Answer
4240 s
Work Step by Step
Let's apply equation 5.1 $T=\frac{2\pi R}{V}$ to find the period of the satellite.
$T=\frac{2\pi R}{V}-(1)$
We know that,
$V=\sqrt {\frac{GM}{R}}-(2)$
(2)=>(1),
$T=2\pi \sqrt {\frac{R^{3}}{GM}}-(3)$
We can write,
$mass = density\times volume=>M=\frac{4}{3}\pi R^{3}\rho-(4)$
(4)=>(3),
$T=2\pi \sqrt {\frac{R^{3}}{G\frac{4}{3}\pi R^{3}\rho}}=\sqrt {\frac{3\pi}{G\rho}}$ ; Let's plug known values into this equation.
$T=\sqrt {\frac{3\pi}{6.67\times10^{-11}Nm^{2}/kg^{2}\times7860\space kg/m^{3}}}\approx4240\space s$
Period of the satellite = 4240 s
The value of $\rho$ is taken from the table 11.1
