MHT CET · Physics · Gravitation
Two satellites of equal mass are launched in circular orbits at heights ' \(R\) ' and ' \(2 R\) ' above the surface of the earth. The ratio of their kinetic energies is ( \(\mathrm{R}=\) radius of the earth)
- A 1:3
- B 3:2
- C 4:9
- D 9:4
Answer & Solution
Correct Answer
(B) 3:2
Step-by-step Solution
Detailed explanation
Kinetic energy of a satellite is given by
\(\mathrm{K} \cdot \mathrm{E}=\frac{\mathrm{GMm}}{2 \mathrm{r}}\)
The two satellites are of the same mass
\( \begin{aligned} & \therefore \mathrm{K} \cdot \mathrm{E} \propto \frac{1}{\mathrm{r}} \\ & \frac{(\mathrm{K} \cdot \mathrm{E})_1}{(\mathrm{~K} \cdot \mathrm{E})_2}=\frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{3 \mathrm{R}}{2 \mathrm{R}}=\frac{3}{2} \end{aligned} \)
\(\mathrm{K} \cdot \mathrm{E}=\frac{\mathrm{GMm}}{2 \mathrm{r}}\)
The two satellites are of the same mass
\( \begin{aligned} & \therefore \mathrm{K} \cdot \mathrm{E} \propto \frac{1}{\mathrm{r}} \\ & \frac{(\mathrm{K} \cdot \mathrm{E})_1}{(\mathrm{~K} \cdot \mathrm{E})_2}=\frac{\mathrm{r}_2}{\mathrm{r}_1}=\frac{3 \mathrm{R}}{2 \mathrm{R}}=\frac{3}{2} \end{aligned} \)
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