MHT CET · Physics · Gravitation
Binding energy of a revolving satellite at height ' \(h\) ' is \(3.5 \times 10^8 \mathrm{~J}\). Its potential energy is
- A \(-3.5 \times 10^8 \mathrm{~J}\)
- B \(-7 \times 10^8 \mathrm{~J}\)
- C \(7 \times 10^8 \mathrm{~J}\)
- D \(3.5 \times 10^8 \mathrm{~J}\)
Answer & Solution
Correct Answer
(A) \(-3.5 \times 10^8 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
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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