**Question:**

As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly $36,000 \mathrm{~km}$ from the surface of the earth. What is the potential due to earth's gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = $6.0 \times$ $10^{24} \mathrm{~kg}$, radius $=6400 \mathrm{~km}$.

**Solution:**

Mass of the Earth, $M=6.0 \times 10^{24} \mathrm{~kg}$

Radius of the Earth, $R=6400 \mathrm{~km}=6.4 \times 10^{6} \mathrm{~m}$

Height of a geostationary satellite from the surface of the Earth,

$h=36000 \mathrm{~km}=3.6 \times 10^{7} \mathrm{~m}$

Gravitational potential energy due to Earth’s gravity at height *h,*

*$=\frac{-\mathrm{G} M}{(R+h)}$*

$=-\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{3.6 \times 10^{7}+0.64 \times 10^{7}}$

$=-\frac{6.67 \times 6}{4.24} \times 10^{13-7}$

$=-9.4 \times 10^{6} \mathrm{~J} / \mathrm{kg}$

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