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Byju's Answer

Standard XII

Physics

Kepler's Law

A planet move...

Question

A planet moves around the sun. At a given point P, it is closest from the sun at a distance $d_{1}$ and has a speed $v_{1}$ . At another point Q, when it is farthest from the sun at a distance $d_{2}$ , its speed will be

$\frac{d_{1}^{2} v_{1}}{d_{2}^{2}}$

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$\frac{d_{2} v_{1}}{d_{1}}$

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$\frac{d_{1} v_{1}}{d_{2}}$

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$\frac{d_{2}^{2} v_{1}}{d_{1}^{2}}$

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Solution

The correct option is C
$\frac{d_{1} v_{1}}{d_{2}}$

Angular momentum remains constant
$m v_{1} d_{1} = m v_{2} d_{2} \Rightarrow v_{2} = \frac{v_{1} d_{1}}{d_{2}}$

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A planet moves around the sun. At a given point P, it is closest from the sun at a distance $d_{1}$ and has a speed $v_{1}$ . At another point Q, when it is farthest from the sun at a distance $d_{2}$ , its speed will be

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A planet moves around the sun. At a given point P, it is closest from the sun at a distance d1 and has a speed v1. At another point Q, when it is farthest from the sun at a distance d2, its speed will be

The correct option is C d1v1d2 Angular momentum remains constant mv1d1=mv2d2⇒v2=v1d1d2

A planet moves around the sun. At a given point P, it is closest from the sun at a distance $d_{1}$ and has a speed $v_{1}$ . At another point Q, when it is farthest from the sun at a distance $d_{2}$ , its speed will be

The correct option is C
$\frac{d_{1} v_{1}}{d_{2}}$

Angular momentum remains constant
$m v_{1} d_{1} = m v_{2} d_{2} \Rightarrow v_{2} = \frac{v_{1} d_{1}}{d_{2}}$