# Satellite period, which equation?

by izforgoat
Tags: equation, period, satellite
 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$$\pi$$r)/$$\sqrt{gr}$$ or the T$$^{2}$$ = (4$$\pi^{2}r^{3}$$)/(GM$$_{earth}$$) where G = 6.67*1O^-7 3. The attempt at a solution K. So when I use the method of going with T = (2$$\pi$$r)/$$\sqrt{gr}$$ I get about 5114 seconds for the period. when I use T$$^{2}$$ = (4$$\pi^{2}r^{3}$$)/(GM$$_{earth}$$) 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?
 Sci Advisor HW Helper P: 4,739 http://upload.wikimedia.org/math/3/5...88515e4825.png Keplers third law (the forumula above is for spherical orbitals)
 Sci Advisor HW Helper Thanks P: 24,989 And your value for G is wrong. Though it's hard to really say until you put units on it.
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?
P: 18
 Quote by Dick 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
 Sci Advisor HW Helper Thanks P: 24,989 G=6.67*10^(-11)*N*m^2/kg^2. Note the exponent.