Question

A body falls through a distance 'h' in a certain amount of time on the earth. Then if the same body is released on another planet having mass and radius twice as that of the earth, the distance through which it falls in the same time is :

• A. $h/2$
• B. $2h$
• C. $h$
• D. $4h$

Answer 1

The correct option is A $h/2$

Assume that body is released from rest.

$g\alpha \frac{M}{R^2}$ acceleration due to gravity on a planet $a=\frac{g}{2}$

And $s=ut+\frac{1}{2}a{t}^2$

$s=\frac{1}{2}a{t}^2$ as $u=0$

$\therefore$ For both cases time is same.

$\therefore s_1=\frac{1}{2}\times g\times t^2=h$

$s_2=\frac{1}{2}\frac{g}{2}\times t^2=\frac{s_1}{2}=\frac{h}{2}$