What is the radius of a geosynchronous satellite?

  • #1
188
0

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 [tex]6.67 \times 10^{-11}\;{\rm N \; m^2 / kg^2}[/tex].
* The mass of the earth is [tex]5.98 \times 10^{24}\;{\rm kg}[/tex].
* The radius of the earth is [tex]6.38 \times 10^{6}\;{\rm m}[/tex].


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

Homework Equations



[tex]T=2\pi\sqrt{\frac{R^3}{GM}}[/tex]

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?
 

Answers and Replies

  • #2
ehild
Homework Helper
15,541
1,907
The formula is correct, do your calculation again. Use the normal form of the numbers.

ehild
 
  • #3
188
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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:

[tex]\frac{86400\times\sqrt{GM}}{2\pi}=\sqrt{R^3}[/tex]

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

[tex]86400^2=4{\pi^2}\frac{R^3}{GM}[/tex]
[tex]\frac{86400^2\times{GM}}{4{\pi^2}}=R^3[/tex]

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 [tex]x/2\pi[/tex] is not interpreted as [tex]x/(2\pi)[/tex], but instead as [tex]\frac{x}{2}\pi[/tex]
 
Last edited:
  • #4
ehild
Homework Helper
15,541
1,907
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 [tex]x/2\pi[/tex] is not interpreted as [tex]x/(2\pi)[/tex], but instead as [tex]\frac{x}{2}\pi[/tex]
Never forget the parentheses in the denominator!!!!:rofl:

ehild
 
  • #5
Borek
Mentor
28,569
3,023
No idea why radius of a satellite should depend on the Earth mass.
 

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