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(D) The initial total energy of the satellite on the surface of the Earth is $E_i = -\frac{GMm}{R}$.The satellite is launched into a circular orbit at

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$\frac{3}{4} mgR$

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(D) The initial total energy of the satellite on the surface of the Earth is $E_i = -\frac{GMm}{R}$.The satellite is launched into a circular orbit at a height $h = R$,so the orbital radius is $r = R + h = 2R$.The total energy of the satellite in the circular orbit is $E_f = -\frac{GMm}{2r} = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}$.The minimum energy required is the change in total energy: $\Delta E = E_f - E_i$.$\Delta E = -\frac{GMm}{4R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{4R} = \frac{3GMm}{4R}$.Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Substituting this into the expression:$\Delta E = \frac{3(gR^2)m}{4R} = \frac{3}{4} mgR$.

$A$. 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 of mass $m$,initially at rest on the Earth,is launched into a circular orbit at a height equal to the radius of the Earth. The minimum energy required is

Similar Questions

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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$A$ satellite of mass $m$,revolving around the Earth of radius $r$,has kinetic energy $E$. Its angular momentum is:

For a satellite orbiting around the earth in a circular orbit,the ratio of potential energy to kinetic energy at the same height is

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