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(A) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.For Earth,$g_e = \frac{GM}{R^2}$.For the new planet,th

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$t \ s$

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(A) The acceleration due to gravity on the surface of a planet is given by $g = \frac{GM}{R^2}$.For Earth,$g_e = \frac{GM}{R^2}$.For the new planet,the mass $M' = 100M$ and the radius $R' = 10R$.Therefore,the acceleration due to gravity on the new planet is $g_p = \frac{G(100M)}{(10R)^2} = \frac{100GM}{100R^2} = \frac{GM}{R^2} = g_e$.Since the height $h$ is the same and the acceleration due to gravity $g$ is the same,the time taken to reach the ground is given by $h = \frac{1}{2}gt^2$,which implies $t = \sqrt{\frac{2h}{g}}$.Since $h$ and $g$ are identical for both,the time taken $t$ remains the same.

When a ball is dropped from a height $h$,it takes $t \ s$ to reach the ground. If the same experiment is done on a different planet having a mass $100$ times the earth's mass and a radius $10$ times the earth's radius,then the time it will take to cover the same height on the new planet is:

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