Help with this problem from Mastering Physics (Gravitation)

[lucphysics]

Homework Statement

If a satellite is in a sufficiently low orbit, it will encounter air drag from the Earth's atmosphere. Since air drag does negative work (the force of air drag is directed opposite the motion), the mechanical energy will decrease. If E decreases (becomes more negative), the radius of the orbit will decrease. If air drag is relatively small, the satellite can be considered to be in a circular orbit of continually decreasing radius.
a) According to v = (GM/r)^1/2,if the radius of a satellite’s circular orbit decreases, the satellite’s orbital speed increases. How can you reconcile this with the statement that the mechanical energy decreases?
b) 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.

Homework Equations

U = -GMm/r

K = 1/2mv^2

E = K+U

[PLAIN]http://media.wiley.com/Lux/82/331282.image4.png[/B]

The Attempt at a Solution

I think (a) can be solved saying that the mechanic energy will decrease due to the negative work done by the air drag, because doing E = K+U the final mechanic energy is negative. (I'm not sure anyway)
But I'm stuck doing (b) , I don't understand where does △r/2 in the velocity equation come from...I don't know how to solve it...any help would be helpful, thanks to all of you who could help me.PS: this isn't homework, it's just a problem I found to prepare final exams
(Sorry if there are any grammar mistakes, I'm not a native English speaker)

 

[Chestermiller]

Have you had calculus? If so, what is the derivative of v with respect to r?

Chet

 

[lucphysics]

Yes, I've made the derivative of v respect r, but what I get it's:
v'= - (1/2)•[(GM/sqrt(GM/r^3)]
I'm pretty sure I've done something wrong because that isn't the solution given in the problem statement

 

[Samy_A]

v'= - (1/2)•[(GM/sqrt(GM/r^3)]

Rewrite this as v' = -(1/2)[(GM/r^3)^1/2], or dv/dr=-(1/2)[(GM/r^3)^1/2].
This derivative is negative, so that if r increases, v decreases. And if r decreases, v increases.

What then is the increase in orbital speed if the radius decreases from r to r - △r, where the positive quantity △r is much less than r?

 

[lucphysics]

So, that means that when r decreases the derivative is positive due to the velocity increasement?

 

[Samy_A]

So, that means that when r decreases the derivative is positive due to the velocity increasement?

I wouldn't formulate it this way.

When r decreases, v increases. That we can deduce from the fact that dv/dr is negative.
Now, you are asked to compute by how much v increases, when r decreases by a (relatively) small positive quantity △r.

You can use the formula that you have derived for dv/dr to give a very good approximation for that increase in orbital speed. That is what you have to do in the first part of b), and it shouldn't be too difficult.

 

[lucphysics]

Ok, thank you. I think I understand it now.

 

[Chestermiller]

Yes, I've made the derivative of v respect r, but what I get it's:
v'= - (1/2)•[(GM/sqrt(GM/r^3)]
I'm pretty sure I've done something wrong because that isn't the solution given in the problem statement

That's because you did the derivative incorrectly. The correct derivative is given in Samy_A's post #4.

Chet