Question

If distance between two objects is reduced by 60 per cent and mass of these two bodies remains same, then what happens to the gravitational force acting on them? With explanation ​

Answer 1

Answer:

The required gravitational force becomes 6.25 times the initial gravitational force.

Explanation:

The gravitational force between two objects is given by,

 $\setlength{\unitlength}{7mm} \begin{picture}(6,6)\thicklines\put(2,2){\circle{14}}\put(8,2){\circle{14}} \put(2,2){\circle*{0.15}} \put(8,2){\circle*{0.15}} \put(2,2){\line(1,0){6}} \put(4,2){\line( - 1,1){0.5}} \put(6,2){\line(1,1){0.5}} \put(4,2){\line( - 1, - 1){0.5}} \put(6,2){\line( 1, -1){0.5}} \put(4.5,0.7){\vector( - 1,0){2.7}} \put(5.4,0.7){\vector(1,0){2.8}}\put(4.75,0.6){ \bf d }\put( 3.3, - 1){ \framebox{ \bf F = \displaystyle \dfrac{Gm_1 m_2 }{ d^2} }}\end{picture}$

where

G denotes the gravitational constant

m₁ , m₂ denote the masses of the objects

d denotes the distance of separation

___________________________

Let the initial distance between the two objects be 'x'

and the gravitational force is F

Now, when the distance between the objects is reduced by 60%

new distance, d' = d - 60% of d

d' = d - 0.6d

d' = 0.4d

let the new force of gravitation be F'

It's given the masses of the two bodies remain same.

The gravitational force is inversely proportional to the square of the distance between the two objects.

 F ∝ 1/d²

   $\boxed{\tt \dfrac{F'}{F}=\dfrac{d^2}{d'^2}} \\\\ \longrightarrow \sf \dfrac{F'}{F}=\dfrac{d^2}{(0.4d)^2} \\\\ \longrightarrow \sf \dfrac{F'}{F}=\dfrac{d^2}{0.16d^2} \\\\ \longrightarrow \sf \dfrac{F'}{F}=\dfrac{1}{0.16} \\\\ \longrightarrow \sf \dfrac{F'}{F}=\dfrac{100}{16} \\\\ \longrightarrow \sf \dfrac{F'}{F}=\dfrac{25}{4} \\\\ \longrightarrow \boxed{\tt F'=6.25 \ F}$

Therefore, the gravitational force becomes 6.25 times when the distance between two objects is reduced by 60% and mass of these two bodies remains same.