Answer:
6862.96871 seconds
Explanation:
M = Mass of Planet
G = Gravitational constant
r = Radius
[tex]\rho[/tex] = Density
T = Rotation period
In this system the gravitational force will balance the centripetal force
[tex]G\frac{Mm}{r^2}=mr\omega^2[/tex]
[tex]\omega=\frac{2\pi}{T}[/tex].
[tex]M=\rho v\\\Rightarrow M=\rho \frac{4}{3}\pi r^3[/tex]
[tex]\\\Rightarrow G\frac{Mm}{r^2}=mr\left(\frac{2\pi}{T}\right)^2\\\Rightarrow \frac{G\rho \frac{4}{3}\pi r^3}{r^3}=\frac{4\pi^2}{T^2}\\\Rightarrow T=\sqrt{\frac{3\pi}{G\rho}}[/tex]
Hence, proved
[tex]T=\sqrt{\frac{3\pi}{6.67\times 10^{-11}\times 3000}}\\\Rightarrow T=6862.96871\ s[/tex]
The rotation period of the astronomical object is 6862.96871 seconds
