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 $$= 6.0 \times 10^{24}\ kg$$; radius of the earth $$=6.4 \times 10^6\, m$$; G = $$6.67 \times 10^{-11}\ Nm^2kg^{-2}$$.

Solution

## Mass of the Earth, $$M=6.0 \times 10^{24}\ kg$$$$m=200\ kg$$ $$R_e=6.4 \times 10^6\ m$$$$G=6.67 \times 10^{-11}\ Nm^2kg^{-2}$$Height of the satellite,$$h=400\ km = 4 \times 10^5\ m$$Total energy of the satellite at height h$$= (1/2)mv^2+\left(-G \dfrac{M_em}{(R_e+h)}\right)$$Orbital velocity of the satellite, $$v=\sqrt{\dfrac{GM_e}{R_e+h}}$$Total energy at height h $$=\dfrac{1}{2}\dfrac{GM_em}{R_e+h}-\dfrac{GM_em}{R_e+h}$$$$Total\ Energy=-\dfrac{1}{2}\dfrac{GM_em}{R_e+h}$$The negative sign indicates that the satellite is bound to the Earth. Energy required to send the satellite out of its orbit = – (Bound energy)$$=\dfrac{GM_em}{2(R_e+h)}$$$$=\dfrac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200}{2(6.4 \times 10^6+4 \times 10^5)}$$$$=5.9 \times 10^9\ J$$If the satellite just escapes from the gravitational field, then total energy of the satellite is zero. Therefore, we have to supply $$5.9 \times 10^9J$$ of energy to just escape it.PhysicsNCERTStandard XI

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