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* 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.

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Hi Janus !

* 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.

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?

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Its principle is the same. GTO is a highly elliptical orbit with perigee of a few hundred KMs and apogee at Geostationary distance of 36000KMs above the Earth.

https://en.wikipedia.org/wiki/Apogee_kick_motor

https://en.wikipedia.org/wiki/Geostationary_transfer_orbit

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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)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?

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.

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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

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