A satellite of mass 200 ,kg revolves around a | vedclass

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(A) The binding energy of a satellite in a circular orbit is defined as the minimum energy required to remove the satellite to infinity. It is equal t

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$5 	imes 10^{15}$

Vedclass · Answer

(A) The binding energy of a satellite in a circular orbit is defined as the minimum energy required to remove the satellite to infinity. It is equal to the negative of the total mechanical energy of the satellite.Total energy $E = -\frac{GMm}{2r}$.Binding energy $BE = -E = \frac{GMm}{2r}$.Given:$G = 6.67 	imes 10^{-11} \,N \cdot m^2/kg^2$$M = 5 	imes 10^{30} \,kg$$m = 200 \,kg$$r = 6.6 	imes 10^6 \,m$Substituting the values:$BE = \frac{(6.67 	imes 10^{-11}) 	imes (5 	imes 10^{30}) 	imes 200}{2 	imes 6.6 	imes 10^6}$$BE = \frac{6.67 	imes 10^{-11} 	imes 10^{33}}{13.2 	imes 10^6}$$BE \approx \frac{6.67 	imes 10^{22}}{13.2 	imes 10^6} \approx 0.505 	imes 10^{16} \approx 5 	imes 10^{15} \,J$.Thus,the correct option is $A$.

$A$ satellite of mass $200 \,kg$ revolves around a planet of mass $5 \times 10^{30} \,kg$ in a circular orbit of radius $6.6 \times 10^6 \,m$. Binding energy of the satellite is .............. $J$

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