A"synchronous" satellite, which always remains above the same point on a planet's equator, is put in orbit about a planet similar to jupiter. this planet rotates once every 5.7 h, has a mass of 1.9 × 1027 kg and a radius of 6.99 × 107 m. given that g = 6.67 × 10−11 n m2 /kg2 , calculate how far above jupiter's surface the satellite must be. answer in units of m. 016 10.0 poi

The solution for this problem is:

v^2 = GM/r 
r=GM/v^2 

but 

velocity = 2pi r / P 

so we can say that

r = GM/(2pi r)^2 / P^2 
r^3/GM = P^2/(4 pi^2) 
r = (P^2 ( GM/(4pi^2)))^(1/3) 

P = 5.8h = 5.8x3600s 

GM = 6.67 × 10^−11 x 1.9 × 10^27 

( (5.8 x 3600)^2 x (6.67e-11 x 1.9e27) / (4 x pi^2) )^(1 / 3) = 111 856 243 m 

now subtract 6.99 × 10^7 m, the answer is: = 41956243 m

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered