Need help with something very odd.

1. Sep 10, 2003

Ok, we are given the known equation that momentum = mass times velocity.

p=mv.

What is dp/dr physically representing?

Given:

r = position.

dr/dt = v.

so

p = mdr/dt.

dp/dr = m/dt ? If so, that remains constant (or zero?) correct?

2. Sep 11, 2003

HallsofIvy

Staff Emeritus
Okay, momentum equals mass times velocity.
(That's basically the definition of momentum rather than a "known equation".)

p= mv.

IF m is a constant then dp/dt= m dv/dt= m a (mass times acceleration).

If m is not a constant (a very important consideration) then
dp/dt= m dv/dt+ dm/dt v.

dp/dt is, of course, "force"- that's basically, the definition of force.

dp/dr= (dp/dt)(dt/dr) (chain rule) or

dp/dr= (dp/dt)/(dr/dt)= (1/v)*force

Assuming constant mass, dp/dr= (m a)/v

I don't know that that "physically represents" anything in particular. I'm pretty sure it doesn't have a specific name.

As far as "dp/dr = m/dt" is concerned, that makes no sense at all- you have an "unattached" dt on the right.

3. Sep 11, 2003

That's what I am all confused about. Whether it represents anything at all, or if it will be a constant. A better way to visualize this is to imagine the mass is in an orbit.

4. Sep 11, 2003

Staff Emeritus
dp/dt= m dv/dt+ dm/dt v.

This is the force law for a rocket. The force dp/dt has rwo components, current mass times acceleration (m dv/dt) and rate of change of mass times current speed (v dm/dt). This latter is due to the exhaust; dm/dt is the rate that reaction mass is being shot out the back. Note that the higher v is, the more that a given dm/dt contributes to the force on the rocket.

5. Sep 11, 2003

Yes, that much I knew. But this is the question:

p =mv.

dp/dt =mdv/dt (m is constant in this case)

dp = mdv

dp/dr = mdv/dr

Now, dv = dr/dt so we get 1/dt

dp/dr = m/dt.

I am just interested in this. The change in momentum with respect to distance, or mass with respect to time. In an orbit (notice this condition) would this quantity whatever it is, be constant, or zero?

6. Sep 11, 2003

Staff Emeritus
You can't just divide out the dr's. derivatives like dr/dt aren't quotients exactly, they are the limits of quotients. It's allowable to multiply through, such as dr/dt = K, yields dr = Kdt ready for integration, but dividing out doesn't make sense.

7. Sep 11, 2003

Alright then.

dp = mdv

What does

dp/dr = mdv/dr represent?

8. Sep 12, 2003

krab

You said dv=dr/dt. That is quite wrong. v=dr/dt (as you said in a previous post). Therefore, dv=d(dr/dt)=d^2r/dt. So you are misunderstanding the math.

You are using v=dr/dt. This is only true for 1-dimensional motion, for example, a mass moving along a straight line. In that case, what Hallsofivy said is right. dp/dr=F/v; in words, dp/dr is the ratio of force applied, to the velocity attained. It's not zero and not constant. For example, if the force is pushing the mass in the direction opposite to its motion, the mass will slow down, stop, and reverse. At the point at which it has stopped, v=0 so dp/dr=F/v is infinite.

If you are thinking of something like a central force problem like planetary motion, and r is the distance between the masses, then the speed v is NOT dr/dt in general. The motion is 2-dimensional so you need another coordinate besides r to locate the mass. Usually use angle theta. Then the velocity has 2 components, dr/dt and
r*d(theta)/dt, and v is the square root of the sum of the squares of these 2 components.

9. Sep 12, 2003