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

 P: 74 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.
 Sci Advisor HW Helper P: 11,952 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...
 P: 74 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.
 Sci Advisor HW Helper P: 11,952 Minimum height for a satellite to remain over the same geographical point Yes.$v=\omega r$.And now u have only one variable,namely 'r'... Daniel.
 P: 74 sorry for my ignorance, but what is the w(or omega) thing stands for?
 Sci Advisor HW Helper P: 11,952 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.
 P: 74 i found the answer for part a of my question to be 35870 km, can you please approve it. thanks alot so far.
 Sci Advisor HW Helper P: 11,952 It seems correct.However,for point b) u'll need another number,or u can use this 35870Km,but indirectly. Daniel.
 P: 74 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.
 P: 74 lol....k thanks
 Sci Advisor HW Helper P: 11,952 Tell me what number you get for velocity... Daniel.
 P: 74 does 3072.6 m/s sounds good?
 Sci Advisor HW Helper P: 11,952 Pretty much so. Daniel.
 P: 74 k, thanks again.

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