Question

A satellite orbits the earth at a height of $3.6\times {10}^6$m from its surface. Compute its (a) kinetic energy, (b) potential energy, (c) total energy. Mass of the satellite $=500$kg, mass of the earth $=6\times {10}^{24}$kg, radius of the earth $=6.4\times {10}^6$, G$=6.67\times {10}^{-11}$N $m^2$ $k{g}^{-2}$.

Answer 1

Here, $r=R+h=6.4\times {10}^6+3.6\times {10}^6={10}^7$m.
Orbital velocity of the satellite around the earth is given by
$v_0=\sqrt{\frac{GM}{(R+h)}}=\sqrt{\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}}{{10}^7}}$
$=\sqrt{6.67\times 6\times {10}^6}m{s}^{-1}$
a. KE of the satellite $=1/2m{v}_e^2$
$=\frac{1}{2}\times 500\times 6.67\times 6\times {10}^6={10}^{10}J$
b. PE of the satellite$=-\frac{GMm}{R+h}$
$=-\frac{6.67\times {10}^{-11}\times (6\times {10}^{24})\times 500}{{10}^7}$
$=-2\times {10}^{10}J$
c. Total energy $=KE+PE={10}^{10}-2\times {10}^{10}=-{10}^{10}J$