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 Post subject: rocket net force
PostPosted: Mon, 18 Jul 2011 10:14:10 UTC 
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Some of the following is more or less Halliday & Resnick, 4th Ed., section 9.8 (not the whole section and I added a lot):

Assume motion in a stright line. Therefore, for position, velocity, acceleratiom, force, momentum, etc., d/dt|vector|=|d/dt(vector)|. This is not true in general for motion along a curved path. Assume we have an idealized rocket of mass m and speed |v| at time t traveling in the forward direction relative to an inertial reference frame. The rocket then instantaneously begins to emit a constant stream of exhaust in the backwards direction.

let:
F_net_ext = net external force = net force on (rocket + exhaust) = sum of gravity, atmospheric drag, etc.
F_net_r = net force on the rocket
P = momentum of (rocket + exhaust)
p = momentum of rocket
u = velocity of exhaust (as it leaves the rocket) pointing backwards.
v_rel = velocity of exhaust (as it leaves the rocket) relative to rocket = u - v = vector pointing back and is constant for an idealized rocket.
Note F_net_ext, F_net_r, P, p, v, u, v_rel, dv, a, etc. are all vetors. Their lengths are |vector|.

In time interval dt the rocket emits a mass -dm. Note that dm < 0 and |dm/dt| = constant for an idealized rocket = mass flow rate of exhaust backwards from the rocket. So at time t+dt we now have a bit of exhaust of mass -dm moving backwards with velocity u and the rocket, now of mass m+dm, moving forward at velocity v+dv. Note dv clearly points forward: the rocket's acceleration, a, points forward.

(1) F_net__ext = dP/dt = [1/dt)][(m+dm)(v+dv)+(-dm)u - mv] = mdv/dt + vdm/dt - udm/dt (note (dv)dm/dt ->0 in the limit of the derivative)

This equation is standard for rockets. There are generally 2 ways it is rewritten:

(2) F_net_ext = mdv/dt + vdm/dt - udm/dt = dp/dt - udm/dt
(3) F_net_ext = mdv/dt + vdm/dt - udm/dt = mdv/dt - (u-v)dm/dt = ma - (v_rel)dm/dt

F_net_ext is not the force that propels the rocket. The force that propels the rocket is an internal force within the (rocket+exhaust) system. If F_net_ext = 0 then P remains constant but the rocket experiences a thrust which changes its momentum. The change in p in time dt is equal and opposite to the momentum, udm, carried away by the exhaust. for a rocket (u-v)dm/dt = thrust = (v_rel)dm/dt = forward pointing vector since u-v points back and dm/dt < 0. The thrust is the rate at which momentum enters the rocket. For an idealized rocket with constant v_rel and constant dm/dt, the thrust is a constant.

This is pretty much the extent of Halliday & Resnick on this subject, but every time I read it I'm struck by the seeming inconsistency that thrust is clearly constant yet the rate of change of the rocket's momentum, dp/dt, is not.

I hope you wont mind if I walk you thru my reasoning. By (3), F_net_ext=0 implies ma=thrust. thrust is constant and points forward implies ma is therefore a forward pointing constant vector. ma is constant and m is decreasing with time implies |a| is increasing with time. Therefore as long as thrust exits the rocket backwards, then the rocket will not only accelerate in the forward direction, but the rate at which it accelerates increases with time. dm/dt < 0 and v always points forward implies the term vdm/dt is a backwards pointing vector which is not constant but grows in length as v grows. At time t_0 when the exhaust initially begins to flow back from the rocket, u is a backwards pointing vector of length |v_rel| since v=0. u then shrinks in length as the rocket moves faster, but v_rel remains constant. At some point |v| = |v_rel| and u=0, and after this |v| > |v_rel| and u points forward, tho still shorter than v by |v_rel|. By (2), F_net_ext=0 implies dp/dt = udmdt. dm/dt<0 implies the term udm/dt points forward until |v|>|v_rel|, then it points back. Therefore dp/dt points forward and then, after |v|>|v_rel|, dp/dt points back.

This is one of the major counterintuitive things about rockets that confuses me (and presumably most people, unless my calculations are incorrect). When |v|>|v_rel|, dp/dt of the rocket points back, meaning the net force on the rocket (not F_net_ext) points backwards, opposite the forward motion of the rocket. The momentum of the rocket always points forward in this example, but it's getting shorter at the rate of dp/dt, and tho the net force on the rocket points back after |v| exceeds |v_rel|, nevertheless the rocket not only continues to accelerate forward, but the rate of forward acceleration increases with time!

Many books and websites claim that, in the absence of external forces such as gravity, friction, etc., the net force on a rocket is equal to the thrust. I've even seen this on some NASA websites. But the net force on a rocket does not seem to be equal to the thrust. Assuming F_net_ext = 0 then F_net_r = dp/dt = udm/dt by (2) = udm/dt + vdm/dt - vdm/dt = (u-v)dm/dt + vdm/dt = thrust + vdm/dt. So the net force on a rocket, absent external forces, does not seem to be equal to the thrust, but rather equal to the thrust + vdm/dt.

Hopefully, at this point, you're saying to yourself "what's the problem?" If so, then there probably is no problem. But I'm bothered by the equations:

(4) F_net_ext=0 implies F_net_r = thrust + vdm/dt = u dm/dt

Both equations seem counterintuitive even tho I've reasoned thru the details. It bothers me that many books and websites set F_net_r = thrust + external forces and ignore the vdm/dt term. It makes me wonder whether eqn (4) is correct. I haven't found either of the equations (4) in Halliday & Resnick or Marion & Thornotn's Classical Dynamics or Goldstein's Classical Mechanics, but I haven't found anything in those texts that disputes eqns (4) either.

I would like to know if eqns (4) are correct? Is there something simple here that I'm missing? Thanks in advance.


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 Post subject: Re: rocket net force
PostPosted: Tue, 19 Jul 2011 11:32:43 UTC 
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inkliing wrote:
Some of the following is more or less Halliday & Resnick, 4th Ed., section 9.8 (not the whole section and I added a lot):

Assume motion in a stright line. Therefore, for position, velocity, acceleratiom, force, momentum, etc., d/dt|vector|=|d/dt(vector)|. This is not true in general for motion along a curved path. Assume we have an idealized rocket of mass m and speed |v| at time t traveling in the forward direction relative to an inertial reference frame. The rocket then instantaneously begins to emit a constant stream of exhaust in the backwards direction.

let:
F_net_ext = net external force = net force on (rocket + exhaust) = sum of gravity, atmospheric drag, etc.
F_net_r = net force on the rocket
P = momentum of (rocket + exhaust)
p = momentum of rocket
u = velocity of exhaust (as it leaves the rocket) pointing backwards.
v_rel = velocity of exhaust (as it leaves the rocket) relative to rocket = u - v = vector pointing back and is constant for an idealized rocket.
Note F_net_ext, F_net_r, P, p, v, u, v_rel, dv, a, etc. are all vetors. Their lengths are |vector|.

In time interval dt the rocket emits a mass -dm. Note that dm < 0 and |dm/dt| = constant for an idealized rocket = mass flow rate of exhaust backwards from the rocket. So at time t+dt we now have a bit of exhaust of mass -dm moving backwards with velocity u and the rocket, now of mass m+dm, moving forward at velocity v+dv. Note dv clearly points forward: the rocket's acceleration, a, points forward.

(1) F_net__ext = dP/dt = [1/dt)][(m+dm)(v+dv)+(-dm)u - mv] = mdv/dt + vdm/dt - udm/dt (note (dv)dm/dt ->0 in the limit of the derivative)

This equation is standard for rockets. There are generally 2 ways it is rewritten:

(2) F_net_ext = mdv/dt + vdm/dt - udm/dt = dp/dt - udm/dt
(3) F_net_ext = mdv/dt + vdm/dt - udm/dt = mdv/dt - (u-v)dm/dt = ma - (v_rel)dm/dt

F_net_ext is not the force that propels the rocket. The force that propels the rocket is an internal force within the (rocket+exhaust) system. If F_net_ext = 0 then P remains constant but the rocket experiences a thrust which changes its momentum. The change in p in time dt is equal and opposite to the momentum, udm, carried away by the exhaust. for a rocket (u-v)dm/dt = thrust = (v_rel)dm/dt = forward pointing vector since u-v points back and dm/dt < 0. The thrust is the rate at which momentum enters the rocket. For an idealized rocket with constant v_rel and constant dm/dt, the thrust is a constant.

This is pretty much the extent of Halliday & Resnick on this subject, but every time I read it I'm struck by the seeming inconsistency that thrust is clearly constant yet the rate of change of the rocket's momentum, dp/dt, is not.

I hope you wont mind if I walk you thru my reasoning. By (3), F_net_ext=0 implies ma=thrust. thrust is constant and points forward implies ma is therefore a forward pointing constant vector. ma is constant and m is decreasing with time implies |a| is increasing with time. Therefore as long as thrust exits the rocket backwards, then the rocket will not only accelerate in the forward direction, but the rate at which it accelerates increases with time. dm/dt < 0 and v always points forward implies the term vdm/dt is a backwards pointing vector which is not constant but grows in length as v grows. At time t_0 when the exhaust initially begins to flow back from the rocket, u is a backwards pointing vector of length |v_rel| since v=0. u then shrinks in length as the rocket moves faster, but v_rel remains constant. At some point |v| = |v_rel| and u=0, and after this |v| > |v_rel| and u points forward, tho still shorter than v by |v_rel|. By (2), F_net_ext=0 implies dp/dt = udmdt. dm/dt<0 implies the term udm/dt points forward until |v|>|v_rel|, then it points back. Therefore dp/dt points forward and then, after |v|>|v_rel|, dp/dt points back.

This is one of the major counterintuitive things about rockets that confuses me (and presumably most people, unless my calculations are incorrect). When |v|>|v_rel|, dp/dt of the rocket points back, meaning the net force on the rocket (not F_net_ext) points backwards, opposite the forward motion of the rocket. The momentum of the rocket always points forward in this example, but it's getting shorter at the rate of dp/dt, and tho the net force on the rocket points back after |v| exceeds |v_rel|, nevertheless the rocket not only continues to accelerate forward, but the rate of forward acceleration increases with time!

Many books and websites claim that, in the absence of external forces such as gravity, friction, etc., the net force on a rocket is equal to the thrust. I've even seen this on some NASA websites. But the net force on a rocket does not seem to be equal to the thrust. Assuming F_net_ext = 0 then F_net_r = dp/dt = udm/dt by (2) = udm/dt + vdm/dt - vdm/dt = (u-v)dm/dt + vdm/dt = thrust + vdm/dt. So the net force on a rocket, absent external forces, does not seem to be equal to the thrust, but rather equal to the thrust + vdm/dt.

Hopefully, at this point, you're saying to yourself "what's the problem?" If so, then there probably is no problem. But I'm bothered by the equations:

(4) F_net_ext=0 implies F_net_r = thrust + vdm/dt = u dm/dt

Both equations seem counterintuitive even tho I've reasoned thru the details. It bothers me that many books and websites set F_net_r = thrust + external forces and ignore the vdm/dt term. It makes me wonder whether eqn (4) is correct. I haven't found either of the equations (4) in Halliday & Resnick or Marion & Thornotn's Classical Dynamics or Goldstein's Classical Mechanics, but I haven't found anything in those texts that disputes eqns (4) either.

I would like to know if eqns (4) are correct? Is there something simple here that I'm missing? Thanks in advance.


No! You can't spit something out from the back with relative velocity pointing forward, so at no point (assuming no external forces) will the force on rocket point backwards.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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 Post subject: Re: rocket net force
PostPosted: Tue, 19 Jul 2011 14:07:10 UTC 
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Joined: Tue, 2 Mar 2010 17:16:55 UTC
Posts: 11
Quote:
No! You can't spit something out from the back with relative velocity pointing forward, so at no point (assuming no external forces) will the force on rocket point backwards.


My post clearly defines v_rel as a backwards pointing constant vector.

"v_rel = velocity of exhaust (as it leaves the rocket) relative to rocket = u - v = vector pointing back and is constant for an idealized rocket."

Can you please explain your response in more detail. I'm pretty sure what your saying is addressed in my original post.

P.S. Please don't quote my entire original post. That's unnessary.


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