# Orbital period of satellite about a nonrotating body

1. May 9, 2012

### mollybethe

1. The problem statement, all variables and given/known data
A satellite is in circular orbit at an altitude of 1000 km above the surface of a nonrotating planet with n orbital speed of 5.3 km/s. The escape velocity for the planet is 11.3 km/s. In this situation the orbital period of the satellite, in minutes, is....?

2. Relevant equations
v(orbital)=√(GM/R+r)
v(escape)=√(2GM/R)
T=2∏R/v

3. The attempt at a solution
Of course I converted everything to SI units first. So v(orbital)=5300m/s and v(escape)=11300m/s and r=10^6m. What I did is solved the two equations for velocity for M and then plugged in what I had. I found that R is 282,028m which seems small, but then using that I find that and the second equation I find the time period in minutes to be 25. According to my answer key the answer should be 35. I don't know if I am wrong or the answer key is wrong. What I don't know how to do is allow for any effect that the fact that the planet is non-rotating may have. Help...am I making a silly mistake? It is after midnight and I have been at this a while.

2. May 10, 2012

### Filip Larsen

Your value for R is not correct. Assuming that you mean GM/(R+r) in the first equation, your equations are correct and I guess something must have slipped solving those equations for R. (Hint: one way is to find GM from the second equation and insert this into the first equation and solve for R).

3. May 10, 2012

### mollybethe

I got it! I left out a times two...I was trying to be smooth with my algebra to make my calculations easier and left out a x 2. I got it. R=784624 so R+r=1.78562x10^6. Using the third equation that gives me 35 minutes. Thank you, at least I knew I was working it out correctly!