Question:

A space shuttle releases a satellite into a circular orbit 780 km above the Earth. How fast must the shuttle be moving (relative to Earth’s center) when the release occurs?

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer:

$7.46 \times 10^3 \textrm{ m/s}$

### Transcript for this Giancoli solution

This is Giancoli Answers with Mr. Dychko. The force of gravity is providing a centripetal force for the satellite when it's in orbit. So the centripetal force is mass times centripetal acceleration; so that's*m v squared*over

*r*and since it's gravity, it's also

*G*— mass of the satellite— mass of the Earth divided by distance to the Earth's center, squared. And we can multiply both sides by

*r*and then that cancels one of the

*r*'s here, then take the square root of both sides and canceling mass of the satellite, we get the speed that the satellite needs to have, is the square root of

*G*mass of the Earth divided by

*r*. So

*r*is the radius of the Earth plus 780 kilometers above the Earth's surface. So that's 6.38 times 10 to the 6 meters— radius of the Earth—plus 780 times 10 to the 3 meters which is 7.16 times 10 to the 6 meters, from the center of the Earth. So

*v*is gonna be square root of 6.67 times 10 to the minus 11 times—mass of the Earth—5.98 times 10 to the 24 kilograms divided by 7.16 times 10 to the 6 meters, and we get 7.46 times 10 to the 3 meters per second is how fast it needs to go, which is, times 3.6, which is about 27,000 kilometers per hour; that's pretty fast.