Class 10 Science Chapter 1 Gravitation Exercise

Gravitation Exercise Class 10

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

 

I	II	III
Mass	m/s2	Zero at the center of the earth
Weight	kg	Measure of inertia
Acceleration due to gravity	N.m2/kg2	Same in the entire universe
Gravitational constant	N	Depends on height

Answer:

I	II	III
Mass	kg	Measure of inertia
Weight	N	Depends on height
Acceleration due to gravity	m/s2	Zero at the center of the earth
Gravitational constant	N.m2/kg2	Same in the entire universe

Q.2. Answer the following questions

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

Answer:

The mass of an object is the amount of matter present in it. It is the same everywhere in the Universe and is never zero. It is a scalar quantity and its SI unit is kg. The weight of an object is the force with which the earth (or any other planet/ moon/star) attracts it. It is directed towards the center of the Earth. The weight of an object is different at different places on the earth. It is zero at the earth’s center. It is a vector quantity. Its SI unit is the newton (N). The magnitude of weight = mg.

The mass of an object will be the same on the Earth and on Mars, but the weight will not be the same because the value of g on Mars is different from that on the Earth.

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

Answer:

(i) Free fall:

Whenever an object moves under the influence of the force of gravity alone, it is said to be falling freely.

(ii) Acceleration due to gravity:

The acceleration produced in a body due to the gravitational force of the earth is called the acceleration due to gravity.

[Note: On the earth’s surface, the value of the acceleration due to gravity is almost uniform. If a body falls from a low altitude, the value of the acceleration due to gravity is almost the same.]

(iii) Escape velocity:

When a body is thrown vertically upward from the surface of the earth, the minimum initial velocity of the body for which the body can overcome the downward pull by the earth and can escape the earth forever is called the escape velocity.

(iv) Centripetal force:

In the uniform circular motion of a body, the force acting on the body is directed toward the center of the circle. This force 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?

Answer: 

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 area swept over per hour by the radius joining the Sun and the planet is the same in all parts of the planet’s orbit.

Kepler's Third Law: The squares of the periodic times of the planets are proportional to the cubes of their mean distances from the Sun.

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

Answer:

We have, v = u + at …..(1)
and s = ut + 12 at2 …..(2)
∴ s = (v – at) t + 12 at2
= vt – at2 + 12 at2
∴ s = vt – 12 at2 …..(3)
As the stone moves upward from A → B,
s = AB = h, t = t1,
a = -g (retardation),
u = u and v = 0
∴ From Eq. (3), h = 0 – 12 (-g)t12
∴ h = 12gt12 …..(4)
As the stone moves downward from B → A,
t = t2, u = 0, s = h and a = g
∴ from Eq. (2), h = 12 gt2 …..(5)
From Eqs. (4) and (5), t12 = t22
∴ t1 = t2 (∵ t1 and t2 are positive)

(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?
Answer:
To pull an object along the floor, it is necessary to do work against the force of friction between the object and the surface of the floor. This force of friction is proportional to the weight, mg, of the object. If the value of g becomes twice its value, the weight of the object and hence the force of friction will become double. Therefore, it will become two times more difficult to pull a heavy object along the floor.

Q.3. Explain why the value of G is zero at the center of the Earth.

Answer:

The value of g changes while going deep inside the earth. It goes on decreasing as we go from the earth’s surface towards the earth’s center.

We shall treat the earth as a sphere of uniform density. If we consider a particle of mass m at point P at a distance (R – d) from the earth’s center, where R is the radius of the earth and d is the depth below the earth’s surface, the gravitational force on the particle due to the earth is

F = where ‘M’ is the mass of the sphere of radius (R – d).

because the outer spherical shell is not effective (Fig. 1.10). In this case, the acceleration due to gravity is

where M is the mass of the earth. Thus, g decreases as d increases. It is less than that at the earth’s surface (GMR2) At the earth’s center, d = R
∴ g = 0.

[Note: The formulae given in the answer are not given in the textbook. The formula density

 =  Mass  volume      is used to find M’.]

Q.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.

Answer:

T=?, where T = period of revolution of a planet around the Sun, M = mass of the Sun, G = gravitational constant, r = radius of the orbit assumed to be circular = distance of the planet from the Sun.
For r = R, T =T1.