Minimum height for a satellite to remain over the same geographical point

AI Thread Summary
To keep a satellite stationary over a specific point on the equator, it must be positioned at approximately 35,870 km above the Earth's surface. The calculations involve equating centripetal and gravitational forces, using the relationship between angular velocity and tangential velocity. The angular velocity can be derived from the Earth's rotation period. The velocity of the satellite at this height is around 3,072.6 m/s. These values ensure that the satellite remains geostationary.
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a.)calculate at what height above the Earth's surface a satellite must be placed if it is to remain over the same gerographical point on the equator of the earth. b.)what is the velocity of such a satellite?

i have no idea how to do the question, but i know it has to do with these equations:
Fcent.=Fgrav.
m(v^2)/r=GMm/(r^2)

because my book doesn't have answers for this question, i don't know the answer for this question.

 
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That's the equation u need to use,but under an equivalent form.Now find "r" by plugging the correct numerical values...

Daniel.

P.S.HINT:relate angular velocity to tangent velocity...
 
dextercioby, can you please show me some work or maybe even tell me the answer. i am stuck becuase i found that there are two variables, "v" and "r", are unknown.
 
Yes.v=\omega r.And now u have only one variable,namely 'r'...

Daniel.
 
sorry for my ignorance, but what is the w(or omega) thing stands for?
 
Angular velocity of the Earth's rotation motion.U can find it knowing the period of rotation ("length" of a mean day) and the value of \pi,which can be approximated to 3.14

Daniel.
 
i found the answer for part a of my question to be 35870 km, can you please approve it. thanks a lot so far.
 
It seems correct.However,for point b) u'll need another number,or u can use this 35870Km,but indirectly.

Daniel.
 
anyways (i have a feeling that you arn't going to do the question for me), thanks very much dextercioby for your quick response, it helps me a great deal.
 
  • #10
lol...k thanks
 
  • #11
Tell me what number you get for velocity...

Daniel.
 
  • #12
does 3072.6 m/s sounds good?
 
  • #14
k, thanks again. :smile:
 
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