JEE Mains · Physics · STD 11 - Gravitation
- JEE Mains 2024
- Hard
- MCQ
A satellite of \(10^3 \mathrm{~kg}\) mass is revolving in circular orbit of radius \(2 \mathrm{R}\). If \(\frac{10^4 \mathrm{R}}{6} \mathrm{~J}\) energy is supplied to the satellite, it would revolve in a new circular orbit of radius _______.
(use \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2, \mathrm{R}=\) radius of earth)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(6 \mathrm{R}\)
Step-by-step Solution
Detailed explanation
Total energy \(=\frac{-\mathrm{GMm}}{2(2 \mathrm{R})}\)
if energy \(=\frac{10^4 \mathrm{R}}{6}\) is added then
\(\frac{-\mathrm{GMm}}{4 \mathrm{R}}+\frac{10^4 \mathrm{R}}{6}=\frac{-\mathrm{GMm}}{2 \mathrm{r}}\)
where \(r\) is new radius of revolving and \(g=\frac{G M}{R^2}\)
\(-\frac{\mathrm{mgR}}{4}+\frac{10^4 \mathrm{R}}{6}=-\frac{\mathrm{mgR}^2}{2 \mathrm{r}} \quad\left(\mathrm{m}=10^3 \mathrm{~kg}\right)\)
\(-\frac{10^3 \times 10 \times \mathrm{R}}{4}+\frac{10^4 \mathrm{R}}{6}=-\frac{10^3 \times 10 \times \mathrm{R}^2}{2 \mathrm{r}}\)
\(-\frac{1}{4}+\frac{1}{6}=-\frac{\mathrm{R}}{2 \mathrm{r}}\)
\(\mathrm{r}=6 \mathrm{R}\)

if energy \(=\frac{10^4 \mathrm{R}}{6}\) is added then
\(\frac{-\mathrm{GMm}}{4 \mathrm{R}}+\frac{10^4 \mathrm{R}}{6}=\frac{-\mathrm{GMm}}{2 \mathrm{r}}\)
where \(r\) is new radius of revolving and \(g=\frac{G M}{R^2}\)
\(-\frac{\mathrm{mgR}}{4}+\frac{10^4 \mathrm{R}}{6}=-\frac{\mathrm{mgR}^2}{2 \mathrm{r}} \quad\left(\mathrm{m}=10^3 \mathrm{~kg}\right)\)
\(-\frac{10^3 \times 10 \times \mathrm{R}}{4}+\frac{10^4 \mathrm{R}}{6}=-\frac{10^3 \times 10 \times \mathrm{R}^2}{2 \mathrm{r}}\)
\(-\frac{1}{4}+\frac{1}{6}=-\frac{\mathrm{R}}{2 \mathrm{r}}\)
\(\mathrm{r}=6 \mathrm{R}\)

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