Acceleration due to gravity and its variation

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Question

If the radius of Earth is 6000 km, what will be the weight of 120 kg body (in kg Wt.) if taken to a height of 2000 km above sea level (i.e., surface of the Earth).

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Solution

With usual notation, R = 6000 km, h = 2000 km
If g and g_h denote the values of acceleration due to gravity on the surface and at a height h, then
$\frac{g_{h}}{g} = \frac{R^{2}}{(R + h)^{2}} = \frac{(6000)^{2}}{(6000 + 2000)^{2}} = \frac{36 \times 10^{6}}{64 \times 10^{6}} or \frac{g_{h}}{g} = \frac{9}{16}$
Thus, the weight at a height of 2000 km
$= (\frac{9}{16}) (120 kg) = 67 .5 kgwt$

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Acceleration due to gravity and its variation

Remember concepts with our Masterclasses.

If the radius of Earth is 6000 km, what will be the weight of 120 kg body (in kg Wt.) if taken to a height of 2000 km above sea level (i.e., surface of the Earth).

With usual notation, R = 6000 km, h = 2000 kmIf g and gh denote the values of acceleration due to gravity on the surface and at a height h, thenghg=R2(R+h)2=(6000)2(6000+2000)2=36×10664×106 or ghg=916Thus, the weight at a height of 2000 km =916(120 kg)=67.5 kgwt

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