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Satellite period, which equation?

by izforgoat
Tags: equation, period, satellite
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izforgoat
#1
Nov23-07, 04:50 PM
P: 16
1. The problem statement, all variables and given/known data

I'm looking for the period of an orbiting object a certain height from the earth's surface, I am given this height. So I have the total radius of 6,5OO,OOO m, g = 9.81 m/s^2 and the mass of earth = 5.98*10^24 kg

Please note that for this problem G is another constant than what it usually is.

2. Relevant equations

Here is where I am confused I do not know whether to use the T = (2[tex]\pi[/tex]r)/[tex]\sqrt{gr}[/tex]

or the T[tex]^{2}[/tex] = (4[tex]\pi^{2}r^{3}[/tex])/(GM[tex]_{earth}[/tex])

where G = 6.67*1O^-7

3. The attempt at a solution

K. So when I use the method of going with T = (2[tex]\pi[/tex]r)/[tex]\sqrt{gr}[/tex]

I get about 5114 seconds for the period.

when I use T[tex]^{2}[/tex] = (4[tex]\pi^{2}r^{3}[/tex])/(GM[tex]_{earth}[/tex])
I get 52.13584223 seconds, which doesn't logically seem right but since G is different I don't know.

does anyone know what the right method is?
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malawi_glenn
#2
Nov23-07, 05:12 PM
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http://upload.wikimedia.org/math/3/5...88515e4825.png

Keplers third law (the forumula above is for spherical orbitals)
Dick
#3
Nov23-07, 05:31 PM
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And your value for G is wrong. Though it's hard to really say until you put units on it.

izforgoat
#4
Nov23-07, 05:31 PM
P: 16
Satellite period, which equation?

thx alot for that clarification. Still it doesn't make sense that its period would be 52 seconds here's my work.

T(secs)^2 = (4(pi^2)(6,500,000^3 m))/((6.67*10^-7 N*m^2/kg^2)(5.98*10^24 kg))

are the units for T in seconds? was it alright that I changed km to m for the radius?
izforgoat
#5
Nov23-07, 05:32 PM
P: 16
Quote Quote by Dick View Post
And your value for G is wrong. Though it's hard to really say until you put units on it.
I'm using a different value for this problem
Dick
#6
Nov23-07, 05:34 PM
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G=6.67*10^(-11)*N*m^2/kg^2. Note the exponent.
Dick
#7
Nov23-07, 05:35 PM
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Quote Quote by izforgoat View Post
I'm using a different value for this problem
Is it an 'alternative universe' problem? Why would you use a different value for G? It's a 'universal constant'.
izforgoat
#8
Nov23-07, 05:39 PM
P: 16
Quote Quote by Dick View Post
Is it an 'alternative universe' problem? Why would you use a different value for G? It's a 'universal constant'.
you could say that. But either way I don't think it would have much difference for this equation than the plug and chug. Right now I want to know if I am calculating everything else right. I assume I am.
Dick
#9
Nov23-07, 05:53 PM
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The reason you getting 52 seconds is because you are putting in a value of G that is 10000 times too large. Other than that you are doing fine.
izforgoat
#10
Nov23-07, 06:06 PM
P: 16
thank you


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