Maharashtra Board Class 10 Science Part – 1 Chapter 1 Gravitation Solution

Maharashtra Board Class 10 Science Part – 1 Solution Chapter 1 – Gravitation

Balbharati Maharashtra Board Class 10 Science Part – 1 Solution Chapter 1: Gravitation. Marathi or English Medium Students of Class 10 get here Gravitation full Exercise Solution.


Maharashtra Class 10


Science Part – 1 Solution






Exercise: –


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

                 i                ii           iii
Mass m/s^2 Zero at the centre
 Weight Kg Measure of the inertia
 Acceleration due to gravity Nm^2/kg^2 Same in theentire universe
Gravitational constant N Depends on heights


Ans: –

            i          ii          iii
Mass kg Measure of the inertia
Weights N Zero at the centre
Acceleration due to gravity m/s^2 Depends on  height
Gravitational constant Nm^2/kg^2 Same in 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?

Ans: – The weight of an object is the multiplication of mass and gravitational constant whereas the mass is only original value of the object.

As the gravitational constant is different in earth so the weight of and object will be different in mars comparing to earth but mass doesn’t depend on gravitation so it will be same.


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

Ans: –

  • Free fall: – With an initial velocity zero when an object moves alone under the force of gravity is known as free falling body.
  • Acceleration due to gravity: – The gravitational force which is acting on any object is known as acceleration due to gravity. It is denoted by g and its unit is m/s^2.
  • Escape velocity: – The minimum initial velocity which needed for any object to get out from the earth gravitational attraction is known as escape velocity. Its value is 11.86 km/s.
  • Centripetal force: – When an object is moving in a circular path then which force acting on it to keep it moving is known as centripetal force.


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

Ans: – Kepler three laws of planetary motion is –

  • The is at one end of foci of the elliptical path of the planet orbit.
  • The joining of the lines between the sun and the planet has equal area and equal time interval.
  • The period of revolution around the sun (T) and distance of the planet from the sun (r) are in a relation of T^2 ~ r^3.

The time period around the sun is T and the average distance between the sun and the planet is r, so according to Kepler law T^2 ~ r^3 or T^2 / r^3 = constant (k).

We know the centripetal force is F=mv^2 / r, where v = distance/time or 2πr / T.

Or, F= m× (2πr/ T) ^2 /r = 4mπ^2 × r^2 / T^2r.

Or, F = 4mπ^2 × r / T or 4mπ^2 × r^3 / T^2 × r^2 (by multiplying both r^2).

Or, F= 4π^2×/kr^2 (as constant k = T^2/r^3).

So from the equation we can say that force is inversely proportional to the average distance of sun and object which is a Newton inverse square law.


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.

Ans: – Let the time needed by the stone to reach the maximum height is t1 and t2 to coming down.

So according to Newton laws of motion v = u – gt (here a= +g as it throws upward)

Or, u= gt (when maximum height reach the velocity will be zero) or t1=u/g.

And when it downward then,S = ut + ½×at^2 or, s= ½×g×t2^2 (a=g,t=t2, u is zero as starting from reset)

So, s= h= gt2^2/2.

The downward velocity be v which is same as initial velocity u. so according to the formula, v^2 = u^2 + 2as = 0 + 2 × g × gt2^2 / 2 = g^2 ×t2^2; or, t2= u/g (v=u).

So upward and downward time are same.


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?

Ans: – when we pull any object along the floor the force due to gravity try to reduce the speed so as this case g is double the previous value so the it will be more difficult to pull a heavy object along the floor.


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

Ans: –At the centre of the earth the mass of is equally distributed to all direction so the net pull of attraction is zero.  So the value of g is zero 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.

Ans: – From Kepler law we know that T2/r^3 is constant or, T^2~ R^3.

If the R is equal to 2R then,(T1/T2) ^2 = 2R^3 / R^3

Or, T1= √8T.


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?

Ans: – As we know that S =ut + 1/2gt^2.

Here S=5m, u=0, g =? t=5s.

So, 5= ½×g× 25, or, g= 2/5 = 0.4.

So the value of g is 0.4 m/s^2.


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

Ans: –From Newton formula we know that g = GM/ r^2.

Here given r(A) = 1/2r(B).

So by questions, g(B) = ½ g(A).

Or, GM(A)/ r(A)^2 = 2×GM(B)/r(B)^2 = 2×GM(B)/2×r(A).

Or, M(B) = 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

Ans: –As the mass of any object remain unchanged so the mass will be 5kg.

The gravity on moon is g(E)/6 or 9.8/6 = 1.63 m/s^2.

So the weight will be 5×1.63 = 8.15N.


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/s

Ans: – From the Newton laws of motion v^2 = u^2 + 2as.

Here given final velocity v= 0, a = -g= -10, s= 500m.

So, o= u^2 + 2 × (-10) × 500, or, u^2 = 10000; or, u =100m/s.

The it will take to come back is t.

As v= u + at; where u=100, a=-10, t=?

So t = 100/10= 10s.

So the total time is (10 + 10= 20) 20s.


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

Ans: –From Newton equation of motion we know, v= u + at.

Here initial velocity u =0, t=1s, a=g=10.

So, v = 10×1=10 m/s.

Let the height of the table is h. so by formula

S= ut + 1/2×gt^2.  S=h.

So, h= 5.

The speed of reaching the ground is 10 m/s and height is 5m.


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

 Use G = 6.7 x 10-11 N m2 kg-2

Ans: –From force of gravitation we know that, F= Gm1×m2 / r^2.

Here given m1= 6 × 10^24 kg, m2= 7.4 × 10^22kg, r=3.84 × 10^5km or 3.84 × 10^8m.

Then,F= (6.7 × 10^-11 × 6×10^24 × 7.4×10^22)/ (3.84×10^8) ^2 = 2×10^20N.

So the gravitational force between the moon and the earth is 2×10^20 N.


g) The mass of the earth is 6 x 1024 kg. The distance between the earth and the Sun is 1.5x 1011 m. If the gravitational force between the two is 3.5 x 1022 N, what is the mass of the Sun?

Use G = 6.7 x 10-11 N m2 kg-2

Ans: –  As we know that gravitational force F=Gm1 m2/ r^2.

Here gien1 m1=6×10^24 kg,m2=?r= 1.5 × 10^11 m and F = 3.5 ×10^22N.

So, m2=Fr^2/Gm1 = (3.5 ×10^22 × (1.5×10^11) ^2)/ (6×10^24 × 6.7×10^-11)

Or, m2= 1.96×10^30 kg.

The mass of the sun is 1.96×10^30 kg.



Here is your solution of Maharashtra Board Class 10 Science Part – 1 Chapter 1 Gravitation

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Updated: December 1, 2021 — 12:54 pm

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