newtonian-gravitynewtonian-mechanicsorbital-motionsatellites
If a satellite is initially orbiting the earth with a constant velocity $v$ on a circular orbit.
What would happen, if the velocity is suddenly increased by 5%?
Would the orbit still be circular?
I guess it will be elliptic, but how to prove this and find the eccentricity of the ellipse?
I think these are two separate questions that should be approached separately.
1) "Why isn't $m$ in in the first equation?" The mass of a body does change the force acting on it. But the mass of a body also changes its acceleration. If you increase the mass of an object it feels a larger force, but it's also harder to move. The equation for gravitational force is $F = \displaystyle GmM/{r^2}$, while the equation for acceleration is $a = F/m$. Glue to two together and you get $a = GM/r^2$.
2) "Why is $m$ in the second equation?" Think about the moon and the earth. The earth is pulling on the moon, but the moon is also pulling on the earth! The two bodies actually orbit around their common center of mass. This is important for the relative velocity: we need to add how fast the earth is orbiting to how fast the moon is orbiting. That's why you have the $m$ term.
The formula $$ \dot{\theta} = \omega = \frac{2A}{Pr^2} $$ is correct; it can also be derived from the specific angular momentum $h$: $$ h = r^2\omega = \sqrt{GMa(1-e^2)} = b\sqrt{\frac{GM}{a}}, $$ with $e = \sqrt{(a^2-b^2)/a^2}$ the orbital eccentricity. However, this doesn't solve the Kepler problem, because both $\omega$ and $r$ depend on $t$ in a complicated way, which isn't specified by the above formula. In other words, the above formula gives you $\omega(r)$, but not $\omega(t)$ and $r(t)$.
Also, note that $\theta$ is the true anomaly, which is the angle between the direction of periapsis and the current position of the body, as seen from the main focal point (where the attracting body is). And $r$ is the distance between the current position and the focal point. If you want to use cartesian coordinates $(x,y)$, it is better to parametrize them using the eccentric anomaly $E$: $$ x = a\cos E,\qquad y = b\sin E. $$
So how to find $E(t)$? For this, we need to introduce another parameter, called the mean anomaly $M$. The mean anomaly increases linearly with time: $$ M(t) = \frac{2\pi}{P}t = \sqrt{\frac{GM}{a^3}}t. $$ From $M(t)$, we can calculate the eccentric anomaly $E(t)$ using Kepler's equation $$ M = E - e\sin E, $$ which you have to solve numerically. Once $E(t)$ is known, $(x(t),y(t))$ follow immediately.
For completeness, the true anomaly can be calculated from the eccentric anomaly: $$ \tan\frac{\theta}{2} = \sqrt{\frac{1+e}{1-e}}\tan\frac{E}{2}, $$ and the distance $r$ to the attracting body at the focal point is $$ r = \frac{a(1-e^2)}{1+e\cos\theta}. $$
Best Answer
To be clear we have to understand that "speed it up" means applying thrust in the direction of the orbit. To be complete we also have to pay attention to how hard and for how long you thrust.
Two natural limiting cases are
$\Delta v$ applied over a time much shorter than the original orbital period (high thrust applied for a short time)
$\Delta v$ applied over a time much longer than the original orbital period (very low thrust applied continuously over the course of many orbits)
In the former case the circular orbit takes on an elliptical character1 with a apogee2 higher than the original orbit and the perigee remaining at the height of the original orbit. This can be observed in the initial burn of outward going Hohmann transfers.
In the latter case you end up with the satellite in a higher and paradoxically slower but still circular orbit. A natural test of this case is the tidal recession of the moon.
For an in-between case—non-negligible thrust applied for a significant fraction of one orbit—you end with a elliptical orbit where the apogee has also been raised and the perigee has also been raised, but not as high.
1 In extreme cases the orbit can be made parabolic or hyperbolic but let's not bother about that.
2 I'm going to use vocabulary that assumes Earth-centered orbits for concreteness, but of course the results apply to all cases.