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(A) The kinetic energy $(KE)$ of a satellite in a circular orbit is given by the formula: $KE = \frac{GM_e m}{2r}$,where $r = R_E + h$.Given: $M_e = 6

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$3$

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(A) The kinetic energy $(KE)$ of a satellite in a circular orbit is given by the formula: $KE = \frac{GM_e m}{2r}$,where $r = R_E + h$.Given: $M_e = 6 	imes 10^{24} \ kg$,$m = 1000 \ kg$,$R_E = 6.4 	imes 10^6 \ m$,$h = 270 \ km = 0.27 	imes 10^6 \ m$,$G = 6.67 	imes 10^{-11} \ Nm^2 \ kg^{-2}$.Calculating $r$: $r = 6.4 	imes 10^6 + 0.27 	imes 10^6 = 6.67 	imes 10^6 \ m$.Substituting the values into the formula:$KE = \frac{6.67 	imes 10^{-11} 	imes 6 	imes 10^{24} 	imes 1000}{2 	imes 6.67 	imes 10^6}$$KE = \frac{6.67 	imes 6 	imes 10^{16}}{2 	imes 6.67 	imes 10^6} = 3 	imes 10^{10} \ J$.Thus,the kinetic energy is $3 	imes 10^{10} \ J$.

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