# Increasing the speed of a circular orbit

What If the velocity of particle moving in a circular orbit has increased , would the particle be no longer in circular orbit or it would go in an orbit with bigger radius?

* assuming your new velocity is less than ## \sqrt{2}## times that of your starting orbital velocity. Otherwise, you will no longer be in a closed orbit but an open-ended one that will carry you away from the mass you are orbiting, never to return.

Nugatory, Merlin3189 and berkeman

* assuming your new velocity is less than ## \sqrt{2}## times that of your starting orbital velocity. Otherwise, you will no longer be in a closed orbit but an open-ended one that will carry you away from the mass you are orbiting, never to return.
Hi Janus !
Thanks for your answer, it is clear and good in my opinion.

One small follow up question: in the (*) note, you give the sqrt(2) threshold limit factor for the increase of the orbital speed. If the increases in a circular orbital speed are below that threshold, one gets ellipses, if the increase factor is exactly sqrt(2) one gets a parabola, and if the increases are above it one gets hyperbolas.. Am I right?

Hi Janus !
Thanks for your answer, it is clear and good in my opinion.

One small follow up question: in the (*) note, you give the sqrt(2) threshold limit factor for the increase of the orbital speed. If the increases in a circular orbital speed are below that threshold, one gets ellipses, if the increase factor is exactly sqrt(2) one gets a parabola, and if the increases are above it one gets hyperbolas.. Am I right?
That's about it. Below escape velocity, you have elliptical orbits (with a circular orbit being a special case of an ellipse). At escape velocity, you have a parabolic trajectory. Above escape velocity, you have a hyperbolic trajectory. The other way to categorize them is by total orbital energy (KE+GPE)
Thus using the equation:
$$E= \frac{mv^2}{2}- \frac{GMm}{r}$$
Circular/elliptical orbits result in a negative values
Escape velocity/parabolic trajectories result in an answer of 0
Hyperbolic trajectories yield positive values.

Thanks Janus ! Thanks for your confirmation !

I do remember some points on this subject in classical mechanics. I guess I will need to review some lecture notes on the topic.

Among some curious things about orbital energy is the following:
It is nice and clear that the mechanical energy E = K+V , suffices to determine the type of orbit. But this encapsulates all the complications from the momentum equation: for example the fact that for non-circular orbits, the kinetic energy, function of the speed, comes from radial and tangential components of velocity. There is this separation on the tangential kinetic energy and the effective potential..
Whatever, it blows my mind that the formula above captures the vector problem, and finally simply the 'signs, 0s and extremes' of E dictate the type of orbit... amazing !

cheers