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 × 10²⁴kg.

Answer 1

(a) Speed of the satellite in its orbit \[v = \sqrt{\frac{GM}{r + h}} = \sqrt{\frac{g r^2}{r + h}}\]

\[\Rightarrow v = \sqrt{\frac{9 . 8 \times \left( 6400 \times {10}^3 \right)^2}{{10}^6 \times \left( 6 . 4 + 2 \right)}}\]

\[ \Rightarrow v = \sqrt{\frac{9 . 8 \times 6 . 4 \times 6 . 4 \times {10}^6}{8 . 4}}\]

\[ \Rightarrow v = 6 . 9 \times {10}^3 m/s = 6 . 9 km/s\]

(b) Kinetic energy of the satellite

\[K . E . = \frac{1}{2} m v^2\]

\[= \frac{1}{2} \times 1000 \times \left( 6 . 9 \times {10}^3 \right)^2 \]

\[ = \frac{1}{2} \times 1000 \times \left( 47 . 6 \times {10}^6 \right)\]

\[ = 2 . 38 \times {10}^{10} J\]

(c) Potential energy of the satellite

\[P . E . = - \frac{GMm}{\left( R + h \right)}\]

\[= - \frac{6 . 67 \times {10}^{- 11} \times 6 \times {10}^{24} \times {10}^3}{\left( 6400 + 2000 \right) \times {10}^3}\]

\[ = \frac{40 \times {10}^{13}}{8400} = - 4 . 76 \times {10}^{10} J\]

(d) Time period of the satellite \[T = \frac{2\pi\left( r + h \right)}{v}\] 

\[= \frac{2 \times 3 . 14 \times 8400 \times {10}^3}{6 . 9 \times {10}^3}\]

\[ = \frac{6 . 28 \times 84 \times {10}^2}{6 . 9}\]

\[ = 76 . 6 \times 10 . 2 s\]

\[ = 2 . 1 h\]