The energy required to take a satellite to a | vedclass

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

Vedclass · Accepted Answer

$3.2 	imes 10^3 \, km$

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(B) The energy $E_1$ required to lift the satellite to height $h$ is the change in potential energy: $E_1 = U_f - U_i = -\frac{GMm}{R+h} - (-\frac{GMm}{R}) = GMm \left( \frac{1}{R} - \frac{1}{R+h} ight) = \frac{GMmh}{R(R+h)}$.The kinetic energy $E_2$ required for a circular orbit at height $h$ is given by $E_2 = \frac{1}{2}mv^2$. Since the orbital velocity $v = \sqrt{\frac{GM}{R+h}}$,we have $E_2 = \frac{1}{2}m \left( \frac{GM}{R+h} ight) = \frac{GMm}{2(R+h)}$.Setting $E_1 = E_2$:$\frac{GMmh}{R(R+h)} = \frac{GMm}{2(R+h)}$.Canceling common terms $GMm$ and $(R+h)$ from both sides:$\frac{h}{R} = \frac{1}{2} \implies h = \frac{R}{2}$.Given $R = 6.4 	imes 10^3 \, km$,we find $h = \frac{6.4 	imes 10^3}{2} = 3.2 	imes 10^3 \, km$.

The energy required to take a satellite to a height $h$ above the Earth's surface (radius of Earth $R = 6.4 \times 10^3 \, km$) is $E_1$,and the kinetic energy required for the satellite to be in a circular orbit at this height is $E_2$. The value of $h$ for which $E_1 = E_2$ is:

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