# What is the radius of a geosynchronous satellite?

1. Dec 13, 2009

### warfreak131

1. The problem statement, all variables and given/known data
Find the radius R of the orbit of a geosynchronous satellite that circles the earth. (Note that R is measured from the center of the earth, not the surface.) You may use the following constants:

* The universal gravitational constant G is $$6.67 \times 10^{-11}\;{\rm N \; m^2 / kg^2}$$.
* The mass of the earth is $$5.98 \times 10^{24}\;{\rm kg}$$.
* The radius of the earth is $$6.38 \times 10^{6}\;{\rm m}$$.

The correct answer is $$4.23\times10^7\;{\rm m}$$, but I get a different answer.

2. Relevant equations

$$T=2\pi\sqrt{\frac{R^3}{GM}}$$

3. The attempt at a solution

Since T is measured in seconds, and there are 86,400 seconds in a day, some simple algebra gives me the answer of 1,994,400,816 m. What am I doing wrong?

2. Dec 13, 2009

### ehild

The formula is correct, do your calculation again. Use the normal form of the numbers.

ehild

3. Dec 13, 2009

### warfreak131

what is the "normal form" for these numbers?

EDIT: I did the calculation a different way, and I got the right answer, but the only thing I did differently was the order that I did the algebra in. I don't know where my error lies.

I normally did:

$$\frac{86400\times\sqrt{GM}}{2\pi}=\sqrt{R^3}$$

and solve for R, but instead, this time I did:

$$86400^2=4{\pi^2}\frac{R^3}{GM}$$
$$\frac{86400^2\times{GM}}{4{\pi^2}}=R^3$$

and it worked..... what did i do wrong the first time?

if you square the first equation, you get the same thing......

2nd edit, nevermind, i know what I did wrong.

this has been a nice lesson for me in making sure I properly place parenthesis in a calculator to ensure teh correct order of operations. because $$x/2\pi$$ is not interpreted as $$x/(2\pi)$$, but instead as $$\frac{x}{2}\pi$$

Last edited: Dec 13, 2009
4. Dec 13, 2009

### ehild

Never forget the parentheses in the denominator!!!!:rofl:

ehild

5. Dec 13, 2009

### Staff: Mentor

No idea why radius of a satellite should depend on the Earth mass.