Problem 11.11

11.11.

For the satelite to remain stationary, we want the acceleration due to gravity to be equal to the centripital acceleration.
$\frac{GM_{J}}{(R_{J}+d)^{2}} ~=~ \frac{v^{2}}{(R_{J}+d)^{2}}$
where R_J is the radius of Jupiter and d is the distance above the planet's surface.
Now, we can substitue in for the velocity.
$v^{2} = r^{2} \omega^{2} ~=~ \frac{4\pi^{2}r^{2}}{T^{2}}$
where the period of the orbit is given by T.
Substituting in yields,
$GM_{J}T^{2} ~=~ 4\pi^{2} (R_{J}+d)^{3}$
Solving for d yields,
$d = 8.92 \times 10^{7} m ~=~ 89,200 km$ above the planet's surface.