Question

A rock is dropped from a distance RE above the surface of the earth, and is observed to have kinetic energy K1 when it hits the ground. An identical rock is dropped from twice the height (2RE) above the earth’s surface and has kinetic energy K2 when it hits. RE is the radius of the earth. what is k2/k1?

Answer 1

Answer:

$\dfrac{K_2}{K_1}=\dfrac{4}{3}$

Explanation:

When an object strikes the ground, the potential energy of the rock gets converted to the kinetic energy.

A rock is dropped from a distance $R_e$ above the surface of the earth and is observed to have kinetic energy $K_1$ when it hits the ground.

For another rock, It is dropped from twice the height $2R_e$ above the earth’s surface and has kinetic energy $K_2$ when it hits.              

For the first rock, $K_1=GMm(\dfrac{1}{R_e}-\dfrac{1}{2R_e})=\dfrac{GMm}{2R_e}$..............(1)

For the second rock, $K_2=GMm(\dfrac{1}{R_e}-\dfrac{1}{3R_e})=\dfrac{2GMm}{3R_e}$..............(2)

Dividing equation (2) by (1) we get :

$\dfrac{K_2}{K_1}=\dfrac{4}{3}$

Hence, this is the required solution.