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In the following fig. If the If Eccentricity is given by (e) = $\frac{c}{a}$

Then find the speed of the planet at perigee.

Option: 1
$V_{p} = \sqrt{\frac{GM}{a}\left ( \frac{1+e}{1-e} \right )}$

Option: 2
$V_{p} = \sqrt{\frac{GM}{a}\left ( \frac{1}{1-e} \right )}$

Option: 3
$V_{p} = \sqrt{\frac{GM}{a}\left ( \frac{1}{1+e} \right )}$

Option: 4
$V_{p} = \sqrt{\frac{GM}{a}\left ( \frac{1-e}{1+e} \right )}$

Answers (1)

best_answer

As we learn

Velocity of planet in terms of Eccentricity -

$V_{a}=\sqrt{\frac{GM}{a}\left ( \frac{1-e}{1+e} \right )}$

$V_{p}=\sqrt{\frac{GM}{a}\left ( \frac{1+e}{1-e} \right )}$

$V_{A}=$ Velocity of planet at apogee

$V_{p}= Velocity of perigee$

- wherein

Eccentricity (e) = $\frac{c}{a}$

$r_{p}=a-c$

$r_{a}=a+c$

Applying consevation of angular momentum at perigee and apogee

$mv_{p}r_{1} = mv_{a}r_{2} = \frac{v_{p}}{v_{a}} = \frac{r_{2}}{r_{2}} = \frac{a+c}{a-c}$

$\frac{v_{p}}{v_{a}} = \frac{1+\frac{c}{a}}{1-\frac{a}{c}} = \frac{1+e}{1-e}.....(1 )$

Apply energy conservation at perigee and apogee

$\frac{1}{2}mv_{p}^2-\frac{GMm}{r_{1}}=\frac{1}{2}mv_{a}^2-\frac{GMm}{r_{2}}$

$v_{p}^2-v_{a}^2=2GM(\frac{1}{r_{1}}-\frac{1}{r_{2}}).....(2)$

$v_{a}=\frac{r_{1}}{r_{2}}v_{p}$ Put in (2)

$v_{p}^2-(\frac{r_{1}}{r_{2}}v_{p})^2=2GM(\frac{1}{r_{1}}-\frac{1}{r_{2}}).....(2)$

After solving we get

$V_{p} = \sqrt{\frac{GM}{a}\left ( \frac{1+e}{1-e} \right )}$

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Sayak

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