A satellite orbits the earth at a height of 400 km above the surf… - Vidyadip

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 $=200 \mathrm{~kg}$; mass of the earth $=6.0 \times$ $10^{24} \mathrm{~kg}$; radius of the earth $=6.4 \times 10^6 \mathrm{~m} ; \mathrm{G}=6.67 \times 10^{-11} \mathrm{~N} \mathrm{~m}^2 \mathrm{~kg}^{-2}$.

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Answer

Mass of the Earth, $M = 6.0 \times 10^{24}kg$ Mass of the satellite, $m = 200kg$ Radius of the Earth, $R_e = 6.4 \times 10^6m$ Universal gravitational constant, $G = 6.67 \times 10^{–11}Nm^2kg^{–2}$ Height of the satellite, $h = 400km = 4 \times 10^5m = 0.4 \times 10^6m$ Total energy of the satellite at height $\text{h}=\Big(\frac{1}{2}\Big)\text{mv}^2+\Big[\frac{-\text{GM}_\text{e}\text{m}}{(\text{R}_\text{e}+\text{h})}\Big]$ Orbital velocity of the satellite, $\text{v}=\Big[\frac{\text{GM}_\text{e}}{(\text{R}_\text{e}+\text{h})}\Big]^\frac{1}{2}$ Total energy of height, $\text{h}=\frac{\big(\frac{1}{2}\big)\text{GM}_\text{e}\text{m}}{(\text{R}_\text{e}+\text{h})}-\frac{\text{GM}_\text{e}\text{m}}{(\text{R}_\text{e}+\text{h})}$$=-\frac{\big(\frac{1}{2}\big)\text{GM}_\text{e}\text{m}}{(\text{R}_\text{e}+\text{h})}$
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)$=\frac{\big(\frac{1}{2}\big)\text{GM}_\text{e}\text{m}}{(\text{R}_\text{e}+\text{h})}$
$=\frac{\big(\frac{1}{2}\big)\times6.67\times10^{-11}\times6\times10^{24}\times200}{(6.4\times10^6+0.4\times10^6)}$
$=5.9\times10^9\text{ j.}$

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