If $g$ is the acceleration due to gravity on the surface of earth, its value at a height equal to double the radius of earth is

Question

Dr. Abubacker · Accepted Answer

(d) Acceleration due to gravity at a height $h$ form the surface of the earth
$\ \ \ \ \ \ \ \ \ \ \ \ g^{'} = g\frac{1}{\left( 1 + \frac{h}{R} ight)^{2}}$
$Given,\ \ h = 2R$
$	herefore\ \ \ \ \ \ \ g^{'} = g\frac{1}{(1 + 2)^{2}}$
$or\ \ \ \ \ \ g^{'} = \frac{g}{9}$

Q. If $g$ is the acceleration due to gravity on the surface of earth, its value at a height equal to double the radius of earth is

Explanation

Reviewed by Dr. Abubacker

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