At what altitude and velocity is an orbit geosynchronous?

In summary, an orbit is geosynchronous when its period is equal to one day (or 23 hours 56 minutes for more precision). By using the formula a = ( (muE*T2) / (4*pi2) ) 1/3, where a is the semi-major axis and muE is the gravitational parameter of the Earth, the altitude of a circular, geosynchronous orbit can be calculated by subtracting the radius of the Earth from the result. The velocity in a geosynchronous orbit can be calculated using the formula Vc = sqrt(muE / r), where r is the radius of the orbit. An alternative formula for calculating the altitude is a = \sqrt[3]{{\frac{{GM
  • #1
Loren Booda
3,125
4
At what altitude and velocity is an orbit geosynchronous? Please include calculations. I am too old for this to be a homework problem - just exercising my mind.
 
Astronomy news on Phys.org
  • #2
Loren Booda said:
At what altitude and velocity is an orbit geosynchronous? Please include calculations. I am too old for this to be a homework problem - just exercising my mind.

You can use my online calculator to solve this: http://orbitsimulator.com/gravity/articles/formula55.html

Just enter 24 hours for P (or 23 hrs 56 minutes + some seconds, Wikipedia for a more exact answer) and 1 Earth Mass for M, and you'll get your semi-major axis (same as distance for a circular orbit). For altitude, subtract the equatorial radius of Earth: 6378 km. For velocity v=d/t, or in a rotating system, v=2*pi*distance/24 hours (or 23 hours 56 min... for extra precision)
 
  • #3
I should have been clearer. I am looking for an analytical solution to the problem with a circular orbit. Thanks for your assist, tony873004.
 
  • #4
The general equation for the orbital period of a satellite in orbit around the Earth:

T=2*pi*sqrt(a3/muE)​

With a equal to the semi-major axis of the orbit (which is the radius r in a circular orbit). The gravitational paramter of the Earth is called muE, and it has the following value:

muE = 3.98600441*105 km3/s2

If you rewrite your equation for a (or r), you get:

a = ( (muE*T2) / (4*pi2) ) 1/3

The orbital period in geosynchronous orbit is:

Tgeo= 86400 seconds (24 hours)​

Substituting all values in the equation for a (or r), you get:

a = 42241.1 km​

Now subtracting the radius of the Earth RE, which is equal to 6378.1 km, yields the altitude h:

h = 35863 km​

For the velocity, there is an equation for the circular velocity in orbit:

Vc = sqrt(muE / r)​

With r equal to 42241.1 km, it should be 3.1 km/s
 
  • #5
How about deriving the altitude of a circular, geosynchronous orbit most simply - without assuming Kepler's Laws?
 
  • #6
My solution was analytic (no numerical methods used). But if you need to eliminate Kepler, this formula contains Newton, but not Kepler:

[tex]a = \sqrt[3]{{\frac{{GMt^2 }}{{4\pi ^2 }}}} - r[/tex]

where a is your altitude, G the gravitational constant, M the mass of the Earth, t the length of a sideral day, and r the radius of the Earth.
 
  • #7
Well stated, but what about deriving it?
 
  • #8
Loren, here are some (hopefully) familiar formulas to assist you:

Gravitational force:
F = G M m / r^2
(here M = Earth's mass, m = orbiting mass)

Centripetal force:
F = m w^2 r
(here w is really "omega", the orbital frequency in radians/sec)

Finally, w vs. period T:
T = 2*pi / w
or
w = 2*pi / T
(where T = 1 day for geosynch. orbit)

To derive r for a geosynch. orbit, first equate the two force expressions with each other. Solve the equation for "r" in terms of the other quantities.

Next, figure out what w should be for T = 1 day ***, and look up G and M. Plug in the numbers to get r, the radius of the orbit. Subtract the Earth's radius if you want the altitude above the Earth's surface.

Regards,

Mark

p.s. edit:
You'll need to work in a consistent set of units, and it's common to use meters, seconds, and kilograms. Eg., 1 day is ___ seconds? Use that number for T, rather than "1".
 
  • #9
This is essentially what Redbelly posted, but I jumped straight to acceleration rather than force:

Start with what we know:

1. Distance = velocity * time
2. Distance in an orbit is 2*pi*a (the circumference formula), where a is semi-major axis, same as radius for a circular orbit

Set them equal to each other:
(1) [tex]v_{circ} \cdot t = 2\pi a[/tex]

Use the circular orbital velocity formula

(2) [tex]v_{circ} = \sqrt {\frac{{GM}}{a}} [/tex]

If you'd like to derive this, you start with formula for centrepital acceleration
(3) [tex]\frac{{v^2 }}{a}[/tex]
and Newton's gravitational formula for acceleration
(4) [tex]\frac{{GM}}{{a^2 }}[/tex]

and set them (3) & (4) equal to each other, then use algebra to isolate v

[tex]
\begin{array}{l}
\frac{{v^2 }}{a} = \frac{{GM}}{{a^2 }}\,\,\,\, \Rightarrow \,\,\,\, \\
\\
v^2 r^2 = GMr\,\,\,\, \Rightarrow \,\,\,\, \\
\\
v^2 r = GM\,\,\,\, \Rightarrow \,\,\,\, \\
\\
v^2 = \frac{{GM}}{a}\,\,\,\, \Rightarrow \,\,\,\, \\
\\
v = \sqrt {\frac{{GM}}{a}} \\
\end{array}
[/tex]

Re-write formula (1), substituting the circular velocity formula for velocity
(4) [tex]\sqrt {\frac{{GM}}{a}} \cdot t = 2\pi a[/tex]

Use algebra to isolate a

[tex]
\begin{array}{l}
\sqrt {\frac{{GM}}{a}} = \frac{{2\pi a}}{t} \\
\\
\frac{{GM}}{a} = \left( {\frac{{2\pi a}}{t}} \right)^2 \\
\\
\frac{{GM}}{a} = \frac{{4\pi ^2 a^2 }}{{t^2 }} \\
\\
GMt^2 = 4\pi ^2 a^3 \\
\\
a^3 = \frac{{GMt^2 }}{{4\pi ^2 }} \\
\\
a = \sqrt[3]{{\frac{{GMt^2 }}{{4\pi ^2 }}}} \\
\end{array}
[/tex]

And since you want altitude, subtract Earth's radius (r)
(5) [tex]a = \sqrt[3]{{\frac{{GMt^2 }}{{4\pi ^2 }}}} - r[/tex]

Plugging in numbers (using meters, kilograms, seconds)
(6.6725985E-11*5.97369125232006E+24*86400^2/(4*pi^2))^(1/3)=42241095.3597612

I used 86400 for t, which is 24 hours. You can get a little more exact if you look up the length of a sideral day (~23h 56m) in seconds
 
  • #10
Aha! It seems that these are what I was looking for. Thanks both.
 
  • #11
Bruce too.
 
  • #12
Yeah, Bruce-Almight first constructed the Arc.
 

What is a geosynchronous orbit?

A geosynchronous orbit is a type of orbit around a celestial body, such as the Earth, in which an object appears to remain stationary in the same position relative to the body's surface.

What altitude is a geosynchronous orbit?

The altitude of a geosynchronous orbit is approximately 22,236 miles (35,786 kilometers) above the Earth's surface.

What is the velocity of a geosynchronous orbit?

The velocity of a geosynchronous orbit is approximately 6,876 miles per hour (11,072 kilometers per hour). This velocity is necessary for an object to maintain a circular orbit at an altitude of 22,236 miles above the Earth's surface.

What is the purpose of a geosynchronous orbit?

Geosynchronous orbits are used for communication, navigation, and meteorological purposes. Satellites in geosynchronous orbits can provide continuous coverage of a specific area on the Earth's surface, making them ideal for these types of applications.

How is a geosynchronous orbit different from a geostationary orbit?

A geostationary orbit is a specific type of geosynchronous orbit in which the object remains stationary above a specific point on the Earth's equator. A geosynchronous orbit, on the other hand, can be at any inclination relative to the equator.

Similar threads

  • Astronomy and Astrophysics
Replies
2
Views
1K
  • Astronomy and Astrophysics
Replies
22
Views
1K
  • Electrical Engineering
Replies
5
Views
765
  • Astronomy and Astrophysics
Replies
8
Views
318
  • Astronomy and Astrophysics
Replies
15
Views
2K
  • Astronomy and Astrophysics
Replies
27
Views
2K
  • Astronomy and Astrophysics
2
Replies
47
Views
3K
  • Astronomy and Astrophysics
Replies
32
Views
3K
  • Astronomy and Astrophysics
Replies
4
Views
1K
  • Astronomy and Astrophysics
Replies
5
Views
2K
Back
Top