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The value of acceleration due to gravity at a height of 4 R_E from surface of earth is (R_E is radius of earth and acceleration due to gravity on the surface of the earth =10 ms^-2)

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0.2 ms^-2

0.3 ms^-2

0.4 ms^-2

3 ms^-2

answer is C.

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Detailed Solution

$g = \frac{GM}{{(R_{E} + 4 R_{E})}^{2}} = \frac{GM}{25 {R_{E}}^{2}} = \frac{10}{25} = 0.4 {ms}^{- 2}$

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