A small body starts falling on to the earth from a distance equal to the radius

Question

A small body starts falling on to the earth from a distance equal to the radius of the earth's orbit. How long will the body take to reach the sun? Express the time in terms of T, period of revolution of the earth round the sun

Options:

A . T4√2

B . T2√2

C . T√2

D . T

Answer: Option A
:
A
Let x be the distance moved by the particle in time t.
Then by conservation of energy

- \frac{G M m}{d} + 0 = - \frac{G M m}{d - x} + \frac{1}{2} m v^{2}

Where M is mass of the sun and d is the initial separation.

v = \sqrt{\frac{2 G M}{d}} \sqrt{\frac{x}{d - x}}

\frac{d x}{d t} = \sqrt{\frac{2 G M}{d}} \sqrt{\frac{x}{d - x}}

d t = \sqrt{\frac{2 G M}{d}} \int_{0}^{d} \sqrt{\frac{x}{d - x}} d x

t = \sqrt{\frac{d}{2 G M}} {[\sqrt{x (d - x)} - d s i n^{- 1} \sqrt{\frac{d - x}{d}}]}_{0}^{- 1} = \sqrt{\frac{d}{2 G M}} (d s i n^{- 1} 1)

\sqrt{\frac{d}{2 G M}} \times d \times \frac{π}{2}

But,

\frac{G M m_{e}}{d^{2}} = m_{e} ω^{2} d

where

m_{e}

= mass of the earth and

ω

= angular speed of earth.

∴ G M = ω^{2} d^{2}

t = \sqrt{\frac{d}{2 ω^{2} d^{2}}} \times d \times \frac{π}{2} = \frac{π}{2 \sqrt{2} ω} = \frac{π}{2 \sqrt{2} \frac{2 π}{T_{e}}} = \frac{T_{e}}{4 \sqrt{2}}

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