A small block of superdense material has a mass of 3 × 10²⁴kg. It is situated at a height h (much smaller than the earth's radius) from where it falls on the earth's surface. Find its speed when its height from the earth's surface has reduce to to h/2. The mass of the earth is 6 × 10²⁴kg.

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Solution

It is given that h is much lesser than the radius of the earth.

Mass of the earth, M_e = 6 × 10²⁴ kg
Mass of the block, M_b = 3 × 10²⁴ kg
Let V_e be the velocity of earth and V_b be the velocity of the block.

Let the earth and the block be attracted by gravitational force.

Thus, according to the conservation law of energy, the change in the gravitational potential energy will be the K.E. of block.
$G M_{e} M_{b} (\frac{1}{R + (\frac{h}{2})} - \frac{1}{R + h}) = (\frac{1}{2}) M_{e} \times V_{e}^{2} + (\frac{1}{2}) M_{b} \times V_{b}^{2}$ ...(1)

As only the internal force acts in this system, the momentum is conserved.
M_eV_e = M_bV_b
$\Rightarrow V_{e} = \frac{M_{b} V_{b}}{M_{e}} . . . (i i)$

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