# Is the velocity of orbiting around the earth exact?

Let's consider a satellite orbiting the earth. From the centripetal acceleration we have $v=\sqrt{g_{earth}r}$. If we know the $r$ and $g$, we can get $v$, which is the velocity needed to orbit around the earth(is this correct?). My question is, what will happen if you exceed that velocity. Will you just leave the orbit after some time? What if you go slower. Will the earth suck you in?

You have a slight misunderstanding in your question. There is no single orbit - there is effectively an infinite number of orbits your satellite could be in, with different radii (and if you include different ellipses)

So if you change your velocity, all that happens is you change your orbit.

If you increase your velocity such that you exceed escape velocity then you will escape Earth altogether.

The Earth is always "sucking" a satellite in - but that is almost the definition of orbiting. If the orbit intersects the atmosphere then atmospheric drag will change the orbit - and eventually the orbit will decay enough that it will fall to Earth.

• Thanks for the answer but I think there is a problem in there. I do think you need a higher speed in order to orbit closer to earth than to orbit in farther orbits (the distance in the denominator is of power 2 in g's equation), so if you have a higher speed than needed for the orbit x, you will be constantly changing orbits and eventually escape the Earth. – khajvah Dec 11 '14 at 11:23
• No. You will change orbit from x and end up in an orbit appropriate for that velocity. Interestingly enough, the relationship between velocity and orbit is complicated - when Soyuz docks with the ISS it needs to be in a lower orbit to catch up with the ISS and then use power to move out. Inner orbits are faster, but increasing velocity moves you to a wider orbit, which goes round the earth more slowly... – Rory Alsop Dec 11 '14 at 11:30
• This still bigs me. Do you think the equation that I used is the right equation to use to find the speed? – khajvah Dec 11 '14 at 11:50
• en.wikipedia.org/wiki/Orbital_speed#Precise_orbital_speed gives the equation for orbital speed as $v = \sqrt{\mu \left(\frac{2}{r} - \frac{1}{a}\right)}$ where $\mu$ is the standard gravitational parameter, $r$ is the distance at which the speed is to be calculated, and $a$ is the length of the semi-major axis of the elliptical orbit. For a perfectly circular orbit (eccentricity = 0), $a=r$. Nobody said this was trivial stuff. :) – user Dec 11 '14 at 12:20

Your equation is for a perfectly circular orbit. A body at a given altitude, moving tangentially to the Earth's surface below, at the speed given by the equation, orbits in a circle.

If it is moving tangentially at a slower speed than that given by your equation, it will be at the high point of an elliptical orbit. If it's moving too slow, the low part of the elliptical orbit will intersect the atmosphere of the Earth, and will slow down due to atmospheric drag, and eventually fall (aerobraking). If it's moving much too slow, the low part of the orbit will intersect the Earth itself (lithobraking).

If it's moving tangentially at a higher speed than the circular orbit speed, it will be at the low point of an elliptical orbit. Higher speeds yield more eccentric ellipses with higher and higher altitudes at the other end, and a very high finite speed will cause the body to leave orbit and not return; this is called escape velocity.

If your body has a non-tangential velocity, that is, a velocity in a direction not exactly perpendicular to gravity's effect, then it's at some point on the ellipse other than the high or low point.

In orbital scenarios, the high point is called the apogee (when you're talking about Earth orbit specifically) or apoapsis (when you're talking about any other body), and the low point is perigee or periapsis.