Binding energy of a revolving satellite at he | vedclass

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(B) The binding energy $(BE)$ of a satellite is defined as the negative of the total mechanical energy $(E)$.$BE = -E = - (K.E. + P.E.) = - (\frac{GMm

Vedclass · Accepted Answer

$-7.0 	imes 10^{8} \,J$

Vedclass · Answer

(B) The binding energy $(BE)$ of a satellite is defined as the negative of the total mechanical energy $(E)$.$BE = -E = - (K.E. + P.E.) = - (\frac{GMm}{2r} - \frac{GMm}{r}) = \frac{GMm}{2r}$.The potential energy $(P.E.)$ of a satellite is given by $P.E. = -\frac{GMm}{r}$.Comparing the two expressions, we see that $P.E. = -2 	imes BE$.Given $BE = 3.5 	imes 10^{8} \,J$.Therefore, $P.E. = -2 	imes (3.5 	imes 10^{8} \,J) = -7.0 	imes 10^{8} \,J$.

Binding energy of a revolving satellite at height $h$ is $3.5 \times 10^{8} \,J$. Its potential energy is

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