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This question already has an answer here:

We know that there is a particular velocity with which a satellite will move in a circular orbit ( orbital velocity ) but if its velocity is more than the orbital velocity but less than the escape velocity then it would move in a elliptical orbit, however but my question is that why it moves in a elliptical orbit when it is moving with velocity more than the orbital velocity?

Further this question can be extended because it indirectly also asks that why is there a specific value of escape velocity and also why the escape velocity is $\sqrt2$ times the orbital velocity?

thanks in advance!

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marked as duplicate by John Rennie gravity Jan 3 '16 at 15:03

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Your assumptions are unfounded. You should start by reading any online introduction to orbital mechanics. $\endgroup$ – Carl Witthoft Jan 3 '16 at 14:06
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    $\begingroup$ if its velocity is less than orbital velocity then it will fall on the planet Not necessarily. It will take and elliptic orbit again, this time smaller than the circular orbit would have been. It will only crash if it at some point during this smaller elliptical orbit, touches the ground. $\endgroup$ – Steeven Jan 3 '16 at 14:09
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why it moves in a elliptical orbit when it is moving with velocity more than the orbital velocity

The escape speed $v_{escape}$ is the minimum speed needed, for the object satellite to not follow an orbit. So, if the speed is higher than circular orbital speed (so it will not be a circle) but lower than escape velocity (so it will follow an orbit), then it will have to follow another kind of orbit, which happens to be elliptical.

enter image description here source: http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit4/orbits.html

why is there a specific value of escape velocity

  • Shoot something away extremely fast and it will fly onwards away from Earth and never come back. If it was shot off fast enough, it might get so far away that it doesn't really feel gravity anymore of any significant amount.

  • Now instead throw something upwards. It will fall back. It has clearly not escaped gravity.

Somewhere in between the large and small velocity, there of course must be a velocity value, which is just the limit, the minimum needed value, to escape.

We call this limit the escape velocity.

why the escape velocity is √2 times the orbital velocity?

The explanation is simply mathematical, so view the derivation. It starts from the energy conservation, where:

$$K_1+U_1=K_2+U_2$$

where point 1 is right after launch (so $v_1$ start speed) and point 2 is at some point infinitly far away. Infinitly far away, the potential energy is zero $U_2=0$. And if the object does escape, then the velocity will still be positive infinitly far away (it will forever continue moving) - exactly on the limit between escaping (where it keeps moving away) and not escaping (where it will return and would have negative velocity), the velocity will be zero $v_2=0$, so on this limit we have $K_2=0$.

$$K_1+U_1=0+0\\ \frac12 mv_1^2-\frac{Gm_{earth} m}{R}=0\\ v_1=v_{escape}=\sqrt{\frac{2Gm_{earth}}{R_{earth}}}$$

The start speed $v_1$ that fulfills this relationship is the speed, that exactly reaches this lower limit - that is the escape velocity $v_1=v_{escape}$.

That this expression is proportional to the circular orbital speed $v=\sqrt{\frac{GM_{earth}}{r}}$ is somewhat a coincidence, since that speed could be derived in a different way (from Neton's 2nd law).

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