A satellite of mass m,initially at rest on th | vedclass

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(D) The initial energy of the satellite on the surface of the Earth is $E_i = K_i + U_i = 0 - \frac{GMm}{R}$.When the satellite is in a circular orbit

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$\frac{3}{4}mgR$

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(D) The initial energy of the satellite on the surface of the Earth is $E_i = K_i + U_i = 0 - \frac{GMm}{R}$.When the satellite is in a circular orbit at height $h = R$,the orbital radius is $r = R + h = 2R$.The potential energy at this height is $U_f = -\frac{GMm}{2R}$.The orbital velocity is $v_0 = \sqrt{\frac{GM}{2R}}$,so the kinetic energy is $K_f = \frac{1}{2}mv_0^2 = \frac{1}{2}m(\frac{GM}{2R}) = \frac{GMm}{4R}$.The final energy is $E_f = K_f + U_f = \frac{GMm}{4R} - \frac{GMm}{2R} = -\frac{GMm}{4R}$.The minimum energy required is $\Delta E = E_f - E_i = -\frac{GMm}{4R} - (-\frac{GMm}{R}) = \frac{GMm}{R} - \frac{GMm}{4R} = \frac{3GMm}{4R}$.Since $g = \frac{GM}{R^2}$,we have $GM = gR^2$.Substituting this,$\Delta E = \frac{3(gR^2)m}{4R} = \frac{3}{4}mgR$.

$A$ 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

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