MCQ
For a satellite if the time of revolution is $T$, then its $K.E.$ is proportional to
- A$\frac{1}{T}$
- B$\frac{1}{T^2}$
- C$\frac{1}{T^3}$
- ✓$T^{-2/3}$
✓
Answer
Correct option: D.
$T^{-2/3}$
d
$\mathrm{kE}=\frac{\mathrm{GMm}}{2 \mathrm{r}}$
$\mathrm{kE}=\frac{\mathrm{GMm}}{2 \mathrm{r}}$
$\mathrm{kE} \propto \frac{1}{\mathrm{r}} \alpha \mathrm{T}^{\frac{-2}{3}}$
$\because\left(\mathrm{T}^{2} \alpha \mathrm{r}^{3}\right)$
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