The radius of its orbit = 8.27 x 10¹³ m
Further explanation
Given
mass Jupiter=1.9 x 10²⁷ kg
T = 16.9 days=1.46 x 10⁶ s
Required
the radius =r
Solution
To hold the moon in its orbit, the gravitational force between two objects (jupiter and moon) will be equal to the centripetal force
[tex]\tt G\dfrac{M.m}{r^2}=m.\dfrac{v^2}{r}\rightarrow v=\dfrac{2.\pi.r}{T}\\\\M=\dfrac{r^3.4\pi^2}{T^2.G}\rightarrow r^3=\dfrac{GMT^2}{4\pi^2}[/tex]
G = 6.67 x 10⁻¹¹ N/m²kg²
Input the value :
[tex]\tt r^3=\dfrac{6.67\times 10^{-11}\times 1.9\times 10^{27}\times (1.46\times 10^6)^2}{4\pi^2}\\\\r^3=6.85\times 10^{27}\rightarrow r=8.27\times 10^{13}[/tex]
