
#1
Nov2307, 04:50 PM

P: 18

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? 



#2
Nov2307, 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) 



#3
Nov2307, 05:31 PM

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P: 25,176

And your value for G is wrong. Though it's hard to really say until you put units on it.




#4
Nov2307, 05:31 PM

P: 18

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? 



#5
Nov2307, 05:32 PM

P: 18





#6
Nov2307, 05:34 PM

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G=6.67*10^(11)*N*m^2/kg^2. Note the exponent.




#7
Nov2307, 05:35 PM

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P: 25,176





#8
Nov2307, 05:39 PM

P: 18





#9
Nov2307, 05:53 PM

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P: 25,176

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.




#10
Nov2307, 06:06 PM

P: 18

thank you



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