The space shuttle releases a satellite into a circular orbit 700 km above the Earth. How fast must the shuttle be moving (relative to Earth) when the release occurs?

Answer 1

First you must find the Force of Gravity usin5g Fg=Gm1/r^2

G=6.67E-11
m1= mass of the earth: 5.9736E24
r= (radius of earth plus the radius of sattelite released)
radius of earth = 6,371,000m + 700,000m = 7,071,000

Fg=(6.67E-11)(5.9736E24)/(7071000^2)
Fg= 7.97

Next you must find the velocity necessary for centripetal acceleration to equal the force of gravity that you just found.

7.97=v^2/r
7.97=v^2/7,071,000
56355870=v^2
take the square root
v=7507 m/s