Question

A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = 6.0×1024 kg; radius of the earth = 6.4 × 106 m; G = 6.67 × 10-11 N m2 kg-2 .

Answer 1

Mass of the Earth, $M$ = $6.0\times 10^24\,kg$
Mass of the satellite, $m$ = $200 \; kg$
Radius of the Earth, $R_e$ = $6.4\times 10^6\;m$
Universal gravitational constant, $G$ = $6.67 × 10^{–11}\, Nm^2kg^{–2}$
Height of the satellite, $h$ =$400\;km$ = $4\times 10^5\;m$ = $0.4\times 10^6\;m$
Total energy of the satellite at height $h$ = $\frac{1}{2}mv^2 + \bigg(\frac{-GM_e m }{ R_e +h}\bigg)$

Orbital velocity of the satellite, $v$ =$\sqrt\frac{GM_e }{R_e +h}$

Total energy of height, $h$ = $\frac{1}{2} m \bigg(\frac{GM_e }{ R_e +h}\bigg) - \frac{GM_e m }{ R_e +h}$= $-\frac{1}{2} \bigg(\frac{GM_e m }{ R_e +h}\bigg)$

The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite.

Energy required to send the satellite out of its orbit = – (Bound energy)

= $\frac{1}{2} \frac{GM_e m }{ (R_e +h)}$

= $\frac{1}{2} \times \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 200 }{ (6.4 \times 10^6 + 0.4 \times 10^6)}$

= $\frac{1}{2} \times \frac{6.67 \times 6 \times 2 \times 10 }{ 6.8 \times 10^6} = 5.9 \times 10^9 J$