Find the weight of an object at a height 6400 | vedclass

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(A) The acceleration due to gravity $g$ at the earth's surface is given by $g = \frac{GM}{R^2}$.The weight of an object at the earth's surface is $W =

Vedclass · Accepted Answer

$5$

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(A) The acceleration due to gravity $g$ at the earth's surface is given by $g = \frac{GM}{R^2}$.The weight of an object at the earth's surface is $W = mg = 20 \, N$.At a height $h = 6400 \, km$ above the surface,the distance from the center of the earth is $r = R + h = R + R = 2R$.The acceleration due to gravity at this height $g_h$ is given by $g_h = \frac{GM}{(R+h)^2} = \frac{GM}{(2R)^2} = \frac{GM}{4R^2} = \frac{1}{4} g$.Therefore,the weight of the object at this height $W_h$ is $W_h = m \cdot g_h = m \cdot (\frac{1}{4} g) = \frac{1}{4} (mg)$.Substituting the value of $W = 20 \, N$,we get $W_h = \frac{1}{4} 	imes 20 \, N = 5 \, N$.

Find the weight of an object at a height $6400 \, km$ above the earth's surface. The weight of the object at the surface of the earth is $20 \, N$ and the radius of the earth is $6400 \, km$. (in $, N$)

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