What is the Period of an Earth Satellite Orbiting 400km Above the Surface?

[Torquescrew]

Homework Statement

Find the speed of an Earth satellite whose orbit is 400km above the Earth's surface. What is the period of the orbit.

Homework Equations

The first part was pretty easy. I used v=[(Gm)/r]^(1/2) to get the answer.

The Attempt at a Solution

For the first part, I dropped the values into the equation to get
(6.673*10^-11)(5.97*10^24)
(6.37*10^6 + 400000)^2

I came up with about 7671 m/s (7.7*10^3 km/s) <--tangential velocity, right?


But for the second part, I couldn't find any useful details in my notes. I figured I'd get the circumference and divide by my velocity, but that didn't work.
2(pi)(6.37*10^6 + 400000)^2
7671

I wound up with some huge number.

I also tried Kepler's 3rd law (t^2=Kr^3), which doesn't help much, because I don't have a constant.
The thing is, I already know what the answer is supposed to be. I just don't know how to get it.

The answer is supposed to be approx 5.6*10^3 s (about 93 minutes).

I know I'm close. I tried to reverse engineer the problem by solving for x
5.6*10^3=[2(pi)x]/7671.02

I wound up with x = 6.836*10^6, which is remarkably close to 6.37*10^6 + 400000 but that seems inconsistent with the whole two-pi-arr-squared thing.

So I guess my real question is, if the radius is right, and if my velocity is right, why should I use 2(pi)r instead of 2(pi)r^2? What did I miss?

 

[nasu]

You are on the right path.
You have a mistake in the formula for the perimeter of the circle.
It's 2*pi*r
Not r^2. (r^2 is for area, r^3 for volume)

 

[Torquescrew]

Good catch, nasu. And thanks for pointing that out.