The ratio of energy required to raise a satel | vedclass

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Question

(A) The energy required to raise a satellite of mass $m$ to a height $h$ above the earth's surface is the change in potential energy:$E_1 = U_h - U_R

Vedclass · Accepted Answer

$\frac{2h}{R}$

Vedclass · Answer

(A) The energy required to raise a satellite of mass $m$ to a height $h$ above the earth's surface is the change in potential energy:$E_1 = U_h - U_R = -\frac{GMm}{R+h} - (-\frac{GMm}{R}) = GMm \left( \frac{1}{R} - \frac{1}{R+h} ight) = \frac{GMmh}{R(R+h)}$.Using $GM = gR^2$,we get $E_1 = \frac{mgR^2h}{R(R+h)} = \frac{mghR}{R+h}$.The total energy of a satellite in a circular orbit at height $h$ is $E_{orbit} = -\frac{GMm}{2(R+h)}$.The energy required to put it into orbit at that height is the difference between the energy in orbit and the energy at the surface:$E_2 = E_{orbit} - U_R = -\frac{GMm}{2(R+h)} + \frac{GMm}{R} = GMm \left( \frac{1}{R} - \frac{1}{2(R+h)} ight) = GMm \left( \frac{2R+2h-R}{2R(R+h)} ight) = \frac{GMm(R+2h)}{2R(R+h)}$.Using $GM = gR^2$,we get $E_2 = \frac{mgR^2(R+2h)}{2R(R+h)} = \frac{mgR(R+2h)}{2(R+h)}$.The ratio is $\frac{E_1}{E_2} = \frac{mghR}{R+h} 	imes \frac{2(R+h)}{mgR(R+2h)} = \frac{2h}{R+2h}$.For $h \ll R$,$R+2h \approx R$,so the ratio is $\frac{2h}{R}$.

$A$. Potential energy of satellite	$I$. Positive
$B$. Total energy of satellite	$II$. Negative
$C$. Kinetic energy of satellite	$III$. Zero
$D$. Gravitational potential energy of satellite at infinity	$IV$. Infinity

The ratio of energy required to raise a satellite to a height $h$ above the earth's surface to that required to put it into the orbit at the same height is ($R=$ radius of earth).

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Match the following columns.
$A$. Potential energy of satellite $I$. Positive
$B$. Total energy of satellite $II$. Negative
$C$. Kinetic energy of satellite $III$. Zero
$D$. Gravitational potential energy of satellite at infinity $IV$. Infinity

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