An earth satellite remains in orbit at a distance of 11550 km from the center of the earth. what speed would it have to maintain? the universal gravitational constant is 6.67 × 10−11 n · m2 /kg2 and the mass of the earth is 5.98 × 1024 kg. 1. 7093.03 2. 7236.86 3. 5374.04 4. 4845.11 5. 5292.47 6. 5876.55 7. 6477.61 8. 7600.32 9. 6223.84 10. 7414.21 answer in units of m/s.

11,550 km has to be changed to 11,550,000 meters

G · m · t² = 4 · π² · r³  we can change that to
t² = (4 · π² · r³) / (G · m )
t^2 = 4*PI^2*r^3 / (G*m)
t^2 = 4*PI^2*(11,550,000)^3 / 6.67*10^-11*5.98*10^24kg
t^2 =  6.083*10^22 / 3.99 * 10^14
t^2 =  152,500,000
t = 12,350 seconds
and its orbital distance it travels is 11,550 * 2*PI = 70,050 kilometers

Therefore, it is traveling at 70,050 km / 12,350 second which equals
5.67 km per second which  is 5,670 meters per second.

Source:
http://www.1728.org/kepler3a.htm

Idon’t even know the answer hehehehe

2x+10=36. x being the number of friends.