Question

Air resistance increase the Velocity of the Satellite
How?
and the Above satement is not Wrong !

Answer 1

due the friction from the resistence, the orbital energy of the satellite decreases. i.e. it becomes more negative. which means the semi-major axis of the satellite i.e. its orbital radius decreases. which means its angular velocity increases according to kepler's 3rd law.
hence, you can think of the satellite as slowly falling in towards the planet with increasing velocity.
In the above case examplanation, I assumed the satellite is initially below the syncronous orbit, hence its angular velocity is greater than that of the planet. Then you can think of it as kind of sliding over the relatively slowly moving earth's atmosphere. Thus satellite's motion is trying to increase the rotation of the earth. Thus in whatever small way, it must increase the rotational energy of the planet. This in turn would be compensated by a decrease in orbital energy of the satellite. This is manifested as a decrease in orbital radius of the satellite and hence increase in its velocity. So suppose the air resistance happens to grow gigantically large, the satellite would spiral in towards the planet faster and faster.

Answer 2

"due the friction from the resistance, the orbital energy of the satellite decreases"
When a satellite is Experiencing Air Resistance, obviously , in opposite direction , this should decrease the velocity of satellite but My BOOK say it Increase with increase in Air Drag .......... Sorry , But can't really understand your Explanation , but Thanks
Suggested me theory ,chapter or links where i could find my answer....

Answer 3

you are having this confusion because you are only thinking about the kinetic energy of the satellite.
First, The Satellite is below synchronous orbit (hence susceptible to air drag) and thus its angular velocity is greater than the planet.
A body in orbit has its energy in two forms : Kinetic energy and Potential energy.
keep in mind that Potential energy of the satellite increases as its distance from the orbit increases.
In a Keplerian orbit, the total orbital energy of both masses comes to be -Gm1m2/2a where a is the semi-major axis of the orbit.
Now we know that when there is friction, obviously energy of a system decreases because of loss of energy as heat dissipation etc.
In this case, there is friction between the planet and the satellite due to to planet's atmosphere dragging against the satellite.
hence the total energy of the 2 bodies will decrease.
The total energy of the 2 bodies = Rotational Energy of the Planet + Total orbital energy of (Planet + satellite)

\[E = 1/2*I*w^2 -G(m1m2)/2a \]

1st term is the rotational energy of the planet where I is moment of Inertia and w is angular velocity
differentiating this with respect to time

\[dE/dt = Iw*dw/dt + G(m1m2)/2a^2*da/dt\]

look at the above equation,

dE/dt < 0 (total energy is decreasing due to energy loss in friction)

dw/dt > 0 (the planet's rotation is increasing because of the drag of the satellite)

hence da/dt should be <0
This means a(orbital radius) is decreasing with time.