Question:
The maximum and minimum distances of a comet from the Sun are \(1.6 \times 10^{12}\) m and \(8.0 \times 10^{10}\) m respectively. If the speed of the comet at the nearest point is \(6 \times 10^{4}\ \text{ms}^{-1}\), the speed at the farthest point is :
The maximum and minimum distances of a comet from the Sun are \(1.6 \times 10^{12}\) m and \(8.0 \times 10^{10}\) m respectively. If the speed of the comet at the nearest point is \(6 \times 10^{4}\ \text{ms}^{-1}\), the speed at the farthest point is :
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Angular momentum about the Sun is conserved, so \(r_{min} v_{max} = r_{max} v_{min}\).
Updated On: Nov 4, 2026
- \(3.0 \times 10^{3}\) m/s
- \(6.0 \times 10^{3}\) m/s
- \(1.5 \times 10^{3}\) m/s
- \(4.5 \times 10^{3}\) m/s
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The Correct Option is A
Solution and Explanation
The gravitational force on the comet is central, so its angular momentum about the Sun is conserved. At the extreme points the velocity is perpendicular to the radius vector:
\(m\,r_{min}\,v_{near} = m\,r_{max}\,v_{far}\).
\(v_{far} = v_{near}\cdot\frac{r_{min}}{r_{max}} = 6 \times 10^{4} \times \frac{8.0 \times 10^{10}}{1.6 \times 10^{12}} = 6 \times 10^{4} \times 0.05 = 3.0 \times 10^{3}\ \text{m/s}\).
Answer: (1) \(3.0 \times 10^{3}\) m/s
\(m\,r_{min}\,v_{near} = m\,r_{max}\,v_{far}\).
\(v_{far} = v_{near}\cdot\frac{r_{min}}{r_{max}} = 6 \times 10^{4} \times \frac{8.0 \times 10^{10}}{1.6 \times 10^{12}} = 6 \times 10^{4} \times 0.05 = 3.0 \times 10^{3}\ \text{m/s}\).
Answer: (1) \(3.0 \times 10^{3}\) m/s
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