- #1

Brad_Ad23

- 502

- 1

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?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Brad_Ad23
- Start date

- #1

Brad_Ad23

- 502

- 1

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

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 970

(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

Brad_Ad23

- 502

- 1

- #4

selfAdjoint

Staff Emeritus

Gold Member

Dearly Missed

- 6,881

- 10

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

Brad_Ad23

- 502

- 1

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

selfAdjoint

Staff Emeritus

Gold Member

Dearly Missed

- 6,881

- 10

- #7

Brad_Ad23

- 502

- 1

Alright then.

dp = mdv

What does

dp/dr = mdv/dr represent?

dp = mdv

What does

dp/dr = mdv/dr represent?

- #8

krab

Science Advisor

- 893

- 3

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.

Now about physics:

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.

Now about orbits:

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

Brad_Ad23

- 502

- 1

as for the dv = dr/dt thing, that was a mistype on my part. I was getting a bit type happy with d's and other things. Sorry for the math error!

Share:

- Replies
- 5

- Views
- 151

- Last Post

- Replies
- 1

- Views
- 706

- Last Post

- Replies
- 0

- Views
- 2K

- Replies
- 40

- Views
- 2K

- Replies
- 85

- Views
- 2K

- Last Post

- Replies
- 6

- Views
- 457

- Replies
- 11

- Views
- 1K

- Replies
- 5

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 1K

- Replies
- 10

- Views
- 1K