Kinetic and potential energy (satellite low orbit)

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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
 

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  • #2
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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.
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
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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
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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
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Okay, thank you :)!
Another question, for the kinetic and potential energy...should I derivate the equations too?
 
  • #6
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Something like a derivative of some equation is probably useful. You don't need it, but it helps.
 
  • #7
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I'm so grateful for your help, thank you!
 
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