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Rocket flight

  • Thread starter Gwilim
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  • #1
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[problem]

State the fundemental equation of motion for a particle of variable mass.

A rocket of initial mass m0 is fired vertically, under the influence of a uniform gravitational field, and expels propellant at a constant relative velocity c downwards. The propellant is completely consumed after a time T, leaving the rocket with a mass mT. Calculate the velocity and position at time t < T. Show that the velocity at time T is independent of the rate of consuming the propellant.

[/problem]

I don't know how to go about solving this. I vaguely recall the principle behind rocket flight; the reaction force against the expulsion of propellant is what drive the rocket upwards. The resultant force will be something like F = f(c) - mg, m of course being a variable and f being some function of c. I have no worked examples for this type of problem to refer to, so I need someone to show me the steps involved.
 

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  • #2
gabbagabbahey
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The fundamental equation is Newton's second law [itex]\textbf{F}_{\text{net}}=\frac{d\textbf{p}}{dt}[/itex]....for constant mass, [itex]\frac{d\textbf{p}}{dt}=m\frac{d\textbf{v}}{dt}[/itex]....when the mass varies with time, you need to use the product rule....what do you get then?
 
  • #3
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F=m(dv/dt)+v(dm/dt), which makes a delightful amount of sense.

What then?
 
  • #4
tiny-tim
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F=m(dv/dt)+v(dm/dt), which makes a delightful amount of sense.

What then?
Force = (rate of) change of momentum …

so how much momentum is the rocket chucking out every second? :wink:
 
  • #5
gabbagabbahey
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F=m(dv/dt)+v(dm/dt), which makes a delightful amount of sense.

What then?
Now apply this to the rocket....what external force(s) are applied to the rocket?...that gives you a value for [itex]\textbf{F}_{\text{net}}[/itex]. What is the mass of the rocket as a function of time [itex]m(t)[/itex]? What is [itex]\frac{dm}{dt}[/itex]?

Plug those values into your fundamental equation [itex]\textbf{F}_{\text{net}}=m\frac{d\textbf{v}}{dt}+\textbf{v}\frac{dm}{dt}[/itex] and you end up with a first order differential equation for the velocity of the rocket [itex]\textbf{v}[/itex]...you know how to solve such an equation right?
 
  • #6
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Now apply this to the rocket....what external force(s) are applied to the rocket?
In this example, just gravity
that gives you a value for [itex]\textbf{F}_{\text{net}}[/itex]. What is the mass of the rocket as a function of time [itex]m(t)[/itex]?
m(t)=m0 + dm/dt
What is [itex]\frac{dm}{dt}[/itex]?
dm/dt= (mT-m0)/T (this will be a negative value)

Plug those values into your fundamental equation [itex]\textbf{F}_{\text{net}}=m\frac{d\textbf{v}}{dt}+\textbf{v}\frac{dm}{dt}[/itex] and you end up with a first order differential equation for the velocity of the rocket [itex]\textbf{v}[/itex]
F=dv/dt(m0 + (mT-m0/T)) + v(mT-m0/T)

you know how to solve such an equation right?
I passed an exam on differential equations a few months ago, so in theory, yes.

If I want to find the point at which the propellant is consumed I would set F=0? But we already know that, it's T. Hold on I'm getting confused again..

Force = (rate of) change of momentum …

so how much momentum is the rocket chucking out every second? :wink:
it's chucking out mass at a rate of (m0-mT/T) units per unit (second?), with a velocity of c..

so c(m0-mT/T)?

Oh oh, Fnet=c(m0-mT/T) - g(m0+(mT-m0/T))?

How would I eliminate mass to get an expression for position/velocity? Divide through by m0+(mT-m0/T)? Seems awfully messy...
 
  • #7
gabbagabbahey
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In this example, just gravity
Well, that's one force...and that'll be [itex]\textbf{F}_{\text{gravity}}=m(t)\textbf{g}[/itex]...but there is also a reaction force from the ejection of propellant [itex]\textbf{F}_{\text{reaction}}=-\textbf{F}_{\text{propellent}}[/itex] (Newton's 3rd Law).

To calculate [itex]\textbf{F}_{\text{propellent}}[/itex], use Newton's 2nd law again:

[tex]\textbf{F}_{\text{propellent}}=m_{\text{propellent}}\frac{d\textbf{v}_{\text{propellent}}}{dt}+\textbf{v}_{\text{propellent}}\frac{dm_{\text{propellent}}}{dt}[/tex]

m(t)=m0 + dm/dt
You mean [itex]m(t)=m_0+\frac{dm}{dt}t[/itex] right?

dm/dt= (mT-m0)/T (this will be a negative value)
Careful, it only ejects mass at this rate for [itex]0\leq t \leq T[/itex], for [itex]t>T[/itex], dm/dt=0....so you can write

[tex]m(t)=\left\{ \begin{array}{lr}m_0+\frac{m_T-m_0}{T}t, & 0\leq t \leq T \\ m_T, & t>T\end{array}[/tex]


If I want to find the point at which the propellant is consumed I would set F=0?
When the propellant is consumed, there will still be a gravitational force of [itex]m_T\textbf{g}[/itex] won't there?

it's chucking out mass at a rate of (m0-mT/T) units per unit (second?), with a velocity of c..

so c(m0-mT/T)?

Oh oh, Fnet=c(m0-mT/T) - g(m0+(mT-m0/T))?
Close, the net force on the rocket should be [itex]F=-m'(t)c-m(t)g[/itex] upward

How would I eliminate mass to get an expression for position/velocity? Divide through by m0+(mT-m0/T)? Seems awfully messy...
Set it equal to [itex]\textbf{F}_{\text{net}}=m(t)\frac{d\textbf{v}}{dt}+\textbf{v}\frac{dm}{dt}[/itex]
 
  • #8
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Well, that's one force...and that'll be [itex]\textbf{F}_{\text{gravity}}=m(t)\textbf{g}[/itex]...but there is also a reaction force from the ejection of propellant [itex]\textbf{F}_{\text{reaction}}=-\textbf{F}_{\text{propellent}}[/itex] (Newton's 3rd Law).

To calculate [itex]\textbf{F}_{\text{propellent}}[/itex], use Newton's 2nd law again:

[tex]\textbf{F}_{\text{propellent}}=m_{\text{propellent}}\frac{d\textbf{v}_{\text{propellent}}}{dt}+\textbf{v}_{\text{propellent}}\frac{dm_{\text{propellent}}}{dt}[/tex]
That comes out as reaction= (mT-m0)(c/T + dv/dt)

so the whole thing is Fnet=dv/dt(mT-m0)+c/T(mT-m0)+(m0+((mT-m0)/T))g

You mean [itex]m(t)=m_0+\frac{dm}{dt}t[/itex] right?
yes

Careful, it only ejects mass at this rate for [itex]0\leq t \leq T[/itex], for [itex]t>T[/itex], dm/dt=0....so you can write

[tex]m(t)=\left\{ \begin{array}{lr}m_0+\frac{m_T-m_0}{T}t, & 0\leq t \leq T \\ m_T, & t>T\end{array}[/tex]
Yes.

When the propellant is consumed, there will still be a gravitational force of [itex]m_T\textbf{g}[/itex] won't there?
Yes.

Close, the net force on the rocket should be [itex]F=-m'(t)c-m(t)g[/itex] upward
right

Set it equal to [itex]\textbf{F}_{\text{net}}=m(t)\frac{d\textbf{v}}{dt}+\textbf{v}\frac{dm}{dt}[/itex]
I'm still not getting it :(

I'll keep working at this. If I do figure it out I'll make a post saying so.
 
  • #9
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hold up, I've got to here:

dv/dt(m(t)-(mT-m0))+dm/dt(v-c)+m(t)g=0

I've got something equal to 0 at least.
 
  • #10
gabbagabbahey
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That comes out as reaction= (mT-m0)(c/T + dv/dt)

so the whole thing is Fnet=dv/dt(mT-m0)+c/T(mT-m0)+(m0+((mT-m0)/T))g
Close, you're missing a negative sign (from the Newton's third law equation) and the propellent is ejected at a constant speed, so dv/dt=0 for the propellent. Therefor Freaction=-(mT-m0)c/T=-m'(t)c


I'm still not getting it :(

I'll keep working at this. If I do figure it out I'll make a post saying so.
Well, you have two equations for Fnet:

(1) [tex]\textbf{F}_{\text{net}}=m(t)\frac{d\textbf{v}}{dt} +\textbf{v}\frac{dm}{dt}[/tex]

(2)[tex]\textbf{F}_{\text{net}}=(-m'(t)c-m(t)g)\textbf{k}[/tex]

(Choosing k to be the upward unit vector)

Set them equal to eachother and you will get an inhomogeneous, linear, first order differential equation for [itex]\text{v}_z(t)[/itex] (the other components of [itex]\text{v}[/itex] have trivial equations and solutions)

Remember, you've already calculated the values of m(t) and m'(t), so you can just plug them into your differential equation....what does your DE become for 0<t<T ? How about for t>T ?
 
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  • #11
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what does your DE become for 0<t<T ?
dv/dt(m0+(mt-m0/T))+v(mT-m0/t)=-(m0+(mT-m0/T))g-c(mT-m0/T)

How do I go about solving that? General solution and particular integral?

I'll look back over my differential equations, but can you show me the steps from here?

How about for t>T ?
Well.. from this point the propellant force dissapears completely, as does dm/dt, and the mass becomes the constant mT so:

dv/dt(mT)=-mTg => dv/dt=-g

Thanks for all the help so far.
 
  • #12
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bump because I still haven't figured this out
 
  • #13
gabbagabbahey
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dv/dt(m0+(mt-m0/T))+v(mT-m0/t)=-(m0+(mT-m0/T))g-c(mT-m0/T)
You really need to be much more careful with your [itex]t[/itex]'s and [itex]T[/itex]'s; the above expression is not correct.

How do I go about solving that? General solution and particular integral?

I'll look back over my differential equations, but can you show me the steps from here?
Once you have the correct expression, manipulate it to get it into the form [itex]\frac{dv}{dt}+f(t)v=g(t)[/itex]. From there, you should be able to solve it without much trouble.

Post you work if you get stuck.
 

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