A planet of mass m moves in an elliptical orb | vedclass

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(A) According to the law of conservation of angular momentum at the perihelion and aphelion points:$m v_1 r_1 = m v_2 r_2$$\Rightarrow v_2 = \frac{v_1

Vedclass · Accepted Answer

$m \sqrt{\frac{2 G M r_1 r_2}{r_1+r_2}}$

Vedclass · Answer

(A) According to the law of conservation of angular momentum at the perihelion and aphelion points:$m v_1 r_1 = m v_2 r_2$$\Rightarrow v_2 = \frac{v_1 r_1}{r_2}$ $(i)$From the law of conservation of total mechanical energy:$-\frac{G M m}{r_1} + \frac{1}{2} m v_1^2 = -\frac{G M m}{r_2} + \frac{1}{2} m v_2^2$Substituting $v_2$ from $(i)$:$-\frac{G M}{r_1} + \frac{1}{2} v_1^2 = -\frac{G M}{r_2} + \frac{1}{2} \left(\frac{v_1 r_1}{r_2}\right)^2$$\frac{1}{2} v_1^2 \left(1 - \frac{r_1^2}{r_2^2}\right) = G M \left(\frac{1}{r_1} - \frac{1}{r_2}\right)$$\frac{1}{2} v_1^2 \left(\frac{r_2^2 - r_1^2}{r_2^2}\right) = G M \left(\frac{r_2 - r_1}{r_1 r_2}\right)$$\frac{1}{2} v_1^2 \frac{(r_2 - r_1)(r_2 + r_1)}{r_2^2} = G M \frac{(r_2 - r_1)}{r_1 r_2}$$v_1^2 = \frac{2 G M r_2}{r_1(r_1 + r_2)}$$v_1 = \sqrt{\frac{2 G M r_2}{r_1(r_1 + r_2)}}$Angular momentum $L = m v_1 r_1 = m \sqrt{\frac{2 G M r_2}{r_1(r_1 + r_2)}} \cdot r_1 = m \sqrt{\frac{2 G M r_1 r_2}{r_1 + r_2}}$

$A$ planet of mass $m$ moves in an elliptical orbit around an unknown star of mass $M$ such that its maximum and minimum distances from the star are equal to $r_1$ and $r_2$ respectively. The angular momentum of the planet relative to the centre of the star is

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