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Question: The maximum and minimum distance of earth from the sun are \({r_1}\) and \({r_2}\) respectively. Wha...

The maximum and minimum distance of earth from the sun are r1{r_1} and r2{r_2} respectively. What will be the distance of earth from the sun when its position vector is perpendicular to the major axis of its orbit?
A. r1+r24\dfrac{{{r_1} + {r_2}}}{4}
B. (r1+r2r1−r2)\left( {\dfrac{{{r_1} + {r_2}}}{{{r_1} - {r_2}}}} \right)
C. 2r1r2r1+r2\dfrac{{2{r_1}{r_2}}}{{{r_1} + {r_2}}}
D. r1+r23\dfrac{{{r_1} + {r_2}}}{3}

Explanation

Solution

We know that earth revolves around the sun in an elliptical orbit. We know that the distance between earth and sun is maximum when the earth is at one of the two end points of the major axis of its orbit and minimum when the earth is at one of the two end points of the minor axis of its orbit. Therefore, to solve this problem, we need to use the formula for the position of the particle moving in an elliptical orbit.

Formula used:
r=l1+ecos⁡θr = \dfrac{l}{{1 + e\cos \theta }}
where, rr is the position of the particle moving in an elliptical orbit, ll is the perpendicular distance of the particle from the focus, ee is the eccentricity of an elliptical orbit and θ\theta is the angle between position vector of earth and the major axis.

Complete step by step answer:
It is given that the maximum distance of the earth from the sun is r1{r_1}.
r=l1+ecos⁡θ ⇒r1=l1−e ⇒1−e=lr1  r = \dfrac{l}{{1 + e\cos \theta }} \\\ \Rightarrow {r_1} = \dfrac{l}{{1 - e}} \\\ \Rightarrow 1 - e = \dfrac{l}{{{r_1}}} \\\
Also, the minimum distance of the earth from the sun is given as r2{r_2}.
r = \dfrac{l}{{1 + e\cos \theta }} \\\ \Rightarrow {r_2} = \dfrac{l}{{1 + e}} \\\ \Rightarrow 1 + e = \dfrac{l}{{{r_2}}} \\\
Now, by adding both these equations, we get
2=lr1+lr2 ⇒2=l(1r1+1r2) ⇒2=l(r1+r2r1r2) ∴l=2r1r2r1+r2 2 = \dfrac{l}{{{r_1}}} + \dfrac{l}{{{r_2}}} \\\ \Rightarrow 2 = l\left( {\dfrac{1}{{{r_1}}} + \dfrac{1}{{{r_2}}}} \right) \\\ \Rightarrow 2 = l\left( {\dfrac{{{r_1} + {r_2}}}{{{r_1}{r_2}}}} \right) \\\ \therefore l = \dfrac{{2{r_1}{r_2}}}{{{r_1} + {r_2}}}
Thus, the distance of earth from sun when its position vector is perpendicular to the major axis of its orbit will be 2r1r2r1+r2\dfrac{{2{r_1}{r_2}}}{{{r_1} + {r_2}}}.

Hence, option C is the right answer.

Note: We have used the fact that earth revolves in the elliptical path around the sun. This fact was discovered by the mathematician Kepler. According to his first law, the planets orbit the sun in ellipses with the sun at one focus. The distance between a planet and the Sun changes as the planet moves along its orbit. Also, the Sun is offset from the center of the planet’s orbit.