# Q: 8.22  As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly $\small 36,000\hspace{1mm}km$  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 $\small =6.0\times 10^2^4\hspace{1mm}kg$, radius $\small =6400\hspace{1mm}km$.

Height of satellite from earth's surface :    $3.6\times 10^7\ m$

Gravitational potential is given by :

$=\ \frac{-GM}{R\ +\ h}$

or                                                       $=\ \frac{-6.67\times 10^{-11}\times 6\times 10^{24}}{3.6\times 10^7\ +\ 0.64\times 10^7 }$

or                                                       $=\ -\ 9.4\times 10^6\ J/Kg$

Thus potential due to earth gravity is   $-\ 9.4\times 10^6\ J/Kg$.

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