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)

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

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

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