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(D) The total energy of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.Initially,the satellite is in an orbit of radi

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(D) The total energy of a satellite in a circular orbit of radius $r$ is given by $E = -\frac{GMm}{2r}$.Initially,the satellite is in an orbit of radius $r_1 = 2R$. Its initial energy is $E_1 = -\frac{GMm}{2(2R)} = -\frac{GMm}{4R}$.Given that an energy $\Delta E = \frac{10^4 R}{6} 	ext{ J}$ is supplied to the satellite,the new total energy $E_2$ will be $E_1 + \Delta E$.$E_2 = -\frac{GMm}{4R} + \frac{10^4 R}{6}$.Let the new radius be $r_2$. Then $E_2 = -\frac{GMm}{2r_2}$.Equating the two expressions for $E_2$:$-\frac{GMm}{4R} + \frac{10^4 R}{6} = -\frac{GMm}{2r_2}$.Using $g = \frac{GM}{R^2}$,we have $GM = gR^2$. Substituting this into the equation:$-\frac{mgR^2}{4R} + \frac{10^4 R}{6} = -\frac{mgR^2}{2r_2}$.Given $m = 10^3 	ext{ kg}$ and $g = 10 	ext{ m/s}^2$,so $mg = 10^4 	ext{ N}$.$-\frac{10^4 R}{4} + \frac{10^4 R}{6} = -\frac{10^4 R^2}{2r_2}$.Dividing by $10^4 R$:$-\frac{1}{4} + \frac{1}{6} = -\frac{R}{2r_2}$.$-\frac{3-2}{12} = -\frac{R}{2r_2} \implies -\frac{1}{12} = -\frac{R}{2r_2}$.$\frac{1}{12} = \frac{R}{2r_2} \implies 2r_2 = 12R \implies r_2 = 6R$.

$A$ satellite of $10^3 \text{ kg}$ mass is revolving in a circular orbit of radius $2R$. If $\frac{10^4 R}{6} \text{ J}$ of energy is supplied to the satellite,it would revolve in a new circular orbit of radius: (use $g = 10 \text{ m/s}^2$,$R = \text{radius of earth}$) (in $R$)

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