The minimum and maximum distances of a planet | vedclass

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Question

(C) According to the law of conservation of angular momentum,the angular momentum of a planet revolving around the Sun remains constant at all points

Vedclass · Accepted Answer

$\frac{v_{0} x_{2}}{x_{1}}$

Vedclass · Answer

(C) According to the law of conservation of angular momentum,the angular momentum of a planet revolving around the Sun remains constant at all points in its orbit.$L = mvr \sin(	heta)$At the perihelion (minimum distance $x_{1}$) and aphelion (maximum distance $x_{2}$),the velocity vector is perpendicular to the radius vector,so $\sin(	heta) = 1$.Thus,$m v_{max} x_{1} = m v_{min} x_{2}$.Given that the minimum speed is $v_{0}$ (which occurs at the maximum distance $x_{2}$),we have $v_{min} = v_{0}$.Substituting these values into the conservation equation:$v_{max} x_{1} = v_{0} x_{2}$$v_{max} = \frac{v_{0} x_{2}}{x_{1}}$.

The minimum and maximum distances of a planet revolving around the Sun are $x_{1}$ and $x_{2}$. If the minimum speed of the planet on its trajectory is $v_{0}$,then its maximum speed will be:

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