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Q. Taking the radius of the earth to be $6400 \,km$, by what percentage will the acceleration due to gravity at a height of $100\, km$ from the surface of the earth differ from that on the surface of the earth?

AIIMSAIIMS 2019Gravitation

Solution:

Here, $R_{e}=600 \,km \,\,\,h=100 \,km$
Acceleration due to gravity at height, $h$
$g'=\frac{g R_{e}^{2}}{\left(R_{e}+h\right)^{2}}$
$g'=\frac{g(6400)^{2}}{(6400+100)^{2}}$
$=g\left(\frac{6400}{6500}\right)^{2}$
Percentage change in acceleration due to gravity
$=\frac{g-g'}{g} \times 100$
$=\left(1-\frac{g'}{g}\right) \times 100 \%$
$=\left[1-\frac{6400}{6500}\right]^{2} \times 100 \%$
$=3.05 \%$