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#Engineering

The largest and the shortest distance of the earth from are r1 and r2. It’s the distance from the sun when it is perpendicular to the major axis of the orbit drawn from the sun.

Option 1)

$\left ( \frac{r1+r2 }{4}\right )$

Option 2)

$\left ( \frac{r1+r2 }{r1-r2}\right )$

Option 3)

$\left ( \frac{2r1r2}{r1+r2} \right )$

Option 4)

$\left ( \frac{r1+r2}{3} \right )$

Answers (1)

As we learnt in

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 the planet at apogee

$V_{p}= Velocity of perigee$

- wherein

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

$r_{p}=a-c$

$r_{a}=a+c$

The position of a particle moving in an elliptical orbit is represented as

$r=\frac{l}{1+e \cos \Theta }$

$l$ is perpendicular distance of particle from focus and e is eccentricity of ellipse

$r_{1}=\frac{l}{1-e}\ \: \: and\ \: \: \: r_{2}=\frac{l}{1+e}$

$\Rightarrow 1-e=\frac{l}{r_{1}}\ \: \: and\ \: \: 1+e = \frac{l}{r_{2}}$

$\Rightarrow 2=l\left ( \frac{1}{r_{1}} + \frac{1}{r_{2}}\right )\Rightarrow l=\frac{2r_{1}r_{2}}{r_{1}+r_{2}}$

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lovekush

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