Question

The earth revolves around the sun along an ellipse. The sun lies in the focus of the ellipse. The largest distance from the sun to the earth is 152.1 million kilometers, and the shortest is 147.1 million kilometers. Find the length of the semi minor axis of the ellipse and the eccentricity of the ellipse. What is the distance between the two foci of the ellipse?

Answer 1

$\begin{aligned} (a)\ \textsf{The length of the semi minor axis} = b\\ \\ \textsf{Major axis} = 2a= (152.1+147.1)milli&on\ km\\ =299.2\ million&\ km\\ \\ \therefore a= 149.6\ million&\ km\\ \therefore f= (149.6-147.1)million\ km\\= 2.5\ million\ km\\ \\ from, ae = f\\ e = \dfrac{2,500,000 km}{149,600,000km} = 0.0167\\ a^2(1-e^2) = b^2\\ b = \sqrt{149,600,000(1-0.0167^2)}\\ b = 149,566,000\ km \end{aligned}$

$\therefore \textsf{The length of the semi minor axis} = 149.566\ million\ km$

(b) It's eccentricity as derived above = 0.0167

$\begin{aligned} (c)\ \textsf{distance }&\textsf{between two foci} = F_1\ to\ F_2\\ &= (ae, 0) \to\ (-ae, 0)\\ &= \sqrt{(2ae)^2+(0)^2}\\ &= 2ae\\ &= 2×149,600,000\ million\ km× 0.0167\\ &= 4,996,640\ km \end{aligned}$