A planet moves around the sun. At a given poi | vedclass

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(C) According to Kepler's second law of planetary motion,the angular momentum of a planet revolving around the sun remains constant.The angular moment

Vedclass · Accepted Answer

$\frac{d_1 v_1}{d_2}$

Vedclass · Answer

(C) According to Kepler's second law of planetary motion,the angular momentum of a planet revolving around the sun remains constant.The angular momentum $L$ is given by $L = mvr \sin(	heta)$. At the points of closest approach (perihelion) and farthest distance (aphelion),the velocity vector is perpendicular to the position vector,so $	heta = 90^\circ$ and $\sin(90^\circ) = 1$.Therefore,the conservation of angular momentum gives:$m v_1 d_1 = m v_2 d_2$Canceling the mass $m$ from both sides:$v_1 d_1 = v_2 d_2$Solving for $v_2$:$v_2 = \frac{v_1 d_1}{d_2}$

$A$ planet moves around the sun. At a given point $P$,it is closest to 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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