Question:
A planet revolves around the sun in an elliptical orbit. The areal velocity is 4 10^16 m^2s^-1. Maximum distance is 4 10^12 m. Find minimum speed.
A planet revolves around the sun in an elliptical orbit. The areal velocity is 4 10^16 m^2s^-1. Maximum distance is 4 10^12 m. Find minimum speed.
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Speed in orbital motion is inversely proportional to radius at extreme points due to conservation of angular momentum.
Updated On: Jun 10, 2026
- 1 10^4 ms^-1
- 2 10^4 ms^-1
- 4 10^4 ms^-1
- 8 10^4 ms^-1
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The Correct Option is B
Solution and Explanation
Areal velocity:
\[
\frac{dA}{dt} = \frac{1}{2}rv
\]
At maximum distance:
\[
v_{\min} = \frac{2}{r_{\max}} \frac{dA}{dt}
\]
Substitute:
\[
v_{\min} = \frac{2 \times 4 \times 10^{16}}{4 \times 10^{12}}
\]
\[
= 2 \times 10^4 \, ms^{-1}
\]
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