How Do You Calculate the Period of a Spacecraft's Elliptical Orbit?

[KayleighK]

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*a^3/2)/ sqrt(GM_E)

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,
M_E=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.

 

[LowlyPion]

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

 

[nlightner]

Hello, as a scientist, I would like to point out that your attempt at the solution is correct. However, there might be a typo in the values provided in the homework statement. The value for the Earth's mass (ME) should be 5.97*10^24 kg instead of 5.97*10^24. This small difference in notation could lead to a significant difference in the final answer.

Furthermore, I would like to suggest that you double-check your calculations using scientific notation. For example, instead of writing 6.67*10-11 m2/kg2, it would be better to write it as 6.67*10^-11 m^3/kg/s^2 to avoid any potential errors.

Overall, your approach and use of Kepler's third law are correct. I hope this helps you in finding the correct answer. If you still have trouble, I suggest seeking help from your teacher or a tutor. Keep up the good work!