Question

A satellite in low earth orbit experiences a small drag force from the earth’s atmosphere. The consequence of this drag force will be-
(A) Potential energy decreases, while kinetic energy increases
(B) Potential energy and kinetic energy decreases, since total energy decreases
(C) Drag force doesn’t have any effect on orbital motion of a satellite
(D) None of the above

Answer 1

Hint: Drag force is a fluid force that opposes the motion of bodies moving in a medium, it is caused when small particles of the fluid hit the surface of the object. A drag force may cause the object to slow down or to heat up, which is a loss of energy. A drag force on a satellite may cause it to follow a spiral path, where it spirals in towards the earth instead of following the circular path.

Step by step solution
Drag force is experienced by low earth orbit (LEO) satellites due to the presence of atmospheric gases. The drag force depends on the kinetic energy (RMS) and the density of the air.
In the high altitudes, the density of the atmosphere is small but the particles have very high energy, paired with a high velocity of satellite they can produce a significant effect.
The drag force creates a significant effect in the motion of a satellite, when the particles collide, they cause the satellite body to heat up, causing some Total energy loss due to friction.
As per common intuition a when a satellite loses energy,
-It’s orbit should get lower. (loss of potential energy)
-Its velocity should decrease. (loss of kinetic energy)
The first part is true, the satellite gets lower in orbit, but its kinetic energy does not decrease.
The reason can be found in this equation-
$v = \sqrt {\dfrac{{GM}}{r}} $
Where v= velocity of the satellite
r= radius of the orbit
M and G are Mass of earth and Universal Gravitational constant, which are both constant quantities, thus,
$v \propto \dfrac{1}{r}$
So if a decrease in potential energy lowers the orbit of the satellite, the velocity increases, and since the Kinetic energy is given by,
$K.E. = \dfrac{1}{2}m{v^2}$
$K.E. \propto {v^2}$
And thus the kinetic energy increases with a decrease in height, even if the total energy and potential energy decrease.

Thus, option (A) and (B) are both correct.
Note Increase in kinetic energy does not mean any addition of energy, it just means that some part of the loss of potential energy is converted to kinetic energy, resulting in increased energy while the total energy still decreases.