(Tricky) Comet Picking up Mass - Differential Equations

Tags:
1. Mar 1, 2016

lowea001

1. The problem statement, all variables and given/known data
A comet in deep space picks up mass as it travels through a large stationary dust cloud. It is subject to a gravitational force of magnitude Mf acting in the direction of its motion. When it entered the cloud, the comet had mass M and speed V. After a time t, it has travelled a distance x through the cloud and its mass is M(1+bx), where b is a positive constant and its speed is v. In the case when f is a non-zero constant, show that $$v = \frac{ft+V}{1+bx}$$
and hence find an expression for x in terms of b, V, f, and t. Then show that it is possible for the comet to move with constant speed.

2. Relevant equations
Differential equations of motion for kinetics, force is rate of change of momentum (Newton's second)

3. The attempt at a solution
So right off the bat I'm pretty sure this doesn't make sense because if the comet were to gain mass the force of gravity would change i.e. it wouldn't remain Mf. But disregarding that, my main issue is proving that the comet can move with constant speed. I know the acceleration should be shown to be an equation that has roots, but it seems pretty messy. Sorry if the solution just really is messy, but I have the feeling there's something I'm missing. Anyway, here's what I've tried:

2. Mar 1, 2016

TeethWhitener

Both velocity and mass are functions of time. What does this do to your expression for $\frac{dp}{dt}$?

[EDIT]: On second thought, this problem is ambiguous. Is M a constant? I was assuming it was the mass of the comet as a function of time (which would make sense). Otherwise I don't think there's a simple solution.

Last edited: Mar 1, 2016
3. Mar 1, 2016

lowea001

Yes M is a constant, its initial mass, but its mass is always changing and at any position x is M (1+bx). Therefore gravity remains constant according to the wording of the question I.e. dp/dt is a constant fM (I know it doesn't make sense but if you replace gravity with contant force it does). Integrating both these with respect to time yields the thing we're trying to show. But assuming that this thing is right, how would I show that it has the potential to be constant? Is the only way to differentiate v after finding x and seeing if the acceleration can be zero?

4. Mar 1, 2016

TeethWhitener

Ah, I see. I was making this much harder.

What do you get when you differentiate v? Try to avoid substituting your big clunky expression for x.

5. Mar 2, 2016

lowea001

Oh don't put in the expression for x? If I partial differentiate the expression with respect to time is that still the acceleration regardless of x? Or can I just treat it as a constant somehow? (Sorry I'm a calculus noob).

6. Mar 2, 2016

TeethWhitener

Yes. Differentiate $v$ with respect to $t$ and set the expression equal to zero. But instead of plugging in your expression for $x$, just notice that $x = x(t)$ and $\frac{dx}{dt} = v$. You're looking for an expression for $v$ that only contains constants.

7. Mar 2, 2016

lowea001

Thank you so much! Finally figured it out.

8. Mar 2, 2016

lowea001

Also, just out of interest, apparently as t -> infinity, v -> a constant. Sounds fun, but how can I show that?

9. Mar 2, 2016

TeethWhitener

To get the actual constant, you'd plug the expression for $x(t)$ into the expression for $v$ and take the limit as $t \rightarrow \infty$. If you just want to show it, it's probably enough to notice that $x \propto \sqrt{t^2} = t$ and then look at the behavior of the limit from there.

10. Mar 2, 2016

epenguin

From my sig "Don't tell us you have got the answer, tell us the answer you have got!"

Also, did you realise you could get the final velocity asked for without ever solving the d.e.?

Which know is of the almost simplest type - it's called "separable variables".

11. Mar 3, 2016

lowea001

$$v = \frac{ft + V}{1+bx} \ (1) \\ Using \ quotient \ rule \ and \ asserting \ acceleration \ can \ be \ zero: \\ \frac{dv}{dt} = \frac{f(1+bx) - \frac{dx}{dt}(ft+V)}{(1+bx)^2} \\ 0 = \frac{f}{1+bx} - \frac{bv(ft + V)}{(1 + bx)^2} \\ 0 = \frac{f - bv^2}{1 + bx} \\ 0 = f - bv^2 \\ v = \sqrt{\frac{f}{b}} \ (a constant) \\ \\ Also: \\ x = \frac{\sqrt{1 + bt(ft+2V)} -1}{b} \ (2) \\$$
If acceleration is non-zero, according to equation (1) and (2) (the latter is found by separable d.e.):
$$\\v = \frac{ft + V}{\sqrt{1 + bt(ft +2V)}} \\ v^2 = \frac{(ft+V)^2}{1 + bt(ft +2V)} \\ v^2 = \frac{f^2t^2 + 2ftV + V^2}{bft^2 +2Vbt + 1} \\ Dividing \ numerator \ and \ denominator \ by \ t^2 \ and \ taking \ the \ limit \ as \ t \ approaches \ infinity: \\ v^2 = \frac{f^2}{bf} \\ v = \sqrt{\frac{f}{b}} \ (as \ t \ approaches \ infinity)$$
My solution. It looks nice therefore it must be right right.

Last edited: Mar 3, 2016
12. Mar 3, 2016

epenguin

I agree with your result for the final v.

But you did not need to solve a differential equation to get this! So understand how you could have got it more easily. Just draw a diagram of this behaviour and you will probably see.

This situation arises quite often. The final state of a system is usually simpler to analyse than how you get there. Getting there might require calculus, possibly analytically unsolvable equations, final or steady states only algebra, or solvable equations.

Then, though I can't follow what you're doing, your calculation is at least unnecessarily complicated. You do not need to differentiate. Just rewrite eq. 1 but instead of v write dx/dt.

V, f and b are constants. The label 'separable variables' more or less tells it all.

This is why I say not just to you "Don't tell us you have got the answer, tell us the answer you have got!" The answers students have got are quite often not as good as they think. And also even when right they tend to be isolated results which they're glad to have got, not tied into anything more general which they could get out of a more extensive discussion so educational point of exercise partly lost, but that's something I take up with the forum colleagues from time to time.

13. Mar 3, 2016

TeethWhitener

They didn't. At least as far as I can tell. They just subbed in $x(t)$ for $x$ and differentiated $v$ with respect to $t$, then subbed in $v$ for $\frac{dx}{dt}$ in the resulting expression.

(emphasis added) This might be part of the problem.

14. Mar 3, 2016

epenguin

I do understand more of it now.

It is a valid method of obtaining the final velocity, but it remains true that as a method of calculating final velocity it is all unnecessary and it can be done in one line as I have hinted.

If eq.(2) is meant to represent x at all times, I think it is not right and is obtained by an invalid argument. So far as I can understand, the final velocity is obtained, and then this is fed into an equation to obtain the velocity at other times, which cannot be right.

.

15. Mar 3, 2016

lowea001

This is what I did to get equation 2. Separable variables in this case gives me an equation for x, not v, no? I mean, the question is solved but as you said I should be (and am) just as interested in the method. So if a one line solution exists, I'd like that.

16. Mar 3, 2016

epenguin

Sorry, eq. 2 is all right, I had the identical one and for some reason didn't recognise it. You haven't written how are you got it.

The way to get final velocity more immediately is to say that if velocity is constant then after a long time it will equal x/t. Equate that with RHS of the equation of #1 (main) and you have your result.

17. Mar 3, 2016

lowea001

Doesn't that just give me equation 2 again?

18. Mar 3, 2016

epenguin

It gives you v = √(f/b) but it gives you immediately from the initial problem without going through the rest.