Std 10 Science Part 1 · Chapter 1

1. Gravitation

Complete solutions to the Exercise, with step-by-step numericals. Maharashtra State Board (Balbharati), Std 10 Science and Technology Part 1.

Exercise

1Study 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

2Answer 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?

Point	Mass	Weight
Meaning	Amount of matter present in the object	Force with which the earth attracts the object
Formula	—	W = mg
SI unit	kilogram (kg)	newton (N)
Type of quantity	Scalar	Vector (directed towards the centre of the earth)
Change with place	Same everywhere	Changes from place to place as g changes
Other	It is a measure of inertia	It becomes zero at the centre of the earth

On Mars: the mass will remain the same, but the weight will be different.

Mass is the amount of matter, which does not change from one planet to another. Weight is W = mg, and g depends on the mass and radius of the planet (g = GM/R²). Mars has a different mass and radius, so its g is different (about 3.7 m/s², less than 9.8 m/s² on the earth). So the object weighs less on Mars.

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

Free fall: When an object moves under the influence of the force of gravity alone, it is said to be falling freely. This motion is called free fall. True free fall is possible only in vacuum.
Acceleration due to gravity: The acceleration produced in a body due to the gravitational force of the earth is called acceleration due to gravity (g). It is directed towards the centre of the earth. Its value on the earth’s surface is about 9.8 m/s².
Escape velocity: The minimum initial velocity with which an object must be thrown upwards so that it overcomes the earth’s gravity and never returns is called escape velocity. v_esc = √2GM/R. For the earth it is 11.2 km/s.
Centripetal force: The force acting on an object moving along a circle, directed towards the centre of the circle, is called centripetal force. Its magnitude is F = mv²/r.

Kepler’s first law: The orbit of a planet is an ellipse with the Sun at one of the foci.

Kepler’s second law: The line joining the planet and the Sun sweeps equal areas in equal intervals of time.

Kepler’s third law: The square of the period of revolution of a planet around the Sun is directly proportional to the cube of the mean distance of the planet from the Sun. T² ∝ r³, i.e. T²/r³ = K (constant).

Kepler’s first and second laws. Near the Sun the planet moves faster, so it covers a longer arc in the same time. (Not to scale)

How they helped Newton: Consider a planet of mass m moving around the Sun in a circular orbit of radius r with speed v and period T.

Centripetal force on the planet, F = mv²r
Speed, v = distancetime = 2πrT
∴ F = mr × 4π²r²T² = 4π²mrT²
Multiplying and dividing by r², F = 4π²mr² × r³T²
By Kepler’s third law, T²r³ = K, so r³T² = 1K
∴ F = 4π²mK × 1r² and 4π²m/K is a constant.
∴ F ∝ 1r²

Thus the force that keeps a planet in its orbit is inversely proportional to the square of its distance from the Sun. Newton identified this force with gravity and stated 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.

Upward motion: initial velocity = u, final velocity v = 0, acceleration a = −g. Let the time taken be t₁.

v = u + at ⇒ 0 = u − gt₁ ⇒ t₁ = ug
v² = u² + 2as ⇒ 0 = u² − 2gh ⇒ h = u²2g

Downward motion: initial velocity = 0, distance = h, acceleration a = g. Let the time taken be t₂.

s = ut + 12at² ⇒ h = 0 + 12g t₂²
∴ t₂² = 2hg = 2g × u²2g = u²g²
∴ t₂ = ug

Since t₁ = t₂ = u/g, the time taken to go up = 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?

When we pull an object along the floor, we have to work against the force of friction.
The force of friction depends on the weight of the object pressing on the floor. Weight W = mg.
If g becomes twice, the weight becomes twice (2mg). So the friction also becomes twice.
Hence we need twice the force, i.e. it becomes two times more difficult to pull the object.

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

g = GM/r². As we go deep inside the earth, only the part of the earth below the object (towards the centre) pulls it. So the value of M in the formula keeps decreasing.
At the centre of the earth, the mass of the earth is spread equally all around the object.
Each part pulls the object towards itself with equal force in opposite directions. These forces cancel each other.
So the net gravitational force on the object is zero. Hence g = 0 at the centre of the earth.

An object at the centre of the earth is pulled equally in all directions.

4Let 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.

By Kepler’s third law, T² ∝ r³, i.e. T²/r³ = constant.

Let the new period at distance 2R be T₂.

T²R³ = T₂²(2R)³
∴ T₂² = (2R)³R³ × T² = 8R³R³ × T² = 8T²
∴ T₂ = √8 T

Hence proved: the period of revolution at distance 2R is √8 T (= 2√2 T).

5Solve 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?

Given: s = 5 m, t = 5 s, u = 0 (falls freely). To find: g

s = ut + 12gt²
5 = 0 + 12 × g × (5)² = 25g2
g = 5 × 225 = 1025 = 0.4 m/s²

The value of g on the planet is 0.4 m/s².

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

Given: R_A = R_B/2, i.e. R_B = 2R_A; g_B = g_A/2. To find: M_B

g_A = GM_AR_A² and g_B = GM_BR_B² = GM_B(2R_A)² = GM_B4R_A²
Since g_B = g_A/2,
GM_B4R_A² = 12 × GM_AR_A²
M_B4 = M_A2 ⇒ M_B = 2M_A

The mass of planet B must be 2M_A.

(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.

Given: m = 5 kg, W_earth = 49 N, g_moon = g_earth/6.

Mass does not change, so mass on the moon = 5 kg
W_moon = mg_moon = m × g_earth6 = W_earth6 = 496 = 8.17 N

On the moon, mass = 5 kg and weight = 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.

Given: s = 500 m, final velocity v = 0, a = −g = −10 m/s². To find: u and total time.

v² = u² + 2as
0 = u² + 2 × (−10) × 500 ⇒ u² = 10000 ⇒ u = 100 m/s
Time to go up: v = u + at ⇒ 0 = 100 − 10t ⇒ t = 10 s
Time to come down = time to go up = 10 s
Total time = 10 + 10 = 20 s

Initial velocity = 100 m/s; time to come back = 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.

Given: u = 0, t = 1 s, a = g = 10 m/s². To find: v and s.

v = u + at = 0 + 10 × 1 = 10 m/s
s = ut + 12at² = 0 + 12 × 10 × (1)² = 5 m

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 kg, respectively. The distance between them is 3.84 × 10⁵ km. Calculate the gravitational force of attraction between the two. Use G = 6.7 × 10⁻¹¹ N m² kg⁻².

Given: m₁ = 6 × 10²⁴ kg, m₂ = 7.4 × 10²² kg, r = 3.84 × 10⁵ km = 3.84 × 10⁸ m, G = 6.7 × 10⁻¹¹ N m² kg⁻².

F = Gm₁m₂r²
F = 6.7 × 10⁻¹¹ × 6 × 10²⁴ × 7.4 × 10²²(3.84 × 10⁸)²
= 297.48 × 10³⁵14.7456 × 10¹⁶
= 20.17 × 10¹⁹ ≈ 2 × 10²⁰ N

The gravitational force between the earth and the moon is 2 × 10²⁰ N.

Always convert km into m before using the formula, because G is in SI units.

(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 × 10⁻¹¹ N m² kg⁻².

Given: m₁ = 6 × 10²⁴ kg, r = 1.5 × 10¹¹ m, F = 3.5 × 10²² N, G = 6.7 × 10⁻¹¹ N m² kg⁻². To find: mass of the Sun, m₂.

F = Gm₁m₂r² ⇒ m₂ = F × r²G × m₁
m₂ = 3.5 × 10²² × (1.5 × 10¹¹)²6.7 × 10⁻¹¹ × 6 × 10²⁴
= 3.5 × 2.25 × 10⁴⁴40.2 × 10¹³ = 7.875 × 10⁴⁴4.02 × 10¹⁴
= 1.96 × 10³⁰ kg

The mass of the Sun is 1.96 × 10³⁰ kg.

★Project: Take weights of five of your friends. Find out what their weights will be on the moon and the Mars.

Method: Measure the mass (m) of each friend on a weighing machine. Then find the weight using W = mg.

On the earth: g = 9.8 m/s², so W = m × 9.8
On the moon: g is 1/6th of the earth’s, so W = earth weight ÷ 6
On Mars: g ≈ 3.7 m/s², so W = m × 3.7

A sample table (use your own friends’ masses):

Friend	Mass (kg)	Weight on earth (N)	Weight on moon (N)	Weight on Mars (N)
1	40	392	65.3	148
2	45	441	73.5	166.5
3	50	490	81.7	185
4	55	539	89.8	203.5
5	60	588	98	222

The mass of each friend stays the same on all three. Only the weight changes, because g is different on each.

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