What is the orbital distance for a satellite with a 125 min period around Earth?

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To determine the orbital distance for a satellite with a 125-minute period around Earth, the relevant equation is T = 2π√(R^3/GM). The calculations involve converting the period from minutes to seconds and ensuring proper unit consistency. The result should yield the radius from the Earth's center, requiring the Earth's radius to be subtracted to find the altitude. A common mistake is miscalculating or misinterpreting units, leading to unreasonable answers. The final altitude should be approximately 106.6 km after accounting for the Earth's radius.
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Homework Statement


At what distance about Earth would a satellite have a period of 125 min?
G=6.673x10^-11
T=125 Min
Mass=5.97x10^24
Radius of Earth=6.38*10^6

Homework Equations


T=2pi*Square root of (R^3/constant*mass)


The Attempt at a Solution


125=2*3.14*Square root of ((6.38x10^6)^3+R^3/6.673x10^-11*5.97x10^24)
I got a negative #... Which i know is wrong!
 
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Put the units in - it will help !
 
I did that, but still got an unreasonable answer: 2.6x10^26
 
T=2pi*sqrt(R^3/GM)
so T/2pi = sqrt(R^3/GM)
Square both sides (T/2pi)^2 = R^3/GM
Finally R = cube root ( GM * (T/2pi)^2 )

Check the units (m^3 kg^01 s^-2 * kg * s^2)^1/3 = (m^3 )^1/3 = m
 
Is the answer 106.6 KM? Thanks so much for the explanations by the way!
 
I think you have a slight mistype there, do you have T in seconds?
Remember that the result is the radius of the orbit from the centre of the Earth - if you want the altitude you need to subtract the radius of the Earth.
 

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