MHT CET · Physics · Gravitation
The distance of the two planets A and B from the sun are \(r_A\) and \(r_B\) respectively. Also \(r_B\) is equal to \(100 r_A\). If the orbital speed of the planet \(A\) is ' \(v\) ' then the orbital speed of the planet \(B\) is
- A \(\frac{v}{10}\)
- B \(\frac{v}{2}\).
- C \(\sqrt{2} \mathrm{v}\)
- D 10 v
Answer & Solution
Correct Answer
(A) \(\frac{v}{10}\)
Step-by-step Solution
Detailed explanation
Orbital or Critical velocity \(v_{\text {orb }}=\sqrt{\frac{\mathrm{GM}}{\mathrm{R}}}\)
From the data given,
\(\mathrm{v}_{\text {orb }_{\mathrm{A}}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}_{\mathrm{A}}}}\)
...(i) and
\(\mathrm{v}_{\text {orb }_{\mathrm{B}}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}_{\mathrm{B}}}}=\sqrt{\frac{\mathrm{GM}}{100 \mathrm{r}_{\mathrm{A}}}}=\frac{1}{10} \sqrt{\frac{\mathrm{GM}}{\mathrm{r}_{\mathrm{A}}}}\)
Dividing equation (i) by (ii)
\(\begin{aligned}
& \frac{v_{\text {orb }_A}}{v_{\text {orb }_B}}=\sqrt{\frac{G M}{r_A}} \times 10 \sqrt{\frac{r_A}{G M}} \\
& v_{\text {orb }_B}=\frac{v_{\text {orb }_A}}{10}=\frac{v}{10} \ldots\left(\text { given }_{\mathrm{orb}_A}=\mathrm{v}\right)
\end{aligned}\)
From the data given,
\(\mathrm{v}_{\text {orb }_{\mathrm{A}}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}_{\mathrm{A}}}}\)
...(i) and
\(\mathrm{v}_{\text {orb }_{\mathrm{B}}}=\sqrt{\frac{\mathrm{GM}}{\mathrm{r}_{\mathrm{B}}}}=\sqrt{\frac{\mathrm{GM}}{100 \mathrm{r}_{\mathrm{A}}}}=\frac{1}{10} \sqrt{\frac{\mathrm{GM}}{\mathrm{r}_{\mathrm{A}}}}\)
Dividing equation (i) by (ii)
\(\begin{aligned}
& \frac{v_{\text {orb }_A}}{v_{\text {orb }_B}}=\sqrt{\frac{G M}{r_A}} \times 10 \sqrt{\frac{r_A}{G M}} \\
& v_{\text {orb }_B}=\frac{v_{\text {orb }_A}}{10}=\frac{v}{10} \ldots\left(\text { given }_{\mathrm{orb}_A}=\mathrm{v}\right)
\end{aligned}\)
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