The maximum and minimum distance of earth from sun are $r_{1}$ and $r_{2}$ respectively What will be the distance of earth from sun when its position vector is perpendicular to the major axis of its orbit

\frac{r_{1} + r_{2}}{4}

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(\frac{r_{1} + r_{2}}{r_{1} - r_{2}})

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\frac{2 r_{1} r_{2}}{r_{1} + r_{2}}

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\frac{r_{1} + r_{2}}{3}

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Solution

The correct option is C $\frac{2 r_{1} r_{2}}{r_{1} + r_{2}}$
For an elliptical orbit with eccentricity $e$ , the position of a particle with perpendicular distance from focus is represented as,

$r = \frac{1}{(l + e c o s θ)}$

The maximum distance is given by

$r_{1} = \frac{1}{(l - e)}$

$1 - e = \frac{l}{r_{1}}$ .........(1)

The minimum distance is given by

$r_{2} = \frac{1}{(l + e)}$

$1 + e = \frac{l}{r_{2}}$ ......... (2)

Adding (1) & (2),

$2 = \frac{l}{r_{1}} + \frac{l}{r_{2}}$

$2 = l [\frac{(r 1 + r 2)}{r 1. r 2}]$

$l = 2 \frac{r_{1} . r_{2}}{(r_{1} + r_{2})}$

The distance when perpendicular to principal axis is given by $\frac{2 r_{1} . r_{2}}{(r_{1} + r_{2})}$ .

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The maximum and minimum distance of earth from sun are r1 and r2 respectively What will be the distance of earth from sun when its position vector is perpendicular to the major axis of its orbit

The maximum and minimum distance of earth from sun are $r_{1}$ and $r_{2}$ respectively What will be the distance of earth from sun when its position vector is perpendicular to the major axis of its orbit