Question

The eccentricity of an asteroid's orbit is 0.0380, and the semimajor axis is 3.06 x 10¹¹ m. The Sun's center is at one focus of the asteroid's orbit. (a) How far from this focus is the other focus in meters? (b) What is the ratio of this distance to the solar radius, 6.96 x 10⁸ m?

Answer 1

Given is:-

Eccentricity of an asteroid's orbit is e=0.0380

the length of the semiMajor axis is  $a = 3.06 \times 10^{11}m$

r- a :Semimajor axis e eccentricity Semiminor : axis

Now if c is the distance from the focus to the center of the ellipse,

So the foci should just be 2a.e meters apart

by putting all the values in above equation, we get

$F_1F_2 = 2 \times (3.06 \times 10^{11}) \times (0.0380)$

$F_1F_2 = 2.32 \times 10^{10}m$ this is the distance between both the foci of the given orbital ellipse.

Part-b

Thus the required ratio is

$Ratio = \frac{2.32 \times 10^{10}}{6.96 \times 10^8} = 33.33$

Thus the required ration of the distance between foci to the solar radius is 33.33