Question 117868
Ok, here I come with a response to part a only.
Parts b, c, and d require your participation.

a)Find the eccentricity, e, of a satellite in an elliptical orbit around the earth.
We are given the distance from one focus (the earth's center) to the ellipse (orbit) as 4,800 miles. This comes from 4,000 miles for the earth's radius plus the apogee (greatest distance from the earth's surface to the satellite) of 800 miles.
We are also given the distance from the same focus (earth's center) to the ellipse (orbit) as 4,200 miles. This comes from 4,000 miles for the earth's radius plus the perigee (the smallest distance from the earth's surface to the satellite) of 200 miles.
First, some definitions related to an ellipse.
Let a = the distance, on the major axis, from the ellipse to the center.
Let c = the distance from one focus to the center.
Now you can see that the distance from the earth's center (focus) to the ellipse at the apogee along the major axis is given by (a + c), and the distance from the same focus to the ellipse at the perigee is given by (a - c).
Since we know these two distance we can calculate a and c.
Why do we want to do that, well...because eccentricity, e, is given by {{{e = c/a}}}
{{{a+c = 4800}}}
{{{a-c = 4200}}} Add these two equations to get...
{{{2a = 9000}}} and...
{{{a = 4500}}} so then...
{{{c = a-4200}}}
{{{c = 4500-4200}}}
{{{c = 300}}}
Eccentricity is:
{{{e = c/a}}}
{{{e = 300/4500}}}
{{{e = 1/15}}} or...
{{{e = 0.067}}}