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Spaceship ejecting mass at right angles

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Homework Statement


A spaceship of mass ##m_0## moves in the absence of external forces with a constant velocity ##v_0##. To change the motion direction, a jet engine is switched on. It starts ejecting a gas jet with velocity u which is constant relative to the spaceship and directed at right angles to the spaceship motion The engine is shut down when the mass of the spaceship decreases to ##m##. Through what angle ##\alpha## did the motion direction of the spaceship deviate due to the jet engine operation?


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The Attempt at a Solution


Since there are no external forces, the momentum is conserved. Let the initial direction of motion be the x-axis. At time t, assume that spaceship rotates by an angle ##\theta## and its mass is m(t). The momentum in x direction is ##m(t)v_0\cos(\theta)##. When it ejects mass of dm at t+dt, let it rotate by angle ##d\theta##. The momentum in x-direction at t+dt is ##(m(t)-dm)v_0\cos(\theta+d\theta)+dm(u\sin\theta+v_0\cos\theta)##. Equating them and using ##\cos(\theta+d\theta)=\cos\theta-\sin\theta d\theta## gives
$$dm u=m(t)v_0d\theta$$
Solving the D.E (limits for ##\theta##: 0 to ##\alpha## and for m: ##m_0## to ##m##),
$$\alpha=(v_0/u)\ln(m/m_0)$$
But this is wrong. The given answer is ##\alpha=(v_0/u)\ln(m_0/m)##. I believe there is a sign error but I am not able to spot it. I am not sure if I approached the problem correctly.

Any help is appreciated. Thanks!
 

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Answers and Replies

  • #2
tiny-tim
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Hi Pranav-Arora! :smile:

ln(1/x) = … ? :wink:
 
  • #3
TSny
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The given answer is ##\alpha=(v_0/u)\ln(m_0/m)##.
Are you sure this is the answer?

I believe there is a sign error but I am not able to spot it.
Hint: What you have called "##dm##" is not quite the differential of ##m(t)##.

I am not sure if I approached the problem correctly.
I think your approach is ok. But there is no loss of generality in choosing the orientation of the x-axis to be the direction of the velocity of the ship at time t. Then take momentum components in the y-direction for t and t + dt.
 
  • #4
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Hi Pranav-Arora! :smile:

ln(1/x) = … ? :wink:
-ln(x). Not sure how it helps though. I get a negative answer if I use this.

Are you sure this is the answer?
Sorry, it is ##(u/v_0)## instead of ##(v_0/u)##. :redface:

I too get ##(u/v_0)##. I wrote it wrong while making the post.
Hint: What you have called "##dm##" is not quite the differential of ##m(t)##.
Is it completely wrong or have I used incorrect notations? :confused:

I guess I should simply write m instead m(t) to avoid confusion.

I think your approach is ok. But there is no loss of generality in choosing the orientation of the x-axis to be the direction of the velocity of the ship at time t. Then take momentum components in the y-direction for t and t + dt.
At time t, momentum in y direction, ##mv_0\sin(\theta)##
At time t+dt, ##(m-dm)v_0\sin(\theta+d\theta)+dm(-u\cos\theta+v_0\sin\theta)##
Equating them and solving gives me the same answer again. I still cannot spot the sign error. :(
 
  • #5
TSny
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Suppose you write ##m_s(t)## for the mass of the ship and you let ##dm_e## be the mass ejected between t and t+dt. What is the relation between ##dm_e## and ##dm_s##?

The mass of the ship at t + dt is ##m_s(t+dt)##. How would you express ##m_s(t+dt)## in terms of ##m_s(t)## and ##dm_s##? How would you express ##m_s(t+dt)## in terms of ##m_s(t)## and ##dm_e##? [Edited]
 
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  • #6
D H
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Since there are no external forces, the momentum is conserved.
Just to be clear, it is the momentum of the spaceship + exhaust cloud system that is conserved. The spaceship's momentum is not a conserved quantity.


I still cannot spot the sign error. :(
The quantity that you designated as dm is the (positive) quantity of mass ejected by the spacecraft during the time interval dt. The mass of the exhaust cloud increases by dm. The mass of the spacecraft? It decreases by dm. That's your sign error.
 
  • #7
tiny-tim
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A spaceship of mass ##m_0## …
… when the mass of the spaceship decreases to ##m##.

Solving the D.E (limits for ##\theta##: 0 to ##\alpha## and for m: ##m_0## to ##m##),
$$\alpha=(v_0/u)\ln(m/m_0)$$
But this is wrong. The given answer is ##\alpha=(v_0/u)\ln(m_0/m)##.
-ln(x). Not sure how it helps though. I get a negative answer if I use this.
no, m < mo

so the given answer ##\alpha=(v_0/u)\ln(m_0/m)## is positive :smile:
 
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  • #8
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Suppose you write ##m_s(t)## for the mass of the ship and you let ##dm_e## be the mass ejected between t and t+dt. What is the relation between ##dm_e## and ##dm_s##?
##dm_s=-dm_e##?
The mass of the ship at t + dt is ##m_s(t+dt)##. How would you express ##m_s(t+dt)## in terms of ##m_s(t)## and ##dm_s##? How would you express ##m_s(t+dt)## in terms of ##m_s(t)## and ##dm_e##? [Edited]
##m_s(t+dt)=m_s(t)-dm_s=m_s(t)+dm_e##?

no, m < mo

so the given answer ##\alpha=(v_0/u)\ln(m_0/m)## is positive :smile:
I know that the given answer is positive because m_0>m but I don't understand why do I get the negative answer.
 
  • #9
D H
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I know that the given answer is positive because m_0>m but I don't understand why do I get the negative answer.
Because you did this:

##m_s(t+dt)=m_s(t)-dm_s=m_s(t)+dm_e##?
Does that expression make any sense? (That's a rhetorical question. The answer is no.)

Here, ##dm_e## and ##dt## are (small) positive quantities, so what you are saying with that expression is that the spacecraft's mass increases from time ##t## to time ##t+dt##.
 
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  • #10
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Does that expression make any sense? (That's a rhetorical question. The answer is no.)

Here, ##dm_e## and ##dt## are (small) positive quantities, so what you are saying with that expression is that the spacecraft's mass increases from time ##t## to time ##t+dt##.
Okay sorry for my foolishness. So the correct relation is ##m_s(t+dt)=m_s(t)-dm_e=m(t)+dm_s##.

If I use the proper signs, I get the correct answer. Thanks for the help everyone! :)
 

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