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(D) The energy required to raise a satellite to a height $h$ is the change in potential energy: $E_1 = U_f - U_i = -\frac{GMm}{R+h} - (-\frac{GMm}{R})

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$2h: R$

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(D) The energy required to raise a satellite to a height $h$ is the change in potential energy: $E_1 = U_f - U_i = -\frac{GMm}{R+h} - (-\frac{GMm}{R}) = GMm \left( \frac{1}{R} - \frac{1}{R+h} \right) = \frac{GMmh}{R(R+h)}$.Using $g = \frac{GM}{R^2}$,we get $GM = gR^2$. Substituting this,$E_1 = \frac{gR^2mh}{R(R+h)} = \frac{mgh}{1 + h/R}$.The energy required to put the satellite into orbit at height $h$ is the kinetic energy $E_2 = \frac{1}{2}mv_0^2$. Since $v_0 = \sqrt{\frac{GM}{R+h}}$,$E_2 = \frac{1}{2}m \left( \frac{GM}{R+h} \right) = \frac{GMm}{2(R+h)}$.Substituting $GM = gR^2$,$E_2 = \frac{gR^2m}{2(R+h)} = \frac{mgR}{2(1 + h/R)}$.Taking the ratio: $\frac{E_1}{E_2} = \frac{mgh / (1 + h/R)}{mgR / (2(1 + h/R))} = \frac{h}{R/2} = \frac{2h}{R}$.

Column-$I$	Column-$II$
$(1)$ Polar satellite	$(a)$ Telecommunication
$(2)$ Geostationary satellite	$(b)$ Spying
	$(c)$ Weather forecasting

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