Maharashtra board class 10 science part 1 chapter 1 gravitation solutions



Question 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 centre

Weight.               Kg.         Measure of inertia

Acceleration    nm2/kg2  same in the entire  due to gravity                          universe

Gravitational      N.              Depends on
constant                                    height.

Answer -:
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.                 Zero at the                                                                center
Acceleration      m/s2.           Depends on
 due to gravity                           height

Gravitational    nm2/ kg2.     Same in the
 constant                                entire universe


Question 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?
Answer : 

         Mass.                                 Weight
1.Mass is the amount      weight is the force
of matter contained       exerted on a body
in a body.                         Such as earth ,the                                                  sun and the moon.

2.  Mass is the intrinsic   weight is extrinsic
property of a body          property of a body.

3.The mass of a body    The weight of body
Cannot be zero.               Can be zero.

4.mass is the measure     weight is the                of inertia.                       Measure of force.

5. The mass of a body   The weight of a body
remains the same          depends on the local
everywhere in the         acceleration due to
Universe.                         gravity where it is                                                 placed.

6. The SI unit of mass    SI unit is Newton (N)
 is kilogram(kg).

7.The mass of a body    The weight of a body
Can be measured           can be measured
Using a beam               using a spring balance
 balanced and                and a weighing
pan balance                    machine.


The mass of an object on the Earth will be same as that on Mars but its weight on both the planets will be different. This is because the weight (W) of an object at a place depends on the acceleration due to gravity of that place i.e. W=mg or W∝gW=mg or W∝g and since the values of acceleration due to gravity on both the planets differ, thus the weight of the object will be different for both the planets.


b. What  are (i) free fall, (ii) acceleration due to gravity (iii) escape velocity (iv) centripetal force ?
Answer :  (i) A body is said to be under free fall when no other force except the force of gravity is acting on it.

(ii) The acceleration with which an object moves towards the centre of Earth during its free fall is called acceleration due to gravity. It is denoted by the letter ‘g’. It is a constant for every object falling on Earth’s surface.

(iii) The minimum velocity required to project an object to escape from the Earth's gravitational pull is known as escape velocity. It is given as:
ve=2gR−−−−    (iv)

The force required to keep an object under circular motion is known as centripetal force. This force always acts towards the centre of the circular path.


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

First Law: The orbits of the planets are in the shape of ellipse, having the Sun at one focus.

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 planets orbit.

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

Newton used Kepler’s third law of planetary motion to arrive at the inverse-square rule. He assumed that the orbits of the planets around the Sun are circular, and not elliptical, and so derived the inverse-square rule for gravitational force using the formula for centripetal force. This is given as:
F = mv2/ r ...(i) where, m is the mass of the particle, r is the radius of the circular path of the particle and v is the velocity of the particle. Newton used this formula to determine the force acting on a planet revolving around the Sun. Since the mass m of a planet is constant, equation (i) can be written as:
F ∝ v2/ r ...(ii)

Now, if the planet takes time T to complete one revolution around the Sun, then its velocity v is given as:
v = 2πr/ T  ...(iii)

where, r is the radius of the circular orbit of the planet
or, v ∝ r/ T ...(iv)  [as the factor 2π is a constant]
On squaring both sides of this equation, we get:
v2 ∝ r2/ T2...(v)

On multiplying and dividing the right-hand side of this relation by r, we get:

v2∝r3T2×1r ...(vi)

According to Kepler’s third law of planetary motion, the factor r3/ T2  is a constant. Hence, equation (vi) becomes:
v2 ∝ 1/ r...(vii)
On using equation (vii) in equation (ii), we get: F∝1r2

Hence, the gravitational force between the sun and a planet is inversely proportional to the square of the distance between them.


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.
Answer :
S = h
u = u
v = 0
a = -g

Let t be the time taken by the ball to reach height h. Thus, using second equation of motion, we have
​−2gh=v2−u2
⇒u=2gh−−−√Now, from first equation of motion, we have
v=u−gtt=ug=2hg−−√   .....(i)​

Now, from first equation of motion, we have v=u-gtt=ug=2hg   .....(i)
For vertical downward motion of the stone:
S = h
u = 0
a = g

Let v' be the velocity of the ball with which it hits the ground. Let t' be the time taken by the ball to reach the ground. Thus, using second equation of motion, we have
​2gh=v'2⇒
v'=2gh−−√Now, from first equation of motion, we have
 v'=u+gt't'=
v'g=2hg−−√    .....

Hence, from (i) and (ii), we observe that the time taken by the stone to go up is same as the time taken by it 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?
Answer :
Let the mass of the heavy object be m. Thus, the weight of the object or the pull of the floor on the object is
W = mg
Now, if g becomes twice, the weight of the object or the pull of the floor on the object also becomes twice i.e. W' = 2mg = 2W
Thus, because of doubling of the pull on the object due to the floor, it will become two times more difficult to pull it along the floor.


Question 3: Explain why the value of g is zero at the centre of the earth.
Answer : At the centre of Earth, the force due to upper half of the Earth will cancel the force due to lower half. In the similar manner, force due to any portion of the Earth at the centre will be cancelled due to the portion opposite to it. Thus, the gravitational force at the centre on any body will be 0. Since, from Newton's law, we know
F = mg
Since, mass m of an object can never be 0. Therefore, when F = 0, g has to be 0. Thus, the value of g is zero at the centre of Earth.


Question 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–√8 T.
Answer : From Kepler's third law of planetary motion, we have
T2∝r3 .....(i)

Thus, when the period of revolution of planet at a distance R from a star is T, then from (i), we have
T2∝R3 .....(ii)

Now, when the distance of the planet from the star is 2R, then its period of revolution becomes T21∝(2R)3orT21∝8R3.....(iii)

Dividing (iii) by (ii), we get
T21T2=8R3 R3⇒
T1=−√ 8T.