How Is Rocket Speed and Height Calculated Under Varying Thrust and Mass?

[John004]

Homework Statement

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)=?

Homework Equations

F = dp/dt

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 I am not really too sure about how to solve the above integral

 

[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.

 

[John004]

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 velociity.

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

 

[haruspex]

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

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.
Sorry about the confusion.

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

 

[John004]

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.

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

 

[haruspex]

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

Please see my later edit.

 

[John004]

Ah

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.
Sorry about the confusion.

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

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?

 

[haruspex]

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?

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.

 

[John004]

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.

I just noticed that you can do the following also
F_ext 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?

 

[haruspex]

(-g + (u/m)α) (-dm/α) = dv
(g/α)(m-m0) + u ln(m0/m) = v, then using m - m0 = -αt

Doesn't that step treat u as constant?

 

[John004]

Doesn't that step treat u as constant?

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

 

[haruspex]

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

No. Both m and u vary over time.

 

[John004]

No. Both m and u vary over time.

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?

 

[haruspex]

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?

Unfortunately your image is sideways, making it hard to read.
I get the same expression for v, except that you can simplify it by recognising that the two ln() functions are the same, just opposite sign.
I end up with
$y=\frac 12 (\beta-g)t^2-(\beta-\frac{u_0\alpha}{m_0})((1-\frac{\alpha}{m_0}t)\ln(1-\frac{\alpha}{m_0}t)-\frac{\alpha}{m_0}t))$
I'll leave you to decide whether that matches your result. If it doesn't, try differentiating both to get back to v.