at the end of 90 secs, the rate of change of mass falls by 16.67% compared to at the end of 1 sec; therefore the average acceleration falls by the same amount. If you want to calculate the final velocity using instantaneous acceleration we have to numerically integrate at intermediate points, say every sec, to get an answer closer to 3466 m/s. You will get a velocity much lower if the final acceleration of 20.83 m/sec^2 is used in v = u + at.
i think that at 10:45 you should hvae made it 1/m so when you intergrated you could have got the natural log of M due to log laws. also what i love to do is take the inverse function of log which is e and subtract mf/m0 to the other side and bring Vf/V0 to the other side. otherwise great video!!
Quick question. I thought I knew what "relative to the rocket means". Is it the speed the gas moves because the rocket is moving, or is it just the speed used to push the rocket? Lastly, it's hard for me to understand how that the speed relative to the earth is different from the speed relative to the rocket. I'm wondering what makes them different 🤔
the speed of the ejected gas/mass relative to the rocket means the speed of the mass if observed by someone present in the rocket. imagine you are in the rocket, traveling at the same speed as the rocket, and are viewing the gas/mass being ejected. whatever speed you view the gas/mass being ejected at is the speed of the gas/mass relative to the rocket/you. the speed of the ejected gas/mass relative to the earth is different from the speed of the gas/mass relative to the rocket as both the mass and rocket are moving at two different velocities when observed from the earth's point of reference. however, if viewed from the rocket's point of reference, the speed of the gas/mass is viewed as the speed relative to the rocket, as whoever in the rocket is observing the gas/mass being ejected would be moving at the same velocity as the rocket, technically making them stationary, when viewed relative to the ejected mass.
i am doing this same math for missile considering the missile is cruising and neglecting the thrust used in lift,But problem is no one is using drag component here,can some one tell where I can learn what I am asking plzz thanks in advance!!
even if you did include the extra dv term in the end it multiplies with a dm term and this is a higher order term because it involves 2 infinitesimal quantities. We neglect all of those higher order contributions.
at 5.06 if dm is negative shouldnt we write +Vfuel*dm. If we write this way Vfuel is positive,dm is negative therefore -Vfdm is positive. Is the momentum of the gas negative or positive in the ground frame ?
How would this differ if the rocket takes off (i.e. accounting for gravity). Is the total thrust force the thrust minus G? How does it influence the v(t)
Hi Ninja of physics its an honor to virtualy meet you can you answer my question please , what if I just add momentums but with signs included ... i mean I take dm as positive and velocity to the right as positive ... then i would have .... MV = ( M - dm)(V+dv) + (V - Vex)dm and the result is Vex*dm = M*dv , the only thing that changed from the original is the sign but in this case dm is positive , but when I integrate I feel awkward because dm is an increase from 0 to the entire mass that is ejected ...... is my observation right ? can i solve problems doing it that way ?
this guy doesn't even understand that the rocket & exhaust are part of the same system, so he's essentially defining COM as m1*v1=m1*v1. then when he wants the rocket to accelerate past its exhaust velocity (which is a provable physical impossibility), he redefines COM as m1*v1=(m1*v1+m1*v1). a simple free body diagram proves he is spamming pseudoscience.
What i see in ur observation everything is right except the case of adding momentum of rocket and fuel's while they are moving in opposite directions. Also the case vex is relative velocity in my case the velocity of fuel becomes vf=vex-v.
@@theglobalsuccess1765 dont pay any attention to papa legba. He is literally an imbecile, and could not pass a high school physics exam if his life depended upon it For EXAMPLE, he has said that C4 would NOT even explode in a vacuum. AND he does not know that combusting hydrogen and oxygen makes water.
thereis gravity in space,its miss conception that is weightless sensation mean as no gravity in object because the escape velocity of it and pull force gravity of earth in balance.
The derivation is mathematically correct, and the result is right but the justification for writing (-Vf) rather than (+Vf) at (4:44) is wrong. The velocity of the fuel mixture with respect to the external observer can easily have the same direction as the velocity of the rocket itself which is why it should be written with a plus sign (+Vfuel) and the term (V-Vex) determines the direction of this velocity. This mistake would have easily been avoided if the vector notation had been used. The reason why there in fact SHOULD be a minus there sign has to do with dm. Actually, the dm that appears in the term (m+dm) is NOT the same as the one in the term Vfuel*dm. While the first one denotes the mass DECREASE of the rocket, the second one is the mass INCREASE of the exhaust cloud. Their absolute value is the same but the signs must be different. So, this equation should have the following form: mV = (m+dm)(V+dV)+Vfuel*(-dm) This is of course mathematically equivalent but has a coherent physical justification behind it. I encourage the Physics Ninja to respond to this comment and justify his reasoning :).
Fascinating. Great video. Is it possible to calculate change in velocity if you don't know the exhaust velocity but do know the engine's thrust and change in mass? For example, an engine consumes fuel at a constant rate and produces a constant 100kN for 60 seconds. The initial mass is 50 tons and its final mass is 30 tons. It seems F=ma (a=F/m) could be used with some careful math. F=ma itself, though, yields two different results depending on what is averaged mass or acceleration (acceleration for average mass a=100/40=2.5 but average acceleration t(0) & t(60) = average(2.00 & 3.33) = 2.67.
Based on your calculations, what is the minimum required acceleration for a vertical rocket-launch??? Also, what is the minimum TWR for a similar rocket??? Thanks. K.S.
CLEARLY, gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites (ON BALANCE); as the stars AND PLANETS are POINTS in the night sky. Consider TIME (AND time dilation) ON BALANCE. By Frank DiMeglio
Random person on RUclips.
You saved me.
thk so much!! Clearly and fully explained, perfect for teaching!! when´s the next step...? Jet engines!!
Thank you so much… you’ve solved so many problems to me, you something like a physics god for me from now ❤
10:45 did you forget to divide by m to get acceleration? correct me if I'm wrong.
Oops, thanks for pointing that out. Yes there should be an m in denominator.
@@PhysicsNinja there should be a letter 'h' in the word 'thrust' too. you're not very smart, are you?
@@papalegba6759 You are just a sad little man. Does that comment you feel smarter?
@@Adeloye1000
@@papalegba6759
Totally not learning this for fun
at the end of 90 secs, the rate of change of mass falls by 16.67% compared to at the end of 1 sec; therefore the average acceleration falls by the same amount. If you want to calculate the final velocity using instantaneous acceleration we have to numerically integrate at intermediate points, say every sec, to get an answer closer to 3466 m/s. You will get a velocity much lower if the final acceleration of 20.83 m/sec^2 is used in v = u + at.
i think that at 10:45 you should hvae made it 1/m so when you intergrated you could have got the natural log of M due to log laws. also what i love to do is take the inverse function of log which is e and subtract mf/m0 to the other side and bring Vf/V0 to the other side. otherwise great video!!
10:47 you forgot to divide by m on the right hand side for the acceleration on the left
You sir are a life-saver!
Pls how was the formula for vfuel derived
Quick question. I thought I knew what "relative to the rocket means". Is it the speed the gas moves because the rocket is moving, or is it just the speed used to push the rocket? Lastly, it's hard for me to understand how that the speed relative to the earth is different from the speed relative to the rocket. I'm wondering what makes them different 🤔
the speed of the ejected gas/mass relative to the rocket means the speed of the mass if observed by someone present in the rocket. imagine you are in the rocket, traveling at the same speed as the rocket, and are viewing the gas/mass being ejected. whatever speed you view the gas/mass being ejected at is the speed of the gas/mass relative to the rocket/you.
the speed of the ejected gas/mass relative to the earth is different from the speed of the gas/mass relative to the rocket as both the mass and rocket are moving at two different velocities when observed from the earth's point of reference. however, if viewed from the rocket's point of reference, the speed of the gas/mass is viewed as the speed relative to the rocket, as whoever in the rocket is observing the gas/mass being ejected would be moving at the same velocity as the rocket, technically making them stationary, when viewed relative to the ejected mass.
i am doing this same math for missile considering the missile is cruising and neglecting the thrust used in lift,But problem is no one is using drag component here,can some one tell where I can learn what I am asking
plzz thanks in advance!!
hi, please could you explain at 3.55 why Vfuel = V-Vex, instead of Vfuel = V+dv-Vex ?
even if you did include the extra dv term in the end it multiplies with a dm term and this is a higher order term because it involves 2 infinitesimal quantities. We neglect all of those higher order contributions.
@@PhysicsNinja great, thank you!
How can dm
dm is a small difference in mass (m_final-m_intial) and will be a negative number because the m_final is smaller then m_initial
yay rocket science
@10:43 There is a slight mistake. you forgot to divide by the mass in the acceleration equation. Thanks for helping alsways!
Yes, sorry about that
at 5.06 if dm is negative shouldnt we write +Vfuel*dm. If we write this way Vfuel is positive,dm is negative therefore -Vfdm is positive. Is the momentum of the gas negative or positive in the ground frame ?
Awesome explanation!
How would this differ if the rocket takes off (i.e. accounting for gravity). Is the total thrust force the thrust minus G? How does it influence the v(t)
could you please explain how you turn the expression to positive in 12:53 ?thanks
he flipped the fraction within the ln() from m/m0 to m0/m
11:26 why do you need it to be a prime
Hi Ninja of physics its an honor to virtualy meet you can you answer my question please , what if I just add momentums but with signs included ... i mean I take dm as positive and velocity to the right as positive ... then i would have .... MV = ( M - dm)(V+dv) + (V - Vex)dm and the result is Vex*dm = M*dv , the only thing that changed from the original is the sign but in this case dm is positive , but when I integrate I feel awkward because dm is an increase from 0 to the entire mass that is ejected ...... is my observation right ? can i solve problems doing it that way ?
this guy doesn't even understand that the rocket & exhaust are part of the same system, so he's essentially defining COM as m1*v1=m1*v1. then when he wants the rocket to accelerate past its exhaust velocity (which is a provable physical impossibility), he redefines COM as m1*v1=(m1*v1+m1*v1). a simple free body diagram proves he is spamming pseudoscience.
What i see in ur observation everything is right except the case of adding momentum of rocket and fuel's while they are moving in opposite directions. Also the case vex is relative velocity in my case the velocity of fuel becomes vf=vex-v.
@@theglobalsuccess1765 dont pay any attention to papa legba. He is literally an imbecile, and could not pass a high school physics exam if his life depended upon it
For EXAMPLE, he has said that C4 would NOT even explode in a vacuum.
AND he does not know that combusting hydrogen and oxygen makes water.
THANKS.
Nice
Where’s the mass at the equation 1 (acceleration equation)
Oops, forgot it
Great 💗💙💬
10:17 smoothh
It'd have been more American if you shooted this video itself
Wait...
thereis gravity in space,its miss conception that is weightless sensation mean as no gravity in object because the escape velocity of it and pull force gravity of earth in balance.
It was all fun and games until calculus showed up
Trust huh😂
wow! You discovered an equation for Trust! Great!
The derivation is mathematically correct, and the result is right but the justification for writing (-Vf) rather than (+Vf) at (4:44) is wrong.
The velocity of the fuel mixture with respect to the external observer can easily have the same direction as the velocity of the rocket itself which is why it should be written with a plus sign (+Vfuel) and the term (V-Vex) determines the direction of this velocity. This mistake would have easily been avoided if the vector notation had been used.
The reason why there in fact SHOULD be a minus there sign has to do with dm. Actually, the dm that appears in the term (m+dm) is NOT the same as the one in the term Vfuel*dm. While the first one denotes the mass DECREASE of the rocket, the second one is the mass INCREASE of the exhaust cloud. Their absolute value is the same but the signs must be different. So, this equation should have the following form:
mV = (m+dm)(V+dV)+Vfuel*(-dm)
This is of course mathematically equivalent but has a coherent physical justification behind it.
I encourage the Physics Ninja to respond to this comment and justify his reasoning :).
Fascinating. Great video. Is it possible to calculate change in velocity if you don't know the exhaust velocity but do know the engine's thrust and change in mass? For example, an engine consumes fuel at a constant rate and produces a constant 100kN for 60 seconds. The initial mass is 50 tons and its final mass is 30 tons. It seems F=ma (a=F/m) could be used with some careful math.
F=ma itself, though, yields two different results depending on what is averaged mass or acceleration (acceleration for average mass a=100/40=2.5 but average acceleration t(0) & t(60) = average(2.00 & 3.33) = 2.67.
Based on your calculations, what is the minimum required acceleration for a vertical rocket-launch??? Also, what is the minimum TWR for a similar rocket??? Thanks. K.S.
It's based upon your rocket weight
To launch a rocket, Total thrust generated should be 30% more to take your rocket in air
@@harshitjajoria9689 30% more than what???
@@kareemsalessi 30% more weight thrust
Like if your rocket's total weight is 70 kg so you need a thrust of 100 kg thrust for take off 👍
@@harshitjajoria9689 According to WHAT ???
Please give the definition of exhaust speed again in details.
very clear
Glad to hear that
more so use leibniz derivative notation for clarity?
Great explanation. Thanks.
thank you sir
This is great
CLEARLY, gravity AND ELECTROMAGNETISM/energy are linked AND BALANCED opposites (ON BALANCE); as the stars AND PLANETS are POINTS in the night sky.
Consider TIME (AND time dilation) ON BALANCE.
By Frank DiMeglio
The idea of positives and negatives is scarier than pennywise
great and the best teacher ever.
How mv/mv=0 it should be 1
Can I ask something ?Why the velocity of the rocket changes from v to v+dv instead of remaining v after the fuel burned ?
The thrust force, which is due to ejection of the gases...gives the rocket an acceleration, and therefore, an increment in its speed!