When the speed of a Satellite travelling in a perfectly circular orbit is increased does the path of the satellite remain circular and the radius is increased OR does the path of the satellite follow an elliptical path after the speed has been changed OR are there some other possibilities that I am missing?
Also, would the answer be different if the speed is instantaneously increased and not over a period of time?
The satellite may be assumed to be a point object and the planet to be perfectly spherical.
Thanks!
Satellite Path – Change in Path of a Satellite on Changing Its Velocity
accelerationnewtonian-gravitynewtonian-mechanicsorbital-motionsatellites
Related Solutions
If an object is moving in a circular motion, its velocity $\vec{v}$ changes. The centripetal acceleration is just a formula that gives you the length of the derivative $\frac{d\vec{v}}{dt}$ which is the acceleration. It must be caused by some force, according to Newton's second law. If you are holding the object with a rope, then it is the tension of the rope, if it is a satelite on a circular orbit, then the force is of gravitational nature.
When the asteroid hits the satellite, $\vec{v}$ changes, while the gravitional force remains the same. So, the force now creates the same acceleration, but now it does not coincide with 'centripetal acceleration' for this speed (which is just a number characterizing the orbit, not the object). This simply means that the object will leave the circular orbit, because its acceleration and speed now correspond to a different trajectory. This trajectory happens to be elliptic/parabolic/hyperbolic depending on the speed. These cases can be distinguished by total energy -- $E<0$, $E=0$, $E>0$ respectively.
Start with your satellite velocity, $v_0$, equal to $\sqrt{GM/r}$ so we get a 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):
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:
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.
Best Answer
Firstly, a perfectly circular orbit is an idealization which simplifies physical scenarios to help build understanding of concepts using easier mathematics. All orbits are elliptical. See: Kepler's First Law
Secondly, a satellite velocity in order to maintain orbit will only depend on the mass of the object being orbited, and the distance of the satellite from that object. That is, $$v_{sat} = \sqrt{\frac{GM_{central}}{R}}$$
If that velocity decreased suddenly, what happens? The satellite does not have enough speed to maintain orbit and it falls inwards until collision.
Conversely, if that velocity increased suddenly, this brings the risk of escaping the orbit and skipping off into empty space as the velocity is too large to maintain stable orbit.