8.19 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.

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Vishal Baghel | Contributor-Level 10

5 months ago

Mass of the Earth, M = 6.0 * 10²⁴ kg

Mass of the satellite, m = 200 kg

Radius of the Earth, $R_{e}$ = 6.4 *10⁶ m

Universal gravitational constant, G = 6.67 * 10^–11 N m² kg^–²

Height of the satellite, h = 400 km = 0.4 $* 10^{6}$ m

Total energy of the satellite at height h = $\frac{1}{2} {m v}^{2}$ + ( $\frac{- G M_{e} m}{R_{e} + h}$ )

Orbital velocity of the satellite, v = $\sqrt{\frac{G M_{e}}{R_{e} + h}}$

Total energy of the satellite at height h = $\frac{m}{2}$ ( $\frac{G M_{e}}{R_{e} + h})$ - ( $\frac{G M_{e} m}{R_{e} + h}$ ) = $-$ $\frac{1}{2} (\frac{G M_{e} m}{R_{e} + h})$

The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite $T y p e e q u a t i o n h e r e .$

Energy require to send the satellite out of its orbit = - (bound energy) = $\frac{1}{2} (\frac{G M_{e} m}{R_{e} + h})$

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