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(D) The energy of a satellite in a circular orbit is given by $E = -\frac{GM_E m}{2r}$, where $r$ is the radius of the circular orbit.The energy to be

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$1000$

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(D) The energy of a satellite in a circular orbit is given by $E = -\frac{GM_E m}{2r}$, where $r$ is the radius of the circular orbit.The energy to be supplied is $\Delta E = E_f - E_i$.$\Delta E = \left( -\frac{GM_E m}{2(3R_E)} ight) - \left( -\frac{GM_E m}{2(1.5R_E)} ight)$$\Delta E = \frac{GM_E m}{2R_E} \left( \frac{1}{1.5} - \frac{1}{3} ight) = \frac{GM_E m}{2R_E} \left( \frac{2}{3} - \frac{1}{3} ight) = \frac{GM_E m}{6R_E}$.Using the relation $g = \frac{GM_E}{R_E^2}$, we have $GM_E = gR_E^2$.Substituting this into the expression for $\Delta E$:$\Delta E = \frac{gR_E^2 m}{6R_E} = \frac{1}{6} mgR_E$.Given $m = 100 \ kg$, $g = 10 \ m/s^2$, and $R_E = 6 	imes 10^6 \ m$:$\Delta E = \frac{1}{6} 	imes 100 	imes 10 	imes 6 	imes 10^6 = 1000 	imes 10^6 \ J$.Comparing this with $\alpha 	imes 10^6 \ J$, we get $\alpha = 1000$.

$A$. 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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