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We often find velocity required to keep a satellite in orbit by the formula $v=\sqrt{\frac{GM}{r}}$ where $v$ is perpendicular to the gravitational force. It is very intuitive that the object will go in a perfectly circular orbit.
But if the speed of the object were to decrease by any reason such as air drag or by firing rockets(to change momentum), I can't visualize how and why the path of the satellite will change. Or initially the speed of object was higher than required speed but not too high to escape. what will happen to the object. I think it will have kind of elliptical orbit but I can't visualize from what I have learned. Can anyone show an animation/diagram of how this would hold true in Einstein's way of looking at gravity(space-time curvature). For me it is getting very non-intuitive. It's not a homework question but speculations about physics so please show support.

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    $\begingroup$ Ehm, why don't you just set up the differential equation yourself and impose the boundary conditions? This is showing insufficient effort, as far as I'm concerned. Furthermore, it reeks of being a homework question. $\endgroup$ – Danu Sep 13 '14 at 17:08
  • $\begingroup$ Please note that, if $v$ is always perpendicular to $g$, it will describe a perfect circular orbit regardless of the speed. $\endgroup$ – André Chalella Sep 13 '14 at 17:10
  • $\begingroup$ @AndréNeves $\overrightarrow {v}$ is not always perpendicular to $\overrightarrow {g}$, especially after a change in $\overrightarrow {v}$. Since the only force on an orbiting object is $\overrightarrow {g}$, then it is responsible for modifying $\overrightarrow {v}$. $\endgroup$ – LDC3 Sep 13 '14 at 17:25
  • $\begingroup$ Now I am completely sure it doesn't look like a homework question and neither it was before. show some support. I study physics from internet out of interest. how can it be a homework question? $\endgroup$ – Suchal Sep 15 '14 at 13:08
  • $\begingroup$ If you are on a circular orbit and change the modulus of your velocity, you change the heigh of the opposite point of the orbit, which becomes an ellipse. $\endgroup$ – DarioP Sep 15 '14 at 14:11
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Start with your satellite velocity, $v_0$, equal to $\sqrt{GM/r}$ so we get a circular orbit:

Circular orbit

Now if we increase the velocity, $v > v_0$, the satellite will move away from the Earth faster than the satellite in a circular orbit, and we get an elliptical orbit that looks like this (I've drawn the original circular orbit dotted):

Elliptical orbit

Alternatively if we decrease the velocity, $v < v_0$, the satellite will move away from the Earth slower than the satellite in a circular orbit, and we get an elliptical orbit that looks like this:

Elliptical orbit

In all cases the orbit is an ellipse with the Earth at one of the focus points. The circle is a special case of an ellipse where the foci coincide.

Solving the equations of motion for the satellite in an elliptical orbit is harder than you probably expect. However there are various convenient equations that describe aspects of the motion. For our purposes the easiest equation to work with is the vis viva equation:

$$ v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right) $$

where $r$ is the distance from the Earth and $a$ is the semi-major axis of the ellipse. If we rearrange this to give the semi major axis we can explain the three diagrams above:

$$ a = \frac{r}{2 - \frac{r}{GM}v^2} $$

or:

$$ a = \frac{r}{2 - \phi} \tag{1} $$

where:

$$ \phi = \frac{r}{GM}v^2 $$

If we start with $v = \sqrt{GM/r}$ then that will make $\phi = 1$, and equation (1) tells us that $a = r$. In other words the semi major axis is equal to the orbital radious so the orbit is a circle.

Now make $v > \sqrt{GM/r}$ as in the second diagram, then $\phi > 1$ and therefore $a > r$. The semi major axis is greater that the distance $r$ we started with so we have an ellipse wider than the circular orbit.

And finally, though it should be obvious now, if we make $v < \sqrt{GM/r}$ as in the third diagram, then $\phi < 1$ and therefore $a < r$. The semi major axis is less that the distance $r$ we started with so we have an ellipse narrower than the circular orbit.

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  • $\begingroup$ Sir I don't know if it's relevant to ask this in comments but can you tell me which software do the users of this site use to produce such diagrams. $\endgroup$ – Suchal Sep 17 '14 at 12:02
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    $\begingroup$ @Suchal: I use the Google Draw app. It's basic, but it's fine for simple diagrams and it's free! :-) $\endgroup$ – John Rennie Sep 17 '14 at 12:52

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