Question:
A planet revolves around the Sun in an elliptical orbit. The areal velocity of the planet is \(4 \times 10^{16}\,\text{m}^2\text{s}^{-1}\). If the maximum distance between the planet and the Sun is \(4 \times 10^{12}\,\text{m}\), then the minimum speed of the planet is:
A planet revolves around the Sun in an elliptical orbit. The areal velocity of the planet is \(4 \times 10^{16}\,\text{m}^2\text{s}^{-1}\). If the maximum distance between the planet and the Sun is \(4 \times 10^{12}\,\text{m}\), then the minimum speed of the planet is:
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Shortcut:
\[
v = \frac{2}{r}\frac{dA}{dt}
\]
Updated On: Jun 10, 2026
- \( 1 \times 10^4\,\text{m s}^{-1} \)
- \( 2 \times 10^4\,\text{m s}^{-1} \)
- \( 4 \times 10^4\,\text{m s}^{-1} \)
- \( 8 \times 10^4\,\text{m s}^{-1} \)
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The Correct Option is B
Solution and Explanation
Concept:
Areal velocity is constant:
\[
\frac{dA}{dt} = \frac{1}{2} r v
\]
Step 1: Use condition at maximum distance At maximum distance \(r_{\max}\), speed is minimum \(v_{\min}\): \[ \frac{dA}{dt} = \frac{1}{2} r_{\max} v_{\min} \]
Step 2: Substitute values \[ 4 \times 10^{16} = \frac{1}{2} \times 4 \times 10^{12} \times v_{\min} \]
Step 3: Solve \[ 4 \times 10^{16} = 2 \times 10^{12} \cdot v_{\min} \] \[ v_{\min} = 2 \times 10^4\,\text{m s}^{-1} \]
Step 1: Use condition at maximum distance At maximum distance \(r_{\max}\), speed is minimum \(v_{\min}\): \[ \frac{dA}{dt} = \frac{1}{2} r_{\max} v_{\min} \]
Step 2: Substitute values \[ 4 \times 10^{16} = \frac{1}{2} \times 4 \times 10^{12} \times v_{\min} \]
Step 3: Solve \[ 4 \times 10^{16} = 2 \times 10^{12} \cdot v_{\min} \] \[ v_{\min} = 2 \times 10^4\,\text{m s}^{-1} \]
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