Beautiful explanation of rocket theory. I love it! Going upwards, as more fuel is burned the mass decreases and by effect, velocity increases making up for the decrease in mass. It's easier to understand it from the point of conservation of momentum and also impulse.
This is very interesting sir. What about water bottle rockets though? What would be different when calculating the boost in speed and/or acceleration if we take into account the pressurized air within the bottles? I would be immensely grateful if you could make a video on that.
Thanks for watching. I suppose the difference in the case you're describing would be that the velocity of the ejected water gets smaller over time as the pressure decreases. Will have a think about this!
What would be the changes to this derivation in the case of a water bottle rocket? Considering the pressurized air inside the water bottle which causes water expulsion and thrust?
Very basic question, Hope you would answer me shortly. In your video at the last part you wrote f which is F/m=gt but the force of gravity being applied on the rocket changes continiously as the rocket is gradually losing mass . so how can you write F/m=g should not it be f=(m-dm)g/m??
All objects in a gravitational field of strength g experience a force per unit mass of g, independent of their mass. The rocket is indeed losing mass over time, this changes its weight but not the applied force per unit mass.
I think that adding dm to the rocket while is burning mass makes the explanation much more complicated. If you are really bothered by negative dm, I would just say that mass at time step I is m+dm and at timestep II you have m for the rocket and dm for the gas. You will arrive to the same equations.
Doing this gives Fdt=mdv-wdm, the same equation that you get if you write the mass at time II as m-dm. This infinitesimal equation is still correct with your redefinition of dm as a positive quantity, but the problem arises when you try to integrate, as the natural log term will end up with the wrong sign. To make the integral work out correctly we need to stick with the convention that dm is the signed change in m.
The last term integrated gives you the finite integral of -wdm between initial mass (lower bound) and final mass (upper bound). This resolves (with Tsiolkovsky 's hypothesis) to -w(ln(mf) - ln(m0)) (Using the fundamental theorem of calculus) --> w(ln(m0/mf)) [using property of logs]. The same as your equation.
@@nicolabombace2004 When you then rearrange for v as we did in the video, you need to move that term over to the other side of the equation and it's going to pick up a minus sign, suggesting that the speed decreases as fuel is burnt.
I think you derivation is wrong in 2 places: 1) mass does not carry momentum so the (-dm), to me, is wrong. The momentum sign comes from the velocity sign. 2) the relative velocity with respect to stationary frame is the vectorial sum of w and V ( you should add dv as well, but that will lead to higher order terms.) . So the actual contribution is v - w . (Positive direction up). I will give you an example here. If w is 0, as if the rocket was to place a particle of gas in its reference frame, the stationary observer will see the rocket moving at v + dv and using your equation the particle at -v. This is a contradiction of the first Newton's law. It derives that the correct equation of motion for a gas particle from stationary observer is v - w
@@nicolabombace2004 Your second point is consistent with the video - I said that it was w-v downwards, which is completely equivalent to v-w upwards. In your example of w=0, my expression gives a velocity of -v downwards, which is the same as v upwards. For the mass element, as demonstrated above, it needs to be -dm simply because dm is a signed change in mass. The minus sign doesn't affect the direction of the momentum because -dm is a positive quantity. The final expression for momentum, -(-dm)(w-v), has an extra minus sign in front because it's pointing downwards.
The equation in the video is correct - if you start with the equation you suggest, you arrive at the incorrect conclusion that the speed of the rocket decreases as fuel is burnt. Take a look through the comments and you'll find an extensive discussion I had with someone else about this same issue!
Hi I had a question regarding the w∫dm/m. Here, based on your earlier implications you assigned m as the initial mass of the rocket at the first step based on the earlier momentum formula. So wouldn't m here be constant and so why would its integral be ln(m)? In general I'm finding it difficult to grasp at which times you mean m as in the m that changes as a function of time according to the fuel ejection rate, and the m that is the mass of the rocket at the first step (which is a constant, while the first m is a function)
The two steps are at an arbitrary time t, and a short time later, t + dt. Because t is arbitrary, m is not really the initial mass when we're looking at the full motion of the rocket - it's the mass at time t and therefore is a function of time. The only constant mass is m₀, which is the actual initial mass, i.e. m₀ is m(t) evaluated at t = 0.
got it, thanks for the answer and the video! i had another question - if i were to substitute values from the real world into this equation, would w be a positive or a negative value?
No problem. The value of w should be positive if the rocket is speeding up - the equation we derived shows that the velocity would be decreasing if w were negative, and this would represent the rocket's thrusters firing in reverse.
Great stuff! Now I know how we can launch our rockets up easily! Thank you very much for the videos!!!
Good to hear from you again! Thanks for watching.
Very interesting, thanks !
Thank you!
Beautiful explanation of rocket theory. I love it! Going upwards, as more fuel is burned the mass decreases and by effect, velocity increases making up for the decrease in mass. It's easier to understand it from the point of conservation of momentum and also impulse.
Thanks for saying so, glad you enjoyed it!
@@DrBenYelverton you're welcome and I'm subscribed already. Cheers.
This is very interesting sir. What about water bottle rockets though? What would be different when calculating the boost in speed and/or acceleration if we take into account the pressurized air within the bottles? I would be immensely grateful if you could make a video on that.
Thanks for watching. I suppose the difference in the case you're describing would be that the velocity of the ejected water gets smaller over time as the pressure decreases. Will have a think about this!
What would be the changes to this derivation in the case of a water bottle rocket? Considering the pressurized air inside the water bottle which causes water expulsion and thrust?
Very basic question, Hope you would answer me shortly. In your video at the last part you wrote f which is F/m=gt but the force of gravity being applied on the rocket changes continiously as the rocket is gradually losing mass . so how can you write F/m=g should not it be f=(m-dm)g/m??
All objects in a gravitational field of strength g experience a force per unit mass of g, independent of their mass. The rocket is indeed losing mass over time, this changes its weight but not the applied force per unit mass.
I think that adding dm to the rocket while is burning mass makes the explanation much more complicated. If you are really bothered by negative dm, I would just say that mass at time step I is m+dm and at timestep II you have m for the rocket and dm for the gas. You will arrive to the same equations.
Doing this gives Fdt=mdv-wdm, the same equation that you get if you write the mass at time II as m-dm. This infinitesimal equation is still correct with your redefinition of dm as a positive quantity, but the problem arises when you try to integrate, as the natural log term will end up with the wrong sign. To make the integral work out correctly we need to stick with the convention that dm is the signed change in m.
The last term integrated gives you the finite integral of -wdm between initial mass (lower bound) and final mass (upper bound). This resolves (with Tsiolkovsky 's hypothesis) to -w(ln(mf) - ln(m0)) (Using the fundamental theorem of calculus) --> w(ln(m0/mf)) [using property of logs].
The same as your equation.
@@nicolabombace2004 When you then rearrange for v as we did in the video, you need to move that term over to the other side of the equation and it's going to pick up a minus sign, suggesting that the speed decreases as fuel is burnt.
I think you derivation is wrong in 2 places:
1) mass does not carry momentum so the (-dm), to me, is wrong. The momentum sign comes from the velocity sign.
2) the relative velocity with respect to stationary frame is the vectorial sum of w and V ( you should add dv as well, but that will lead to higher order terms.) . So the actual contribution is v - w . (Positive direction up). I will give you an example here.
If w is 0, as if the rocket was to place a particle of gas in its reference frame, the stationary observer will see the rocket moving at v + dv and using your equation the particle at -v. This is a contradiction of the first Newton's law. It derives that the correct equation of motion for a gas particle from stationary observer is v - w
@@nicolabombace2004 Your second point is consistent with the video - I said that it was w-v downwards, which is completely equivalent to v-w upwards. In your example of w=0, my expression gives a velocity of -v downwards, which is the same as v upwards. For the mass element, as demonstrated above, it needs to be -dm simply because dm is a signed change in mass. The minus sign doesn't affect the direction of the momentum because -dm is a positive quantity. The final expression for momentum, -(-dm)(w-v), has an extra minus sign in front because it's pointing downwards.
cheers m8
Well made video but I suppose there is one inaccuracy the terms in the first equation should be
mv + Fdt = (m - dm)(v + dv) - (dm)(w - v)
The equation in the video is correct - if you start with the equation you suggest, you arrive at the incorrect conclusion that the speed of the rocket decreases as fuel is burnt. Take a look through the comments and you'll find an extensive discussion I had with someone else about this same issue!
Hi I had a question regarding the w∫dm/m. Here, based on your earlier implications you assigned m as the initial mass of the rocket at the first step based on the earlier momentum formula. So wouldn't m here be constant and so why would its integral be ln(m)?
In general I'm finding it difficult to grasp at which times you mean m as in the m that changes as a function of time according to the fuel ejection rate, and the m that is the mass of the rocket at the first step (which is a constant, while the first m is a function)
The two steps are at an arbitrary time t, and a short time later, t + dt. Because t is arbitrary, m is not really the initial mass when we're looking at the full motion of the rocket - it's the mass at time t and therefore is a function of time. The only constant mass is m₀, which is the actual initial mass, i.e. m₀ is m(t) evaluated at t = 0.
got it, thanks for the answer and the video! i had another question - if i were to substitute values from the real world into this equation, would w be a positive or a negative value?
No problem. The value of w should be positive if the rocket is speeding up - the equation we derived shows that the velocity would be decreasing if w were negative, and this would represent the rocket's thrusters firing in reverse.
thank you!!