What will happen to the gravitational force of attraction between two objects if the mass of each is doubled and the distance between them is also doubled?

Last updated at May 30, 2019 by Teachoo

NCERT Question 6 What happens to the force between two objects, if (i) the mass of one object is doubled? (ii) the distance between the objects is doubled and tripled? (iii) the masses of both objects are doubled? We know that gravitational force between two objects is given by, F = 𝐺𝑀𝑚/𝑟^2 where G = Gravitational constant M = Mass of object 1 m = Mass of object 2 r = Distance between the two objects When mass of one object is doubled Let Mass of Object 1 be doubled New Mass of Object 1 = 2M Thus, New Force = (𝐺 × 2𝑀 × 𝑚)/𝑟^2 = 2𝐺𝑀𝑚/𝑟^2 = 2 × Old Force ∴ If mass of one object is doubled, the force is also doubled Distance between object is doubled and tripled Distance is doubled So, New Distance = 2r New Force = 𝐺𝑀𝑚/(2𝑟)^2 = 𝐺𝑀𝑚/(4𝑟^2 ) = 1/4 × Old Force Distance is tripled So, New Distance = 3r New Force = 𝐺𝑀𝑚/(3𝑟)^2 = 𝐺𝑀𝑚/(9𝑟^2 ) = 1/9 × Old Force Therefore, When distance is doubled, Force becomes 𝟏/𝟒 times of Old Force When distance is tripled, force becomes 𝟏/𝟗 times of Old Force (iii) When mass of both objects is doubled New Mass of Object 1 = 2M New Mass of Object 2 = 2m Thus, New Force = (𝐺 × 2𝑀 × 2𝑚)/𝑟^2 = 4𝐺𝑀𝑚/𝑟^2 = 4 × Old Force ∴ If mass of both objects is doubled, the force becomes four times

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Option 3 : The force would be halved

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10 Questions 10 Marks 10 Mins

CONCEPT:

Newton's law of gravitation states that any two bodies having masses (m1 and m2) keeping at a distance (r) from each other exerts a Force of attraction on each other.
- This force is directly proportional to the masses of bodies and inversely proportional to the square of the distance between them.

\(\Rightarrow F \propto \frac{M_1M_2}{R^2} \Rightarrow F = \frac {GM_1M_2}{R^2} \)

Where G = 6.674 × 10-11 m3Kg-1s-2 is a universal constant

The dimension of the force is MLT-2 and the SI unit is Newton (N).

CALCULATION:

Given: Mass of the one body (M1) = M, Mass of the another body (M2) = 2M, initial distance = R, Final distance = 2R,F = Initial Force, F' = Final force

M1 = M, M2 = 2M, R' = 2R

\(\Rightarrow F \propto \frac{M_1M_2}{R^2}\)

\(F'=\frac{GM_1M_2}{R'^2} =\frac{G M_1(2M_2)}{(2R)^2}=\frac {2GM_1M_2}{4R^2}=\frac {1}{2}F\)

Hence, the Force would be halved is the correct answer.

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