The value of acceleration due to gravity on the Earth at a distance of 29,000 km from the surgace is `0.3 ms^(-2)`. The value of acceleration due to gravity at the same height on a planet whose mass is `66.70 xx 10^(22)` kg and diameter is 8700 km is _______ `ms_(-2)`.
`("Take" G = 6.67 xx 10^(-11) N m^(2) kg^(-2))`

Options

A. 0.05
B. 0.04
C. 0.06
D. 0.09

Answer

B

Explanation

`g = (GM)/(R + h)^(2) = (6.67 xx 10^(-11) xx 66.7 xx 10^(22))/([(4350 + 29000) xx 10^(3)]^(2))`
`= (444.889 xx 10^(11))/((33350)^(2)xx10^(6))=0.04 m s^(-2)`