Linear velocity of satellite I have a problem from a sample Final, I believe they have incorrect answers. If not, I would appreciate if someone could show me why the answer I am getting is wrong, and if so, how to get the correct answer. 
Question: An Earth's satellite with a mass of 500 kg orbits the Earth in 180 minutes. What is the linear speed of the satellite? ($M_\mathrm E = 5.98\times 10^{24}\ \mathrm{kg}$, $R_\mathrm E = 6.37\times 10^6\ \mathrm m$).
My solution:
$$m = 500\ \mathrm{kg}$$
$$T = 180\ \mathrm{min} = 10\,800\ \mathrm s$$
$$v = \frac{2\pi R}{T}$$
$$F_\mathrm c = ma_\mathrm c$$
$$F_\mathrm c = F_\mathrm g = G{\frac{M_\mathrm Em}{R^2}}$$
$$a_\mathrm c = \frac{v^2}{R}$$
So...
$$G{\frac{M_\mathrm Em}{R^2}} = m\frac{v^2}{R}$$
Since we only know the radius of Earth, we need to calculate the distance of the satellite $R$ from Earth's center: $R = R_\mathrm E + h$, where $h$ is the distance of the satellite from Earth's crust (assuming Earth was a perfect sphere).
After some algebraic manipulations to the above equations, we get:
$$G{\frac{M_\mathrm E}{R}} = v^2 = \frac{4\pi^2 R^2}{T^2}$$ 
$$R^3 = G\frac{M_\mathrm ET^2}{4\pi^2}$$
$$R = \left[G\frac{M_\mathrm ET^2}{4\pi^2}\right]^{\frac{1}{3}}$$
When I plug in the numbers:
$$R = \left[\frac{(6.67\times 10^{-11}\ \mathrm{m^3\ kg^{-1}\ s^{-2}})(5.98\times 10^{24}\ \mathrm{kg})(10\,800\ \mathrm s)^2}{4\pi^2}\right] = 1.06\times 10^7\ \mathrm m$$
Then it should be as simple as plugging $R$ back into $v = \frac{2\pi R}{T}$ and that gives:
$$v = \frac{2\pi(1.06\times 10^7\ \mathrm m)}{10\,800\ \mathrm s} = 6\,167\ \mathrm{\frac{m}{s}}$$ ($6\,145\ \mathrm{m/s}$ if using exact numbers)
Now the answer the sample exam gives is: $5227\ \mathrm{m/s}$. I do not see how this can be true. Any assistance is appreciated. 
 A: You're answer is correct.
However, you don't need to calculate $r$. You aren't asked to calculate it, and there's no need to do so. Alternatively, it can be eliminated.
I'll start with Kepler's third law, which says (in modern parlance) that the product of the square of a satellite's mean motion and the cube of the satellite's semi-major axis length are a constant:
$$n^2a^3 = G(M+m) \tag{1}$$
where $n$ is the mean motion ($n=2\pi/P$ where $P$ is the orbital period), $a$ is the semi-major axis length, $G$ is the universal gravitational constant, and $M$ and $m$ are the masses of the two bodies.
In this problem, $m$ is about 22 orders of magnitude smaller than is $M$, we can quite safely ignore the satellite's mass. The product $GM_E \equiv \mu_E$ for the Earth is known to many more places of accuracy than is either the gravitational constant or the Earth's mass, so it's better to use $\mu_E=398600.4418\,\text{km}^3/\text{s}^2$:
$$n^2a^3 = \mu_E\tag{2}$$
For an object in a circular orbit, each of the the orbital radius $r$, angular velocity $\omega$, and speed (magnitude of velocity) $v$ are constant. The orbital radius $r$ is the semi-major axis length $a$, the angular velocity $\omega$ also is the mean motion $n$, and the speed is $$v=r\omega=an \tag{3}$$
Combining equations (2) and (3) results in $$v=\left(\mu_E n\right)^{1/3} = \left(\frac{2\pi\mu_E}P\right)^{1/3}\tag{4}$$
Plugging in the values $\mu_E=398600.4418\,\text{km}^3/\text{s}^2$ and $P=180\,\text{minutes}$ results in
$$v=6143.71833\,\text{m}/\text{s}\tag{5}$$
Note that the above assumes the 180 minutes is exact.

Aside: The reason $\mu_E$ is known to so many more places that $GM_E$ is simple. The gravitational parameter $\mu_E$ can be determined by observing the orbits of the large number of satellites humanity has placed in orbit about the Earth since the late 1950s. In fact, the best way to "weigh" the earth (i.e., estimate $M_E$) is to divide the well-observable $\mu_E$ by the poorly observable gravitational constant $G$.
