How Do You Calculate the Orbital Period of a Satellite?

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SUMMARY

The discussion focuses on calculating the orbital period of an Earth satellite with a perigee height of 100 nautical miles and an apogee height of 600 nautical miles. The key equations used include the period formula, Period = (2*pi/sqrt(mu)) * A^(2/3), and the semi-major axis calculation, rp + ra = 2A. The calculated semi-major axis A was found to be 648.2 km, leading to an initial period estimate of 0.744 seconds, which raised concerns about unit accuracy. The conversation clarified that the gravitational parameter mu incorporates Earth's mass and suggested converting to MKS units for consistency.

PREREQUISITES
  • Understanding of orbital mechanics and satellite dynamics
  • Familiarity with gravitational parameters, specifically mu for Earth
  • Knowledge of converting between nautical miles and kilometers
  • Basic proficiency in using mathematical equations related to orbits
NEXT STEPS
  • Learn about the Standard Gravitational Parameter for Earth (mu_Earth)
  • Study the conversion between different unit systems, particularly MKS and AU
  • Explore the implications of the semi-major axis on satellite orbits
  • Investigate the use of canonical units in astrodynamics
USEFUL FOR

Astronomy students, aerospace engineers, and anyone involved in satellite orbit calculations will benefit from this discussion.

oneredballoon
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Homework Statement



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.

Homework 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)
rp is radius of periapsis
ra is radius of apoapsis
A is the semi-major axis of an elliptic orbit.


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!
 
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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?
 
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.
 
oneredballoon said:
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.

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.
 
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.
 

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