Satellite period, which equation?

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SUMMARY

The discussion centers on calculating the orbital period of a satellite at a height of 6,500,000 meters above Earth's surface using two equations: T = (2πr)/√(gr) and T² = (4π²r³)/(GMearth). The correct gravitational constant for this problem is G = 6.67 x 10-11 N·m²/kg², not the incorrect value of 6.67 x 10-7. The user initially calculated a period of 5114 seconds using the first equation and 52.14 seconds using the second, which was deemed incorrect due to the wrong G value. The correct approach confirms that the period should be calculated using the proper gravitational constant.

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Homework Statement



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.

Homework Equations



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

or the T^{2} = (4\pi^{2}r^{3})/(GM_{earth})

where G = 6.67*1O^-7

The Attempt at a Solution



K. So when I use the method of going with T = (2\pir)/\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?
 
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And your value for G is wrong. Though it's hard to really say until you put units on it.
 
thx a lot 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?
 
Dick said:
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
 
G=6.67*10^(-11)*N*m^2/kg^2. Note the exponent.
 
izforgoat said:
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'.
 
Dick said:
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.
 
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
thank you
 

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