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(D) Let $M$ be the mass of the Earth,$R$ be the radius of the Earth,and $r$ be the orbital radius of the satellite.Using the principle of conservation

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$\sqrt{v_e^2 - 2v_0^2}$

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(D) Let $M$ be the mass of the Earth,$R$ be the radius of the Earth,and $r$ be the orbital radius of the satellite.Using the principle of conservation of energy,the total energy at the orbital position equals the total energy at the Earth's surface.Initial energy at distance $r$: $E_i = -\frac{GMm}{r}$.Final energy at the surface $(r=R)$: $E_f = \frac{1}{2}mv^2 - \frac{GMm}{R}$.Equating $E_i = E_f$: $-\frac{GMm}{r} = \frac{1}{2}mv^2 - \frac{GMm}{R}$.Rearranging for $v^2$: $v^2 = \frac{2GM}{R} - \frac{2GM}{r}$.We know that the escape velocity $v_e = \sqrt{\frac{2GM}{R}}$,so $v_e^2 = \frac{2GM}{R}$.We also know the orbital speed $v_0 = \sqrt{\frac{GM}{r}}$,so $v_0^2 = \frac{GM}{r}$,which implies $2v_0^2 = \frac{2GM}{r}$.Substituting these into the equation: $v^2 = v_e^2 - 2v_0^2$.Therefore,$v = \sqrt{v_e^2 - 2v_0^2}$.

$A$. 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

$A$ satellite is revolving around the Earth with orbital speed $v_0$. If it stops suddenly,the speed with which it will strike the surface of the Earth would be: ($v_e =$ escape velocity of a particle on the Earth's surface)

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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

The total energy of a circularly orbiting satellite is

$A$ satellite of earth of mass $m$ is moved from an orbital radius of $2R$ to $3R$. Calculate the minimum work done.

An artificial satellite moving in a circular orbit at a distance $h$ from the centre of the Earth has a total energy $E_0$. Then,its potential energy is

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