- #1

- 10

- 0

Thread moved from the technical physics forums, so no HH Template is shown.

Hello, I have this problem statement : "Due to air drag, the radius of a satellite’s circular orbit decreases from r to r - △r, where the positive quantity △r is much less than r. The mass of the satellite is m. Show that the increase in orbital speed is △v = +(△r/2)[(GM/r^3)^1/2]; that the change in kinetic energy is △K = + (GMm/2r^2); that the change in gravitational potential energy is △U = -2△K= - (GMm/r^2)△r; and that the amount of work done by the force of air drag is W = - (GMm/2r^2)△r."

Well, I did the derivative for the velocity equation and I got v' = -(1/2)[(GM/r^3)^1/2], but another way to solve it could be using ""r - △r = (r - △r/r)r = (1-△r/r)r"", my problem is that I don't know how to apply that to the velocity equation.

Doing the derivative of the velocity I get the given results in the problem statement for the kinetic and potential energy, but if instead of doing the derivative...which other method could I use to get the same results?

△K = + (GMm/2r^2)

△U = -2△K= - (GMm/r^2)△r

Well, I did the derivative for the velocity equation and I got v' = -(1/2)[(GM/r^3)^1/2], but another way to solve it could be using ""r - △r = (r - △r/r)r = (1-△r/r)r"", my problem is that I don't know how to apply that to the velocity equation.

Doing the derivative of the velocity I get the given results in the problem statement for the kinetic and potential energy, but if instead of doing the derivative...which other method could I use to get the same results?

△K = + (GMm/2r^2)

△U = -2△K= - (GMm/r^2)△r