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In a cosmic event, suppose a planet heavier than the earth with mass KM (K > 1) and radius K'R (K' > 1) passes through a path near the earth (M and R are the mass and radius of earth).At what closest distance from surface of planet, we are in danger of being thrown into space:

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${[\frac{KGM}{g}]}^{1 / 2} - K^{'} R$

${[\frac{2 KGM}{g}]}^{1 / 2} - \frac{1}{2} K^{'} R$

${[\frac{KGM}{g}]}^{1 / 2} - \frac{1}{2} K^{'} R$

${[\frac{KGM}{2 g}]}^{1 / 2} - \frac{1}{2} K^{'} R$

answer is D.

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Detailed Solution

We will be thrown into space, if weight mg is equal to gravitational force due to the planet. If y is the closest distance.

$\begin{array}{l} \frac{GMm}{R^{2}} = mg = \frac{G (KM) m}{{(K^{'} R + y)}^{2}} \\ {(K^{'} R + y)}^{2} = \frac{KGM}{g} \\ y = {(\frac{KGM}{g})}^{1 / 2} - K^{'} R \end{array}$

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