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.
When can a natural satellite possibly be in polar orbit around it's primary?
Well, normally you don't expect to see this. As explained in Why are our planets in the solar system all on the same disc/plane/layer? and the questions linked to it (such as Accretion disk physics - Stellar formation), we expect that when clouds of gas and dust condense into stars, planets, and moons, everything will be confined to a disk with the same angular momentum (or at least the same direction of the vector).
If you do see a polar orbit, or more generally a highly inclined* orbit, then it is an indication that one of the following probably occurred:
- The smaller body was captured after the main body formed,
- The smaller body's orbit was altered by (usually gravitational) interactions with other bodies, or
- The larger body's spin was changed after the formation of the system.
Are there any such observations on record?
Why, yes indeed! In our own Solar System, my favorite example (there are many others) is Neptune's moon Triton. It is in a retrograde orbit, and the general consensus is that it falls into category (1) above. It was orbiting the Sun in the outskirts of the Solar System in what was probably an eccentric orbit, it got too close to Neptune, and it's close approach to Neptune completely by chance had it on the side of the planet such that it's orbit ended up going opposite to Neptune's spin.
While the planets in our Solar System aren't particularly inclined with respect to one another or the Sun's spin, this is not true of some other planetary systems we have observed. The past few years have seen a small explosion of measurements of exoplanetary systems' "spin-orbit misalignment" using the Rossiter–McLaughlin effect. For a nice light summary of this effect, you can take a look at [1], especially the figures.
Since that paper was written, many more exoplanets have been discovered, and many more astronomers have taken spin-orbit alignment data. The general consensus is that most systems are aligned, but there are some notable retrograde and polar orbits. It remains to be seen what this implies for our models of planetary formation and migration. There may even be an explanation for these inclined systems in (uninteresting) observer bias or (interesting) stellar physics, as discussed in [2], which also gives a list of measured inclinations for systems with relatively precise data.
* Inclination is defined as the angle between the main body's spin axis and the orbiting body's orbital axis. $0^\circ$ means perfect alignment in an equatorial orbit, $90^\circ$ is for a polar orbit, and $180^\circ$ is for a retrograde orbit (you are back in the equatorial plane, but going around the "wrong" way).
[1] Winn, 2006, "Exoplanets and the Rossiter-McLaughlin Effect."
[2] Winn et al., 2010, "Hot Stars with Hot Jupiters Have High Obliquities."
Best Answer
Assuming that the Sun's rays are parallel near the earth and that $BC$ is a straight line
The satellite has to go through a distance $2BC$
$$2\sqrt{R_E^2-(R\theta)^2}\tag1$$
if $\theta$ is small
For an orbit of radius $R$ the speed $v$ of the satellite is from
$$\frac{v^2}{R} = \frac{GM_E}{R^2}$$
$$v= \sqrt{\frac{GM_E}{R}}\tag 2$$
where $M_E$ is the mass of the earth.
Combining (1) and (2), the time in the shadow is
$$T = 2\sqrt{\frac{R_E^2R-R^3\theta^2}{GM_E}}\tag3$$
for $R\theta \lt R_E$, otherwise there is no shadow to go through.
If $$k=\frac{R_E}{R}$$ then (3) can be written as $$T =\sqrt{\frac{4R_E^3}{GM_E}\times\frac{k^2-\theta^2}{k^3}}\tag4$$ or $$T =1610\sqrt{\frac{k^2-\theta^2}{k^3}}\tag5$$
By considering the time taken for the satellite to go through an angle $\phi$
An accurate formula valid for all $k$ and $\theta$ can be found
$$T = 1610 \frac{\sin^{-1}((k^2-(\sin\theta)^2)^{0.5})}{k^{3/2}}\tag6$$
There is only a small difference between (5) and (6).