# Homework Help: Rocket motion under gravity

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1. Jan 21, 2017

### John004

1. The problem statement, all variables and given/known data
A rocket with initial mass of m0. The engine that can burn gas at a rate defined by m(t)=m0-αt, and expel gas at speed (relative to the rocket) of u(t)=u0-βt. Here, m0, α, u0, and β are all constants. Assume the lift-off from ground is immediate

a) The rocket speed v(t)=?

b) The rocket height (from ground) y(t)=?

2. Relevant equations
F = dp/dt

3. The attempt at a solution

So F = dp/dt → -mg = u dm/dt + m dv/dt since dm/dt = -α → -g = (-α/m)u + dv/dt

→ ∫(-g + α(u0 - βt)/(m0 - αt) dt) = v

so Im not really too sure about how to solve the above integral

2. Jan 21, 2017

### haruspex

You need to be careful using dp/dt =mdv/dt+vdm/dt. That equation is as though the mass can magically change without the total momentum changing. E.g. consider a car moving at speed v, not burning fuel, just leaking it. With no external force, dp/dt =0. Since dm/dt<0, you would conclude that the car must be accelerating. In short, it is only correct in a reference frame where the gained or lost mass now has zero velocity.

Last edited: Jan 21, 2017
3. Jan 21, 2017

### John004

I don't see what the problem is using dp/dt in the way i have here is?

4. Jan 21, 2017

### haruspex

The lost mass, the burnt fuel, does not have zero velocity in the reference frame.

Edit: Maybe your final equation is ok, though. I haven't checked that. I will now.

Edit 2: My mistake. I misread this line: u dm/dt + m dv/dt as v dm/dt + m dv/dt.

To solve the integral, expand the (u0-βt)/(m0-αt) using partial fractions.

Last edited: Jan 21, 2017
5. Jan 21, 2017

### John004

So then how can i take into account the changing mass? I thought your argument was just about when there is zero external force

6. Jan 21, 2017

### haruspex

7. Jan 21, 2017

### John004

Ah
ah ok, i see. anyways, i was still using the equation without knowing why it was right. So it is valid because "u" is measured relative to the rocket?

8. Jan 21, 2017

### haruspex

On this forum, we often see dp/dt = d(mv)/dt = mdv/dt+vdm/dt, using the product rule. That's why I misread what you posted.
I'm not sure how you figured out to use mdv/dt+udm/dt, where u is the relative velocity of the exhaust. Maybe you did it using the frame of reference of the instantaneous velocity of the rocket.

9. Jan 21, 2017

### John004

I just noticed that you can do the following also
Fext dt = u dm + m dv, since dm/dt = -α
-mg dt = -uαdt + m dv
-g dt + (u/m) α dt = dv
(-g + (u/m)α)dt = dv. Then using the same relation from before

(-g + (u/m)α) (-dm/α) = dv
(g/α)(m-m0) + u ln(m0/m) = v, then using m - m0 = -αt
-gt + u ln (m0/m) = v
Isn't this contradictory to my previous result?

10. Jan 21, 2017

### haruspex

Doesn't that step treat u as constant?

11. Jan 22, 2017

### John004

It is a constant with respect to m isn't it?

12. Jan 22, 2017

### haruspex

No. Both m and u vary over time.

13. Jan 22, 2017

### John004

Not sure if you worked out the whole problem, but if you did, could you take a look at what I got and let me know if it's ok?

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14. Jan 22, 2017

### haruspex

$y=\frac 12 (\beta-g)t^2-(\beta-\frac{u_0\alpha}{m_0})((1-\frac{\alpha}{m_0}t)\ln(1-\frac{\alpha}{m_0}t)-\frac{\alpha}{m_0}t))$