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 = 6.0×10^{24} kg; radius of the earth = 6.4 × 10^{6 }m; G = 6.67 × 10^{–11} N m^{2} kg^{–2}

Asked by Pragya Singh | 1 year ago | 122

Height of the satellite, h = 400 km = 4 × 10^{5} m

Mass of the Earth, M = 6.0 × 10²⁴ kg

Mass of the satellite, m = 200 kg

Radius of the Earth, R_{e} = 6.4 × 10^{6} m

Total energy of the satellite at height h = kinetic energy + potential energy

= (\( \dfrac{1}{2}\))mv² + [\( \dfrac{-G_mM_e}{(R_e + h)}\) ]

Orbital velocity of the satellite, v = \( \sqrt{\dfrac{GM_{e}}{R_{e}+h}}\)

Total energy of height, h = \( \dfrac{1}{2}\frac{GM_{e}m}{R_{e}+h}-\dfrac{GM_{e}m}{R_{e}+h}\)

Total Energy= \( - \dfrac{1}{2}\dfrac{GM_{e}m}{R_{e}+h}\)

The negative sign indicates that the satellite is bound to the earth. This is known as the binding energy of a satellite.

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

=\( \dfrac{1}{2}\dfrac{GM_{e}m}{R_{e}+h}\)

Putting values of all terms.

=\( \dfrac{6.67 \times 10^{-11}\times 6\times 10^{24}\times 200}{2(6.4\times 10^{6}+4\times 10^{5})}\)

= 5.9 × 10^{9} J

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