# At what altitude and velocity is an orbit geosynchronous?

1. Jul 8, 2008

### Loren Booda

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.

2. Jul 8, 2008

### tony873004

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. Jul 8, 2008

### Loren Booda

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. Jul 9, 2008

### Bruce-Almight

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. Jul 9, 2008

### Loren Booda

How about deriving the altitude of a circular, geosynchronous orbit most simply - without assuming Kepler's Laws?

6. Jul 9, 2008

### tony873004

My solution was analytic (no numerical methods used). But if you need to eliminate Kepler, this formula contains Newton, but not Kepler:

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

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. Jul 9, 2008

### Loren Booda

Well stated, but what about deriving it?

8. Jul 9, 2008

### Redbelly98

Staff Emeritus
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. Jul 9, 2008

### tony873004

This is essentially what Redbelly posted, but I jumped straight to acceleration rather than force:

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) $$v_{circ} \cdot t = 2\pi a$$

Use the circular orbital velocity formula

(2) $$v_{circ} = \sqrt {\frac{{GM}}{a}}$$

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

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

$$\begin{array}{l} \frac{{v^2 }}{a} = \frac{{GM}}{{a^2 }}\,\,\,\, \Rightarrow \,\,\,\, \\ \\ v^2 r^2 = GMr\,\,\,\, \Rightarrow \,\,\,\, \\ \\ v^2 r = GM\,\,\,\, \Rightarrow \,\,\,\, \\ \\ v^2 = \frac{{GM}}{a}\,\,\,\, \Rightarrow \,\,\,\, \\ \\ v = \sqrt {\frac{{GM}}{a}} \\ \end{array}$$

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

Use algebra to isolate a

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

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

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. Jul 10, 2008

### Loren Booda

Aha! It seems that these are what I was looking for. Thanks both.

11. Jul 10, 2008

### tony873004

Bruce too.

12. Jul 10, 2008

### Loren Booda

Yeah, Bruce-Almight first constructed the Arc.