Question

If the distance between two bodies becomes 8 times more than the

usual distance .The force F becomes​

Answer 1

Let the initial distance be r

then initial force,

${ \red{ \boxed{ \rm{ \: F\: initial \: = \: G \frac{m1 \times m2}{ {r}^{2} }}}}}$

when the distance between two bodies becomes 8 times the initial distance

final distance = 8r

final force,

${ \red{ \boxed{ \rm{ \: F \: final \: = \: G \frac{m1 \times m2}{ {(8r)}^{2} }} }}}$

${ \rm{ \: F\: final \: = \: G\frac{m1 \times m2}{64 \: {r}^{2} } }}$

${ \rm{ \: F\: final \: = \frac{1}{64} (G \frac{m1 \times m2}{ {r}^{2} } )}}$

${ \green{ \boxed{ \boxed{ \rm{ \: F\: final \: = \frac{1}{64} \times \: F \: initial}}}}}$

The force becomes one sixty forth times of the initial force