How Does That Satellite Stay in Orbit (while I just fall to the ground)?

In order to understand how satellites stay in orbit, we need to first look at gravity. Everything that mass has a gravitational field. Right now, your computer's gravitational field is pulling on you, and your gravitational field is pulling on the computer. However, since gravity is a relatively weak force, unless you're an extremely large object like a planet, you can't feel this pull. We do however feel the pull of the Earth on us.

Everytime you jump into the air, the Earth's gravitational field pull's you back down. If you use more force you can jump higher, but you always return to the Earth. The force that the Earth's gravity pulls a specific object with is expressed in the formula F=mg, where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s at the Earth's surface). However, that is a specific case at the surface of the planet. The full formula is:

F=GmM/R^2

where F is force, G is the universal gravitational constant (6.67 x 10-11 Nm2/kg2), m is the mass of an the small object, M is the mass of the planet, and R is the distance from the center of the planet to the object.

At the Earth's surface, neglecting all friction, an object could be put into orbit. However the speed would have to be 8 km/s (5 mi/s or 18,000 mi/h). We discover this by combining two formulae. The object in orbit would have to have a centripetal force (F=mv2/r) that counter-balances the gravitational force (F = mg). The two have a common factor of "F" so we combine them as mg = mv2/r. The m's can then be divided out, giving us g = v2/r. Then we multiply both sides by "r", giving us gr = v2. Finally, by taking the squareroot of both sides, we get :

v = (gr)1/2

Interpretation of this formula shows us that the velocity required for orbit is equal to the squareroot of the distance from the object to the center of the Earth times the acceleration due to gravity at that distance. At the surface this would be (6.4 x 106 m * 9.8 m/s2)1/2, or (6.272 x 107)1/2, or approximately 7,920 m/s. This agrees with our previous number of 8 km/s. (Isn't great when somebody else does the math for you?)

However, satellites don't orbit at the surface (they could though). Going back to our other formula for the force of gravity on an object from any height, we can go through the process of rewriting the formula again. This time I think I'll just tell you it:

v=(GM/R)1/2

So if we wanted to put a satellite in a circular orbit at 500 km above the surface, it would need a speed of ((6.67 x 10-11 * 6.0 x 1024)/(6900000))1/2, or (58 x 106)1/2, or 581/2 x 103, or:

7615.77 m/s

As you can see, this number is considerably smaller. If you use a larger speed, you can get an elliptical orbit, but don't use too much. If you do, you just may escape from the Earth's gravitational pull.