What is the orbital period of a satellite in a low-Earth orbit?

  • Thread starter Thread starter Masschaos
  • Start date Start date
  • Tags Tags
    Orbit Period
Click For Summary
SUMMARY

The orbital period P of a satellite in a low-Earth orbit can be approximated using the formula P = C(1 + 3h/[2R⊕]), where h is the elevation above Earth's surface and R⊕ is the Earth's radius. The derivation involves the semi-major axis a = h + R⊕ and utilizes the equation P² = (4π²)/(GM) * a³, with G being the gravitational constant and M the mass of the Earth. The discussion highlights the application of Taylor series to simplify the expression for small values of h compared to R⊕, leading to a linear approximation that is sufficient for practical calculations.

PREREQUISITES
  • Understanding of gravitational physics and orbital mechanics
  • Familiarity with the Taylor series expansion
  • Knowledge of the gravitational constant (G) and Earth's mass (M)
  • Basic algebra and calculus skills for manipulating equations
NEXT STEPS
  • Study the derivation of Kepler's laws of planetary motion
  • Learn about the implications of orbital mechanics in satellite design
  • Explore the applications of Taylor series in physics
  • Investigate the effects of altitude on satellite orbital periods
USEFUL FOR

Students of physics, aerospace engineers, and anyone interested in satellite dynamics and orbital mechanics.

Masschaos
Messages
30
Reaction score
0

Homework Statement


Consider a satellite in a circular, low-Earth orbit; that is, its
elevation above the Earth’s surface is h ≪ R⊕. Show that the orbital period P for such a satellite is approximately P=C(1+ 3h/[2R⊕]).


Homework Equations


P2 = (4pi2)/(GM) * a^3. (G - gravitational constant, M - mass of the Earth (in this case) and a = semi-major axis)


The Attempt at a Solution


Well, the semi-major axis will be: a = h + R⊕.
I've also picked up that a useful representation of a will be: a = R⊕(1 + h/R⊕)
This means our equation because P2 = (4pi2)/(GM) * (R⊕(1 + h/R⊕))^3.
Now we just want P, so P = (2pi/√GM) * (R⊕(1 + h/R⊕))^(3/2).

This obviously doesn't leave me with much. I've picked up from a few lectures that it may have something to do with Taylor series, but I'm severely stumped.
 
Physics news on Phys.org
Good guess. It is taylor series. (1+x)^(3/2)=1+3x/2+3x^2/8+... for x small. Truncate to the linear term. Does that help?
 
Oh yes, I do see. That is almost exactly what is required.
I don't suppose you'd know any reason for keeping it to the linear term?
Perhaps because h << R the other terms become negligible. I think that is reasonable!
Thank you very much.
That was much simpler than I thought.
 

Similar threads

Replies
10
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
964
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 10 ·
Replies
10
Views
980
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K