Question

A satellite of mass 1000 kg is supposed to orbit the earth at a height of 2000 km above the earth's surface. Find (a) its speed in the orbit, (b) is kinetic energy, (c) the potential energy of the earth-satellite system and (d) its time period. Mass of the earth = 6 × 10²⁴ kg.

Answer 1

(a) Speed of the satellite in its orbit
$v=\sqrt{\frac{\mathrm{G}M}{r+h}}=\sqrt{\frac{g{r}^2}{r+h}}$
$$\begin{gathered} \Rightarrow v=\sqrt{\frac{9.8\times {\left(6400\times {10}^3\right)}^2}{{10}^6\times \left(6.4+2\right)}} \\ \Rightarrow v=\sqrt{\frac{9.8\times 6.4\times 6.4\times {10}^6}{8.4}} \\ \Rightarrow v=6.9\times {10}^3\mathrm{m}/\mathrm{s}=6.9\mathrm{km}/\mathrm{s} \end{gathered}$$

(b) Kinetic energy of the satellite
$\mathrm{K}.\mathrm{E}.=\frac{1}{2}m{v}^2$
$$\begin{gathered} =\frac{1}{2}\times 1000\times {\left(6.9\times {10}^3\right)}^2 \\ =\frac{1}{2}\times 1000\times \left(47.6\times {10}^6\right) \\ =2.38\times {10}^{10}\mathrm{J} \end{gathered}$$

(c) Potential energy of the satellite
$\mathrm{P}.\mathrm{E}.=-\frac{\mathrm{GM}m}{\left(\mathrm{R}+h\right)}$
$$\begin{gathered} =-\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}\times {10}^3}{\left(6400+2000\right)\times {10}^3} \\ =\frac{40\times {10}^{13}}{8400}=-4.76\times {10}^{10}\mathrm{J} \end{gathered}$$

(d) Time period of the satellite
$T=\frac{2\mathrm{\pi }\left(r+h\right)}{\mathrm{v}}$
$$\begin{gathered} =\frac{2\times 3.14\times 8400\times {10}^3}{6.9\times {10}^3} \\ =\frac{6.28\times 84\times {10}^2}{6.9} \\ =76.6\times 10.2\mathrm{s} \\ =2.1\mathrm{h} \end{gathered}$$