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On rocket thrust

by calculus_jy
Tags: rocket, thrust
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calculus_jy
#1
Oct12-08, 05:43 AM
P: 56
recently i came across quesiton:
a rocket car movves on str horz track. half of car mass is propellent, during run, propellent consumed at constant rate and ejected at constant nozzle velocity
the answer says that force is constant but accel is increasing

however i got a major question:
by defitiion thrust of rocket is -dm/m=dv/u where u is the exit speed so rearranging
-udm/dt=mdv/dt,
but thrust of rocket is -udm/dt, let this be T(constant)

however F=dp/dt=mdv/dt+vdm/dt=T+vdm/dt
however v is changing for rocket due to acceleration and dm/dt is constant and T is constant, does that not mean F is not constant on rocket car ????
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Rake-MC
#2
Oct12-08, 06:47 AM
P: 328
I think you're thinking about it way too hard. F = ma, as mass decreases, acceleration is increasing proportionatly. Does the question imply it's a 1:1 ratio?
D H
#3
Oct12-08, 07:39 AM
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F=dp/dt = ma + mdot*v (F=ma assumes constant mass or zero velocity) is a frame-dependent quantity in a variable mass problem. calculus_jy, you chose the frame in which the rocket was initially at rest; an Earth-fixed frame. Think about it from the perspective of rocket that starts from the surface of the Earth and goes to Mars. Using an Earth-centered frame as the basis for calculating thrust when the rocket is about to perform a maneuver near Mars doesn't make much sense.

One choice of frames does make sense, always: The inertial frame whose origin is instantaneously co-located and co-moving with the rocket. This is called the rocket's instantaneous rest frame. This frame is equally meaningful for a rocket on a track and a rocket around Mar. Rocket thrust is always assess in this frame.

In this frame, the simplistic form of Newton's 2nd law, F=ma, once again applies because v=0. While position and velocity are relative in Newtonian mechanics, acceleration is absolute. All inertial observers will see the rocket as having the acceleration calculated in the rocket's instantaneous rest frame.

HallsofIvy
#4
Oct12-08, 07:53 AM
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On rocket thrust

Quote Quote by Rake-MC View Post
I think you're thinking about it way too hard. F = ma, as mass decreases, acceleration is increasing proportionatly. Does the question imply it's a 1:1 ratio?
No, F is NOT equal to ma in this problem. F= ma is true for constant mass, which is not the case here. F= dp/dt (p= momentum) or F= d(mv)/dt= vdm/dt+ mdv/dt. If dm/dt is a constant, then m is a linear function of time: m= at+ b. If force is also constant, then
(at+ b)dv/dt+ av= F gives dv/dt= (F- av)/(at+ b) or dv/(F-av)= dt/(at+b). Integrating both sides, (-1/a)ln(F- av)= (1/a)ln(at+ b)+ C so F- av= C'/(at+b), not a constant.
Rake-MC
#5
Oct12-08, 08:36 AM
P: 328
Okay this is beyond me..
D H
#6
Oct12-08, 09:07 AM
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P: 15,152
Quote Quote by HallsofIvy View Post
If force is also constant, then (at+ b)dv/dt+ av= F gives dv/dt= (F- av)/(at+ b) or dv/(F-av)= dt/(at+b).
Halls, you are implicitly assuming that force is constant in the track's rest frame. This is the same mistake the OP made. Force is constant in the rocket's rest frame. I'll let the OP take it from there.
calculus_jy
#7
Oct12-08, 07:16 PM
P: 56
so what you are saying is that from rocket frame, F= constant, from earth frame F is not constant?? then do they have to specifiy in question wheter they are from rocket frame or from the earth frame??? i am a bit confused...in addition, in the rocket frame the observer will conclude that the his frame is stationary, but experiencing a force!??
D H
#8
Oct12-08, 09:10 PM
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P: 15,152
Some things, like kinetic energy, are always frame dependent. Do your homework questions specify the frame? While it might be better if they did do so, they usually don't. The frame can be inferred from the context of the problem.

Similarly, force is obviously frame dependent in a variable mass situation. The frame in which force is referenced can be inferred from the context. In this case, you are given that the relative exhaust velocity is constant, making the force constant in the rocket's rest frame.
calculus_jy
#9
Oct12-08, 11:45 PM
P: 56
1) can the rocket frame acutally say they are experiencing force but no acceleration since there frame is at rest? such that net F>0 v=0??
2) becuase rocket is acceleating frame, is the galilean transformation of u-v=U actually viable??
where u is velocity of gas relative to earth
v velocity of velocity of rocket relative to earth
U velocity of gas relative to rocket
3) why do most website give the formula
net force on rocket (without gravity) =mdv/dt=-udm/dt
whist the defition of force is dp/dt?giving mdv/dt+vdm/dt=-udm/dt +vdm/dt
D H
#10
Oct13-08, 05:28 AM
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P: 15,152
Quote Quote by calculus_jy View Post
1) can the rocket frame acutally say they are experiencing force but no acceleration since there frame is at rest? such that net F>0 v=0??
2) becuase rocket is acceleating frame, is the galilean transformation of u-v=U actually viable??
The rocket frame and the rocket's rest frame are two different things. The rest frame is the inertial frame whose origin is instantaneously co-located and co-moving with the rocket's center of mass and whose axes are instantaneously co-aligned with some rocket-based set of axes. If you can compute the rocket's acceleration in this frame (which you can), you have just computed the rocket's acceleration in all inertial frames because acceleration is invariant in Newtonian mechanics.


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