Orbital Period Units Question

1. Sep 1, 2011

oneredballoon

1. The problem statement, all variables and given/known data

An earth satellite is observed to have a height of perigee of 100 n mi and a height of apogee of 600 n mi. Find the period of the orbit.

2. Relevant equations

(1) Period = (2*pi/sqrt(mu)) * A^(2/3)
(2) rp+ra = 2A

Where:
Mu is the Standard Gravitational Parameter 4X10^5 (km^3/s^2)
A is the semi-major axis of an elliptic orbit.

3. The attempt at a solution
I converted nautical miles to km, and used equation (2) to obtain A=648.2 km.
(185.2km+1111.2km)=2A; A =648.2km

I then plugged my value of A into the Period equation to obtain an answer of .744.
I can't quite figure out the units for this number. It looks like it would be seconds, but that seems like a pretty unreasonable solution if it is.

I know orbital periods are usually calculated in AU, but if I calculate in AU, I don't know what to do with the constant Mu, which is in km^3/s^2.

Any help would be greatly appreciated! Thanks!

2. Sep 1, 2011

SteamKing

Staff Emeritus
An astronomical unit (AU) is the distance from the earth to the sun, and it is not a measure of time.

In your original calculations, what happened to the earth when you were figuring the dimensions of the orbit?

3. Sep 1, 2011

oneredballoon

I should have phrased that better; I meant that the distances are usually in AU, not the period.

I'm not sure if I understand what you mean by "what happened to the earth"...the Earth's mass is taken into account in Mu which is G*M where G is the gravitational constant, and M is the mass of the earth.

4. Sep 1, 2011

Staff: Mentor

Does 648 km look like a reasonable semimajor axis for a satellite circling the Earth? What's the diameter of the Earth in km?

AU distance units are commonly used when the object is orbiting the Sun. The assumption then is that you'll be using the Sun's gravitational parameter, $\mu_{Sun}$, and time units of about 58.13 days (yes, I know it looks odd!).

Since your satellite is orbiting the Earth you could use the appropriate canonical units: The distance unit DU is the radius of the Earth, the time unit TU is $\sqrt{\frac{DU^3}{\mu_{Earth}}}$, about 806.82 seconds. Alternatively, simply convert everything to mks units and proceed. Unless you're specifically doing work in astrodynamics, you'll probably find the latter option preferable.

5. Sep 1, 2011

oneredballoon

Thank you for the response!

My problem was that I hadn't taken into account the radius of the earth. (I also typed the formula above incorrectly.)

Interestingly enough, this was a homework problem for my Astrodynamics class.