from the star, its period of revolution will be √8 T.

According to Kepler’s third law:

T²/R² = constant

For first orbit:

T²/R² = k

For second orbit:

Distance (=2R)

Let new period be (T1)

T₁² ÷ (2R)³ = k

T₁² ÷ (8R)³ = k

Since both are equal,

T²/R³ = T1²/8R³

T₁² = 8T²

T₁ = √8T

Hence proved.

5. Solve the following examples.

a. An object takes 5 s to reach the ground from a height of 5 m on a planet. What is the value of g on the planet?

We use the equation of motion:

s = ut + ½ gt²

Since the object is dropped from rest:

u = 0

So the equation becomes:

s = ½ gt²

Given

s = 5 m

t = 5 s

u = 0

Step 1: Write the formula

s = 1/2 gt²

Step 2: Substitute the values

5 = 1/2 × g × (5)²

Step 3: Solve the square

5² = 25

So,

5 = 1/2 × g × 25

Step 4: Simplify

5 = 25g ÷ 2

Multiply both sides by 2:

10 = 25g

Step 5: Find g

g = 10 ÷ 25

g = 0.4 m/s²

Answer: g = 0.4 m/s²

b. The radius of planet A is half the radius of planet B. If the mass of A is Mᴀ, what must be the mass of B so that the value of g on B is half that of its value on A?

We know the formula for acceleration due to gravity:

g = GM ÷ R²​

Given

Radius of planet A = Rᴀ

Radius of planet B = Rʙ

Given that:

Rᴀ = Rʙ ÷ 2

or

Rʙ = 2Rᴀ

Mass of planet A = Mᴀ

Let mass of planet B = Mʙ

Also given:

gʙ = gᴀ ÷ 2

Step 1: Write formula for planet A

gᴀ = GMᴀ ÷ Rᴀ²

Step 2: Write formula for planet B

gʙ = GMʙ ÷ Rʙ²

Since Rʙ = 2Rᴀ,

Rʙ² = (2Rᴀ)²

= 4Rᴀ²

So,

gʙ = GMʙ ÷ 4Rᴀ²

Step 3: Use the condition

Given:

gʙ = gᴀ ÷ 2

Substitute formulas:

GMʙ ÷ 4Rᴀ² = ½ × GMᴀ ÷ Rᴀ²

Step 4: Cancel common terms

G and Rᴀ² cancel from both sides.

So,

Mʙ ÷ 4 = Mᴀ ÷ 2

Step 5: Solve for Mʙ

Multiply both sides by 4:

Mʙ = 4 × Mᴀ ÷ 2

Mʙ = 2Mᴀ

Answer: Mass of planet B = 2Mᴀ

c. The mass and weight of an object on earth are 5 kg and 49 N respectively. What will be their values on the moon? Assume that the acceleration due to gravity on the moon is 1/6th of that on the earth.

We know that:

W = mg

where

W = weight

m = mass

g = acceleration due to gravity

Given

Mass on Earth = 5 kg

Weight on Earth = 49 N

Gravity on Moon = 1/6 of gravity on Earth

Step 1: Find the mass on the Moon

Mass does not change from place to place.

So,

Mass on Moon = 5 kg

Step 2: Find the weight on the Moon

Weight depends on gravity.

Given:

gₘ = gₑ ÷ 6

So,

Weight on Moon = Weight on Earth ÷ 6

= 49 ÷ 6

= 8.17 N

Answer: Mass on Moon = 5 kg

Weight on Moon = 8.17 N

d. An object thrown vertically upwards reaches a height of 500 m. What was its initial velocity? How long will the object take to come back to the earth? Assume g = 10 m/s²

We use the equation of motion:

v² = u² + 2as

Given

Maximum height, s = 500 m

Final velocity at highest point, v = 0 m/s

Acceleration due to gravity, g = 10 m/s²

Since gravity acts downward,

a = -10 m/s²

Step 1: Find the initial velocity

Use the formula:

v² = u² + 2as

Substitute the values:

0² = u² + 2(-10)(500)

0 = u² - 10000

u² = 10000

u = √10000

u = 100 m/s

Step 2: Find the time taken to reach the highest point

Use the formula:

v = u + at

Substitute the values:

0 = 100 + (-10)t

0 = 100 - 10t

10t = 100

t = 10 s

Step 3: Find total time to come back to Earth

Time to go up = 10 s

Time to come down = 10 s

Total time = 10 + 10

Total time = 20 s

Answer: Initial velocity = 100 m/s

Time taken to come back to Earth = 20 s

e. A ball falls off a table and reaches the ground in 1 s. Assuming g = 10 m/s², calculate its speed on reaching the ground and the height of the table.

We use the equation of motion:

v = u + at

Given

Time taken, t = 1 s

Acceleration due to gravity, g = 10 m/s²

Initial velocity, u = 0 m/s

(because the ball falls from rest)

Step 1: Find the speed on reaching the ground

Use the formula:

v = u + at

Substitute the values:

v = 0 + (10 × 1)

v = 10 m/s

Step 2: Find the height of the table

Use the equation:

s = ut + ½ at²

Substitute the values:

s = 0 × 1 + 1/2 × 10 × (1)²

s = 0 + 5 × 1

s = 5 m

Answer: Speed on reaching the ground = 10 m/s

Height of the table = 5 m

f. The masses of the earth and moon are 6 × 10²⁴ kg and 7.4 × 10²² respectively. The distance between them is 3.84 × 10⁵ km. Calculate the gravitational force of attraction between the two.

We know the formula for gravitational force:

F = Gm₁m₂ ÷ r²

Where:

• G = 6.67 × 10⁻¹¹ N m²/kg²

• m₁=6 × 10²⁴ (mass of Earth)

• m₂=7.4 × 10²² kg (mass of Moon)

• Distance r = 3.84 × 10⁵ km

First convert distance into metres:

3.84 × 10⁵ km = 3.84 × 10⁵ × 10³ m

r = 3.84 × 10⁸ m

Now substitute the values:

F = (6.67 × 10⁻¹¹) (6 × 10²⁴)(7.4 × 10²²) ÷ (3.84 × 10⁸)²

Step 1: Multiply masses

6 × 7.4 = 44.4

10²⁴ × 10²² = 10⁴⁶

So,

(6 × 10²⁴) (7.4 × 10²²) = 44.4 × 10⁴⁶

Now multiply by G:

6.67 × 44.4 = 296.148

10⁻¹¹ × 10⁴⁶ = 10³⁵

Numerator:

296.148 × 10³⁵

Step 2: Square the distance

(3.84 × 10⁸)²

= 3.84² × 10¹⁶

= 14.7456 × 10¹⁶

Step 3: Divide

F = 296.148 × 10³⁵ ÷ 14.7456 × 10¹⁶

F ≈ 20.08 × 10¹⁹

F ≈ 2.0 × 10²⁰ N

Answer: 2 × 10²⁰ N

g. The mass of the earth is 6 × 10²⁴ kg. The distance between the earth and the Sun is 1.5 × 10¹¹ m. If the gravitational force between the two is 3.5 × 10²² N, what is the mass of the Sun? Use G = 6.7 x 10⁻¹¹ N m2 kg⁻²

We use the formula of gravitational force:

F = Gm₁m₂ ÷ r²

Here we have to find the mass of the Sun.

Let the mass of the Sun be M.

So,

F = GmM ÷ r²

Rearranging the formula for M:

M = Fr² ÷ Gm

Given

F = 3.5 × 10²² N

m = 6 × 10²⁴ kg (mass of Earth)

r = 1.5 × 10¹¹ m

G = 6.7 × 10⁻¹¹ N m² kg⁻²

Step 1: Square the distance

r² = (1.5 × 10¹¹)²

= 1.5² × 10²²

= 2.25 × 10²²

Step 2: Calculate the numerator

Fr² = (3.5 × 10²²)(2.25 × 10²²)

Multiply numbers:

3.5 × 2.25 = 7.875

Multiply powers:

10²² × 10²² = 10⁴⁴

So,

Fr² = 7.875 × 10⁴⁴

Step 3: Calculate the denominator

Gm = (6.7 × 10⁻¹¹)(6 × 10²⁴)

Multiply numbers:

6.7 × 6 = 40.2

Multiply powers:

10⁻¹¹ × 10²⁴ = 10¹³

So,

Gm = 40.2 × 10¹³

Step 4: Divide

M = (7.875 × 10⁴⁴) ÷ (40.2 × 10¹³)

Divide numbers:

7.875 ÷ 40.2 ≈ 0.196

Subtract powers:

10⁴⁴ ÷ 10¹³ = 10³¹

So,

M ≈ 0.196 × 10³¹

Convert into standard scientific notation:

0.196 × 10³¹ = 1.96 × 10³⁰

Answer: 1.96 × 10³⁰ kg

Note: If you have any additional questions or feedback, please leave them in the comment section below. We will try to answer them and update this blog accordingly as soon as possible.