# Period of Elliptical Orbit

## Homework Statement

Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is 400 km above the earth's surface; at the high point, or apogee, it is 4000 km above the earth's surface.

1. What is the period of the spacecraft's orbit?

## Homework Equations

Kepler's 3rd Law: T=(2*pi*a3/2)/ sqrt(GME)

where a=semi-major axis

## The Attempt at a Solution

So the first thing I did was find the semi-major axis (value a of the eqn above):
(1/2)*(4000+400)=2200 km or 2.2*106 m

Then I plugged it into the equation along with the following constants:
G=6.67*10-11 m2/kg2,
ME=5.97*1024kg

T=(2*pi*2.2*10(6)3/2)/ sqrt(6.67*10-11*5.97*1024)= 1027 seconds

I checked the back of the book and the answer is wrong. I have no clue what I am doing wrong....:( Any help would be greatly appreciated.

## Answers and Replies

LowlyPion
Homework Helper
400 km above the earth's surface; at the high point, or apogee, it is 4000 km above the earth's surface.

Perigee = 6380 + 400 in km
Apogee = 6380 + 4000 in km