Gravitational force on a telescope orbiting earth

[logan3]

Homework Statement

A large telescope of mass 8410 kg is in a circular orbit around the earth, making one revolution every 927 minutes. What is the magnitude of the gravitational force exerted on the satellite by the earth?
$M_E = 6.0x10^{24} kg$
$m_s = 8410 kg$
$T_s = 927 min = 55,620 s$
$G = 6.67x10^{-11} Nm^2/kg^2$

Homework Equations

$T^2 = \frac {4{\pi}^2 r^3}{GM_E} \Rightarrow r = \sqrt[3]{\frac {T^2 GM_E}{4\pi^2}}$
$F_G = \frac {GM_E m_s}{r^2}$

The Attempt at a Solution

$r = \sqrt[3]{\frac {T^2 GM_E}{4\pi^2}} = \sqrt[3]{\frac {(55,620 s)^2 (6.67x10^{-11} Nm^2/kg^2)(6.0x10^{24} kg)}{4\pi^2}} = 3.1532x10^7 m$

$F_G = \frac {GM_E m_s}{r^2} = \frac {(6.67x10^{-11} Nm^2/kg^2)(6.0x10^{24} kg)(8410 kg)}{(3.1532x10^7 m)^2} = 3385 N$

Is there a simpler equation to get the radius? Am I doing it right?

Thank-you

 

[gneill]

Your results look good. I don't know if there's a simpler way to get to the radius unless you memorize some fact about another radius and period. For example, if an object could orbit at the Earth's surface it would have a period of 84.5 minutes. Then knowing the Earth's radius you could set up the ratio:
$$\left( \frac{927}{84.5} \right)^2 = \left( \frac{r}{6378 km} \right)^3$$
and solve for r.