What is the radius of a geosynchronous satellite?

[warfreak131]

Homework Statement

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.

Homework Equations

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

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?

 

[ehild]

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

ehild

 

[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 the correct order of operations. because x/2\pi is not interpreted as x/(2\pi), but instead as \frac{x}{2}\pi

 

[ehild]

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

Never forget the parentheses in the denominator!

ehild

 

[Borek]

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