What happens to the force of gravity if you double the mass of one of the objects What happens if you double the distance between them?

What happens to the force of gravity if you double the mass of one of the objects What happens if you double the distance between them?

Text Solution

Solution :  (i) The gravitational force between two objects is directly proportional to the product of masses of the two object .So if one of the objcts is doubled , then the force also gets doubled (it becomes 2 times ). <br> (ii) the gravitational force between two objects is inversely proporational to the square of distance between them. <br> (a) it the distance between two objects is doubled (made 2 times) the force becomes `(1/2)^(2) or 1/4` (one - fourth ) . <br> (b) it the distance between two objects is triped (made 2 times) the force becomes `(1/3)^(2) or 1/9` (one - ninth ) . <br> The gravitational force between two objects is directly proportional to the product of their masses. So, if the masses of both the objects are doubled (made 2 times each), the force between them will become `2 xx 2`= 4 times

Option 3 : The force would be halved

What happens to the force of gravity if you double the mass of one of the objects What happens if you double the distance between them?

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CONCEPT:

  • Newton's law of gravitation states that any two bodies having masses (mand 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 M= 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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