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Rocket With Varying Mass

  • Thread starter embphysics
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Homework Statement


Consider a rocket traveling in a straight line subject to an external force Fext acting along the
same line.

a) Show that the equation of motion is [itex]m\dot{v}=-\dot{m}v_{ex} + F^{ext}[/itex] (1)

b) Specialize to the case of a rocket taking off vertically (from rest) in a (constant) gravitational
field g, so the equation of motion becomes

[itex]m \dot{v} = -\dot{m} v_{ex} -mg[/itex] (2)

Assume that the rocket ejects mass at a constant rate, [itex]\dot{m}=-k[/itex] (where k is a positive constant), so that [itex]=m = m_0 - kt[/itex]. Solve the equation for v as a function of t.

(c) Using the rough data from problem 3.7, find the space shuttle's speed two minutes into flight, assuming (what is nearly true) that is travels vertically up during this period and that g does not change appreciably. Compare with the corresponding result if there were no gravity. (d) Describe what would happen to a rocket that was designed so that the first term on the right of equation (2) was smaller than the initial value of the second.


Homework Equations





The Attempt at a Solution


I was successful in solving part (a). Now I am trying to integrate (2), of which I successfully derived when I derived equation (1) on my own. Here is my work:

[itex]m \dot{v} = -\dot{m} v_{ex} -mg[/itex] Since we are dealing with objects of mass, we can safely assume that [itex]m \ne 0[/itex], whereby we can divide by m:

[itex] \dot{v} = - \frac{\dot{m}}{m} v_{ex} -g[/itex].

Integrating with respect to time,

[itex]v(t) = -v_{ex} \int_?^?\frac{\dot{m}}{m} +v_0 - gt[/itex] At first I suspected the limits of integration to be m0 to m, where m0 is the initial mass and m is the mass at some time t. This didn't seem correct, however. Isn't this integration integrating over the amount of that has left the rocket? And if we were to integrating with the lower bound m0, would that imply that initially all of the mass of the rocket has gone into fuel? I am having difficulty interpreting the limits of integration. If someone could be so kind as to help me, I would certainly appreciate it.

Thanks y'all.
 
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Answers and Replies

  • #2
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You can integrate over m, but I think it is much easier to integrate over time.

And if we were to integrating with the lower bound m0, would that imply that initially all of the mass of the rocket has gone into fuel?
It does not matter how much of the initial m0 is fuel, as long as the rocket has enough fuel to work during those two minutes (you can assume this to be true). You know the total mass of the rocket as function of time and you know its rate of change, that is sufficient for the integral.
 
  • #3
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[itex]m = m_0 - kt[/itex]
since you know what m is, replace it in the integral. Also, its given that dm/dt =-k. this simply means dm=-k*dt. Replace both of these in the integral. Now you just have to integrate it over some time interval.

I hope it helps!!!

Edit: mfb is also saying the same thing. I hadn't refresh my page before replying!!!!
 
  • #4
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Hi embphysics

Welcome to PhysicsForums!!!

At first I suspected the limits of integration to be m0 to m, where m0 is the initial mass and m is the mass at some time t. This didn't seem correct, however. Isn't this integration integrating over the amount of that has left the rocket?

Why do you think the limits of integration are not correct ? The integration is done with respect to mass 'm' of the system (Mass of rocket +mass of unspent fuel) , not the mass which has left the rocket .

And if we were to integrating with the lower bound m0, would that imply that initially all of the mass of the rocket has gone into fuel? I am having difficulty interpreting the limits of integration. If someone could be so kind as to help me, I would certainly appreciate it.

m = Mass of rocket(doesn't change over time) + mass of unspent fuel (changes over time)
m0 = Mass of rocket + Initial mass of fuel

Note that fuel burning stops when 'kt' equals the mass of the fuel .

Does this help?
 
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  • #5
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I must be missing something. Let's forget the limits of integration for a moment. I integrating both sides of the differential equation with respect to time.[itex]\int \frac{\dot{m}}{m} dt = \int \frac{-k}{m_0-kt}dt[/itex]. Everything is in terms of t. Evidently when you do the integration, you arrive at [itex]\ln(\frac{m_0}{m})[/itex], which I do see how is possible.
 
  • #6
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I must be missing something. Let's forget the limits of integration for a moment. I integrating both sides of the differential equation with respect to time.[itex]\int \frac{\dot{m}}{m} dt = \int \frac{-k}{m_0-kt}dt[/itex]. Everything is in terms of t. Evidently when you do the integration, you arrive at [itex]\ln(\frac{m_0}{m})[/itex], which I do see how is possible.
your integral doesn't seem correct. it should be [itex]\ln(\frac{m_0-kt}{m_0})[/itex]. check the calculation maybe you missed a minus(-) sign.

Also note that the expression I got here makes things lucid, when the rocket ejects the mass out its speed increases. You can check this by your final v(t) expression, gravity tries to decrease the speed but mass ejection tries to speed it up.
 
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  • #7
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I must be missing something. Let's forget the limits of integration for a moment. I integrating both sides of the differential equation with respect to time.[itex]\int \frac{\dot{m}}{m} dt = \int \frac{-k}{m_0-kt}dt[/itex]. Everything is in terms of t. Evidently when you do the integration, you arrive at [itex]\ln(\frac{m_0}{m})[/itex], which I do see how is possible.
You should get ln(m/m0) .

[itex]\dot{v} = - \frac{\dot{m}}{m} v_{ex} -g[/itex].
This becomes [itex]\dot{v} = - v_{ex} ln(\frac{m}{m_0}) -g[/itex]

Note : There is a minus sign in the first term of the R.H.S .

-ln(m/m0) = ln(m0/m)

So,essentially you have [itex]v_{f} = v_{ex} ln(\frac{m_0}{m}) -gt[/itex]
 
  • #8
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Shouldn't there be a constant that represents the initial velocity?
 
  • #9
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Shouldn't there be a constant that represents the initial velocity?
The rocket starts from rest.
 
  • #10
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Shouldn't there be a constant that represents the initial velocity?
yeah, but it would be zero for the given case
 
  • #11
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See, I am not sure all of this is correct. In a later problem, in the textbook, they want you to integrate the function that we are now trying to determine. In the book, they say it should be[itex]y(t)=v_{ex} - \frac{1}{2} gt^2 - \frac{mv_{ex}}{k} \ln(\frac{m_0}{m(t)})[/itex].

Tanya Sharma, in your v(t) function, there is no a negative appended to the natural log; yet there is suppose to be one. What is going on!?
 
  • #12
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  • #13
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Yes, but that changes the natural log term, which is different than in this function:

[itex]y(t)=v_{ex} - \frac{1}{2} gt^2 - \frac{mv_{ex}}{k} \ln(\frac{m_0}{m(t)})[/itex]

This function y(t), is the integral of the function v(t) that we are now trying to find.
 
  • #14
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Yes, but that changes the natural log term, which is different than in this function:

[itex]y(t)=v_{ex} - \frac{1}{2} gt^2 - \frac{mv_{ex}}{k} \ln(\frac{m_0}{m(t)})[/itex]

This function y(t), is the integral of the function v(t) that we are now trying to find.
it seems you have again messed up in the integration part. you can check this simply taking the derivative of y(t) function, if it doesn't match v(t) function then your y(t) function isn't correct.

it should be the following result
[itex]y(t)=y_{o} + v_{o}t -\frac{v_{ex}m(t)}{k} - \frac{1}{2} gt^2 + v_{ex}t \ln(\frac{m_0}{m(t)})+\frac{v_{ex}m_{o}}{k}\ln(m(t))[/itex]

the above result includes initial position and initial velocity
 
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