Question

The eccentricity of an asteroid's orbit is 0.0442, and the semimajor axis is 1.12 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

Answer:

a. $d=99x10^{8}m$

b. $r=14.22 R_s$

Explanation:

The eccentricity of an asteroid's is 0.0442 so

a.

to find the focus distance between both focus is

$d=2*e*a$

$e=1.12x10^{11}m\\a=0.0442$

So replacing numeric

$d=2*1.12x10^{11}m*0.0442=9900800000$

$d=99x10^{8}m$

b.

Now to find the ratio of that distance between the solar radius and the distance

$r=\frac{d_1}{d_s}$

$r=R_s*\frac{99x10^8m}{6.96x10^8m}$

$r=14.22 R_s$