When a satellite in a circular orbit around the earth enters the atmospheric region, it encounters small air resistance to its motion. Then:
(This question have multiple correct option)
A) Its kinetic energy increases
B) Its kinetic energy decreases
C) Its angular momentum about the earth decreases.
D) Its period of revolution about the earth increases.

Hint: When there is air drag force acting on a satellite it decreases its mechanical energy and due to this change in mechanical energy there is also a change radius of orbit in which the satellite is moving. Due to change in radius of orbit there is a change in kinetic energy, potential energy and angular momentum of earth.

Complete step by step solution:
When a satellite in a circular orbit around the earth enters the atmospheric region, it experiences a drag force on it due to encounter of small air resistance to its motion, this drag force is always opposite in direction of motion so, there is a negative work done of that drag force, which causes decrease of mechanical energy and this energy is converted in heat energy.
We know that, $K.E.=\dfrac{GMm}{r}$, $P.E.=\dfrac{-GMm}{r}$,
Where, $K.E.$ is kinetic energy, $G$is universal gravitation constant, $M$ is the mass of the earth, $m$ is the mass of satellite, $r$ is the radius of orbit, $P.E.$ is potential energy of satellite.

When its mechanical energy decreases then satellites go into a spiral path with decreasing radius of orbit. We can see that its kinetic energy and potential energy is inversely proportional to the radius of orbit. So, its kinetic energy increases with decrease in radius of orbit, but its potential energy decreases because of negative sign in its formula.

Also, for angular momentum $L$, $L=\sqrt{GMr}$
$\Rightarrow L\propto \sqrt{r}$
So, decrease in radius of orbit is also the cause of decrease in angular momentum of the satellite.

Hence, option (A) and option (C) is correct.

Note: We have to remember that potential energy is negative and in decrease it also decreases. Also its time of revolution also depends on the radius of orbit. i.e. $T\propto {{r}^{\dfrac{3}{2}}}$, So the decrease in radius of orbit decreases in angular momentum.