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Q. The value of $g$ at a height equal to half the radius of the earth from the earth's surface is

COMEDKCOMEDK 2014Gravitation

Solution:

Acceleration due to gravity at a height $h$
from the earth's surface is $g_{h} = \frac{gR^{2}_{E} }{\left(R_{E} +h\right)^{2}}$
where g is acceleration due to gravity on the earth's surface.
At $ h =\frac{R_{E}}{2} ,g_{h} = \frac{gR^{2}_{E}}{\left( R_{E} +\frac{R_{E}}{2}\right)^{2}} = \frac{4}{9}g$