# Kinetic and potential energy (satellite low orbit)

1. Dec 15, 2015

### lucphysics

• 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

2. Dec 15, 2015

### Staff: Mentor

I don't understand that approach, and the units do not match, so the formula cannot be right (probably just a typo however). What is the goal of that transformation?

3. Dec 15, 2015

### lucphysics

Instead of use r as the unknown radius, use r - △r because the statement given says:
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.

But maybe I'm wrong and the only way to demonstrate that the variation of velocity is:
△v = +(△r/2)[(GM/r^3)^1/2]
Is by using derivation.

Ps: this problem is from the book of Sears Zemansky University Physics Vol. I

4. Dec 15, 2015

### Staff: Mentor

You can calculate the velocity at radius $r-\Delta r$ and at radius $r$ independently, of course, but approximating this to first order is very similar to taking the derivative.

5. Dec 15, 2015

### lucphysics

Okay, thank you :)!
Another question, for the kinetic and potential energy...should I derivate the equations too?

6. Dec 15, 2015

### Staff: Mentor

Something like a derivative of some equation is probably useful. You don't need it, but it helps.

7. Dec 15, 2015

### lucphysics

I'm so grateful for your help, thank you!