This equation tells us that the speed of the rocket depends only the exhaust speed and essentially the fraction of the fuel mass/rocket mass! Pretty cool. To actually work out say, the acceleration of the rocket, we would need to consider the time derivative of the v, apply that to the fixed rocket mass, which should give us the thrust. Rocket science is cool!
How did you suddenly make dm negative? In your equation above you assumed that delta m was positive when applying conservation of momentum, as you wrote that the rocket mass of m+delta m splits into two parts.
excellent question - we didn't really assume it was positive. We just didn't give it a sign. In the next part I assumed that the mass would be decreasing i.e. it would be negative if so. You could still not assign a sign though and carry through the working out without the negative sign. I hope this helps!
I'm trying to derive a differential equation that completely determines the acceleration of the rocket, but I'm stuck. the change in velocity only depends on the initial and final mass. So, without knowledge of the mass rate of change or mass function, is it impossible to determine the acceleration, right?
You can write this way, it wouldn't change the answer as the initial and final masses are both positive. In general in maths, people often write it as ln(x) from what I remember as you cannot really integrate that function when x is 0 and hence you are always taking either negative values, or positive values.
So when we begin integrating it is the same as summation, adding up all the individual bits of fuel, then just knowing that the integral of 1/x is ln(x) should help! Let me know if something doesn't make sense!
This equation tells us that the speed of the rocket depends only the exhaust speed and essentially the fraction of the fuel mass/rocket mass! Pretty cool. To actually work out say, the acceleration of the rocket, we would need to consider the time derivative of the v, apply that to the fixed rocket mass, which should give us the thrust. Rocket science is cool!
i like you
Amazing!
This is why I love calculus.
Its powerful!
conservation law is also amazing!
I agree! It's at the heart of every physics theory out there.
Thanks!! that was really clean, want more like this
More to come!
How did you suddenly make dm negative? In your equation above you assumed that delta m was positive when applying conservation of momentum, as you wrote that the rocket mass of m+delta m splits into two parts.
excellent question - we didn't really assume it was positive. We just didn't give it a sign. In the next part I assumed that the mass would be decreasing i.e. it would be negative if so. You could still not assign a sign though and carry through the working out without the negative sign. I hope this helps!
Very interesting equation ❤️ i needed that. Thank you 😊
Anytime! Thanks for the comment!
I'm trying to derive a differential equation that completely determines the acceleration of the rocket, but I'm stuck. the change in velocity only depends on the initial and final mass. So, without knowledge of the mass rate of change or mass function, is it impossible to determine the acceleration, right?
excellent! Yes, we'd need a function for dm/dt.
@@zhelyo_physics Thank you! You're very kind and responsive, and I truly appreciate it.
Hi, you probably get asked this a lot but what spec do you teach- I do OCR A?
Excellent! I teach OCR A. I have actually filmed the entire OCR A spec here: ruclips.net/p/PLSygKZqfTjPC3hJ7nRSnnXTw3tI_o67dR
Sorry I’m not quite get what Ve really is.
its the exhaust speed right, relative to the rocket or relative to the ground?
Relative to the rocket : )
@@zhelyo_physics Thanks! Sir🙏
Amazing! love it..
Glad to hear Pavan! Thanks for your comment!
You could also say F = dP/dt
For sure! After some analysis of the initial and final momentum you would reach exactly the same physics.
Hello! Just out of interest what university did you go to?
Sure, I did Maths and Physics at Manchester University.
@@zhelyo_physics Additionally, I just have to say how motivating it is for me to see you talk about Physics with such enthusiasm and such detail.
thanks a lot! Much appreciated!
Isn’t it v=veln(m2/m1)
The negative sign flips the Ln. Hope this helps!
Cool
Glad you enjoyed!
is it weird that im a freshmen in highschool and im procrastinating doing geometry homework by learning how to do this
Shouldn’t the integral of 1/m be ln|m| as in the absolute value of m?
You can write this way, it wouldn't change the answer as the initial and final masses are both positive. In general in maths, people often write it as ln(x) from what I remember as you cannot really integrate that function when x is 0 and hence you are always taking either negative values, or positive values.
You lost me when the calculus started
So when we begin integrating it is the same as summation, adding up all the individual bits of fuel, then just knowing that the integral of 1/x is ln(x) should help! Let me know if something doesn't make sense!