The ratio of the energy required to raise a satellite the upto a height h above the earth of radius R to that the kinetic energy of the satellite into that orbit is:

R: h

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h: R

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R: 2h

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2h: R

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Solution

The correct option is D 2h: R
Energy required to raise the satellite to height h
$Δ U = G m_{e} m (\frac{1}{R} - \frac{1}{R + h}) = G m_{e} \frac{m h}{R (R + h)}$
For moving satellite stationary is orbit
$\frac{m v^{2}}{(R + h)} = \frac{G m_{e} m}{(R + h)} m V^{2} = \frac{G m_{e} m}{R + h} K . E = \frac{1}{2} m V^{2} = \frac{1}{2} \frac{G m_{e} m}{R + h} \frac{Δ U}{K E} = \frac{h 2 (R + h)}{R (R + h)} = \frac{2 h}{R}$

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The ratio of the energy required to raise a satellite the upto a height h above the earth of radius R to that the kinetic energy of the satellite into that orbit is:

The correct option is D 2h: REnergy required to raise the satellite to height hΔU=Gmem(1R−1R+h)=GmemhR(R+h)For moving satellite stationary is orbitmv2(R+h)=Gmem(R+h)mV2=GmemR+hK.E=12mV2=12GmemR+hΔUKE=h2(R+h)R(R+h)=2hR