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Binding energy of a revolving satellite at height ' $h$ ' is $3.5 \times 10^8 \mathrm{~J}$. Its potential energy is
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Verified Answer
The correct answer is:
$-3.5 \times 10^8 \mathrm{~J}$
Concept: Binding energy $=+$ (work done to assemble the planet plus satellite system)
$\therefore \mathrm{BE}=+\mathrm{W}$
We know, potential energy change is related as:
$\mathrm{U}=(-\mathrm{W})$
So the potential energy of the system is
$\mathrm{U}=-\left(\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{r}+\mathrm{h}}\right)=-\mathrm{W}=-3.5 \times 10^8 \mathrm{~J}$
$\therefore \mathrm{BE}=+\mathrm{W}$
We know, potential energy change is related as:
$\mathrm{U}=(-\mathrm{W})$
So the potential energy of the system is
$\mathrm{U}=-\left(\frac{\mathrm{GM}_{\mathrm{m}}}{\mathrm{r}+\mathrm{h}}\right)=-\mathrm{W}=-3.5 \times 10^8 \mathrm{~J}$
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