Why does a velocity that gives gravity's centripetal acceleration make you stay in orbit?

Explanations of basic orbital mechanics that I can find all go,

$$\frac{v_{satellite}^2}{r} = a_c = \frac{F_{gravity}}{M_{planet}} = \frac{(\frac{G M_a M_s}{r^2})}{M_{planet}}$$

, and so you get a velocity,

$$v_{satellite}=\sqrt{\frac{GM_{planet}}{r}}$$

. So you need that velocity to orbit earth at that radius. But it doesn't answer alot of questions like:

How much can you increase or decrease your velocity for this v and r relationship to stop?

What changes when you reach those points which makes you fly away or crash into the earth?

Why would increasing v of the satellite effect something perpendicular like the ac?

Why does having a centripetal acceleration equal the force from gravity, 9.81, prevent you from crashing?

• Because you keep falling at the planet (or other primary) and missing. Cue Douglas Adams quote. – dmckee Jun 5 '16 at 22:24
• Effectively a duplicate of physics.stackexchange.com/q/9049. – dmckee Jun 5 '16 at 22:25
• Ok, but what about the 3rd and 4th question? And what does a rocket that is trying to get out of orbit do? And will something orbit at 1m above the surface of the earth if it is going fast enough? – BoddTaxter Jun 5 '16 at 22:46
• Yes it would orbit 1m above the Earth if it went fast enough, if there was no atmosphere and the Earth was absolutely smooth and spherical. – Peter R Jun 5 '16 at 23:07
• A lot of questions. Perhaps too many. It suggests to me that you really need tuition rather than a single answer to a specific question. You cite the equations but you don't seem to understand them. I think you need to study a textbook or an online tutorial video. – sammy gerbil Jun 7 '16 at 2:32