The gravitational potential at a point above the surface of earth is $-5.12 \times 10^7 \mathrm{~J} / \mathrm{kg}$ and the acceleration due to gravity at that point is $6.4 \mathrm{~m} / \mathrm{s}^2$. Assume that the mean radius of earth to be $6400 \mathrm{~km}$. The height of this point above the earth's surface is :
Solution
<p>$-\frac{G M_E}{R_E+h}=-5.12 \times 10^{-7}$ .... (i)</p>
<p>$\frac{G M_E}{\left(R_E+h\right)^2}=6.4$ ..... (ii)</p>
<p>By (i) and (ii)</p>
<p>$\Rightarrow h=16 \times 10^5 \mathrm{~m}=1600 \mathrm{~km}$</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Kepler's Laws
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