Question

A satellite of mass 1000 kg is suposed to orbit the earth at a height of 2000 km above the earth 's surface. Find (a) its speed in the orbit, (b) its kinetic energy, (c) the potential energy of the earth - satellite system and (d) its time period. Mass of the earth $=6\times {10}^{24}$ kg.

Answer 1

$V=\sqrt{\frac{GM}{r+h}}=\sqrt{\frac{g{r}^2}{r+h}}=\sqrt{\frac{9.8\times {(6400\times {10}^3)}^2}{{10}^6\times (6.4+2)}}=\sqrt{\frac{9.8\times 6.4\times 6.4\times {10}^6}{8.4}}=6.9\times {10}^{3}m/s=6.9km/s$
$\left(b\right)K.E.=\frac{1}{2}m{v}^2=\frac{1}{2}\times 1000\times (47.6\times {10}^6)=2.38\times {10}^{10}j\left(c\right)P.E.=\frac{GMm}{-(R+h)}=-\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\times {10}^3}{(6400+2000)\times {10}^3}=\frac{40\times {10}^{13}}{8400}=-4.76\times {10}^{10}J\left(d\right)T=\frac{2\pi (r+h)}{V}=\frac{2\times 3.14\times 8400\times {10}^3}{6.9\times {10}^3}=\frac{6.28\times 84\times {10}^2}{6.9}=76.6\times 10.2sec=2.1hours$