Energy to achieve constant acceleration - F=ma and 1/2mv^2

[zanick]

Thanks Dale, I'm not ignoring conservation of energy, I'm just trying to understand it and apply it to this thought experiment. I seem to be ignoring the KE of the exhaust gases because I am not fully understanding the relation. Yes, I have seen it mentioned and I know the math works out...but I am still coming back to a constant thrust being applied and therefor a constant acceleration of the rocket.
I see your point though... I'm creating thrust, but the same amount of KE goes into the exhaust. but to fully understand it, are we saying you can't have constant thrust unless you have enough energy to feed both the exhaust and the rocket forces? Then there is the fact that the mass of the rocket is going down, increasing the acceleration... I am getting there , but still not quite. I am fighting the intuition that the energy is going up 2x to velocity, but not squared. help me get there so I can conceptually get it... thanks!

another way I could look at it, is if we had a fixed energy release rate feeding the rocket and exhaust, would acceleration go down proportional to velocity?

In your frustration of helping me here, you reference my attachment to my son as well as your possible lack to me, I will offer what was told to me about explaining complex ideas to others... the more you understand, the easier it is to explain in the most simple terms. When my son doesn't understand something in science, I blame myself for not knowing it well enough to simply it even further if he doesn't get it after my help... so I try again. It forces me to learn the topic better. I appreciate the effort to help me with this example...… its been a long time since I studied it , that I'm sure ill never know as much as you and the group have forgotten! So, thanks!

Yes, except as the mass of the rocket diminishes the acceleration actually increases.

Every problem that anyone has with energy conservation in a rocket is due to failure to consider the kinetic energy of the exhaust. 100%.

I encourage you to actually do the math. Work it out, including both the change in energy of the rocket and the exhaust.

In fact it is the case, and yet energy is conserved.

Yes.

You have been told multiple* times: You are neglecting the change in the KE of the exhaust. That is where the extra energy comes from.

All that remains is for you to work it out mathematically for yourself so that you can understand and believe the answers you have already received.

* I count 9 separate posts where someone has told you that you need to consider the change in KE of the exhaust and 0 posts where you did so. This is very frustrating to respondents and discourages people from helping you in the future. Put yourself in our shoes: When talking with your son, how frustrating do you find it when you have to repeat yourself literally 9 times? And presumably you are more attached to your son than we are to you.

 

[jbriggs444]

I'm creating thrust, but the same amount of KE goes into the exhaust.

No, it does not.

For a fixed fuel burn rate, you have a fixed budget of input power. The more that goes into adding to the kinetic energy of the rocket, the less is available to add to the kinetic energy of the expelled exhaust.

If the rocket is moving faster than 1/2 of its exhaust velocity, you might want to think about whether speaking of adding KE to the exhaust is sensible.

 

[zanick]

No, it does not.

For a fixed fuel burn rate, you have a fixed budget of input power. The more that goes into adding to the kinetic energy of the rocket, the less is available to add to the kinetic energy of the expelled exhaust.

If the rocket is moving faster than 1/2 of its exhaust velocity, you might want to think about whether speaking of adding KE to the exhaust is sensible.

can you explain further?
so, like with cars on earth, if power is kept constant, acceleration goes down proportional to velocity and this means in space the same is true? ….. net thrust (force) would have to go down with velocity as well, correct?

 

[jbriggs444]

can you explain further?
so, like with cars on earth, if power is kept constant, acceleration goes down proportional to velocity and this means in space the same is true? ….. net thrust (force) would have to go down with velocity as well, correct?

In space you carry your reaction mass with you. On Earth you use the road. It is an entirely different proposition. Using the one to reason about the other is not leading you toward correct conclusions.

No, thrust does not go down with increasing velocity if you carry your reaction mass with you.

 

[sophiecentaur]

The rocket is not suitable to help you understand the basics of this. If the ship were being pulled by a long rope on a winch and a constant tension were maintained them the input Power (Force X velocity) would be increasing all the time. A rocket engine would need to provide the same ever increasing Power.

 

[zanick]

I think you might be looking at it backwards: the speed of the reaction mass relative to the rocket is the same every time.

But the speed of the reaction mass relative to the starting point (Earth?) is different every time because the rocket's speed is changing. To an observer on Earth, the first reaction projectile moves to the left and the rocket moves to the right. The second reaction projectile moves to the left a little slower. After a while, the reaction projectiles are actually moving to the right even while accelerating the rocket to the right.

so kind of following the analogy .comparison, of pushing bowling balls out the back of a boat (or the rocket)…. and my comment about the fixed power providing a decreasing force and therefor acceleration as velocity goes up. . I know I was not focused on the KE of the bowling balls leaving the ship, but I still have a problem not seeing how what was confirmed to be "constant acceleration " for the constant force, not accelerating the rocket at a constant rate. I know now this can't be, but still can't understand why... can you do a energy inventory for me so I can put this one to bed? thanks!

 

[jbriggs444]

so kind of following the analogy .comparison, of pushing bowling balls out the back of a boat (or the rocket)…. and my comment about the fixed power providing a decreasing force and therefor acceleration as velocity goes up. . I know I was not focused on the KE of the bowling balls leaving the ship, but I still have a problem not seeing how what was confirmed to be "constant acceleration " for the constant force, not accelerating the rocket at a constant rate. I know now this can't be, but still can't understand why... can you do a energy inventory for me so I can put this one to bed? thanks!

@PeroK did one earlier.

But let us say that we have a rocket with mass M together with a bowling ball of mass m. Together the two have mass m+M.

Let us say that the rocket is traveling at an arbitrary velocity $v_r$.

The rocket hurls the mass m out the rear at velocity $v_e$ relative to the center of mass of the rocket+bowling ball system.

By conservation of momentum, if mass m has relative velocity $v_e$ going rightward from the center of mass, mass M must have relative velocity $v_e\frac{m}{M}$ leftward relative to the center of mass.

None of this is dependent at all on $v_r$.

You have asked for an energy accounting. How much kinetic energy has been added to the system? You anticipate that the amount of energy added to the system depends on $v_r$. Let us see whether that expectation holds up.

The initial kinetic energy of the system is given by $$KE_i = \frac{1}{2}(m+M)v_r^2$$

The final kinetic energy of the system is given by $$KE_f = \frac{1}{2}M(v_r+v_e\frac{m}{M})^2 + \frac{1}{2}m(v_r-v_e)^2$$

Let us evaluate those squares yielding $$KE_f = \frac{1}{2}M(v_r^2 + 2v_rv_e\frac{m}{M} + v_e^2\frac{m^2}{M^2}) + \frac{1}{2}m(v_r^2 - 2v_rv_e + v_e^2)$$

We can distribute the M into the left hand sum and the m into the right hand sum yielding $$KE_f = \frac{1}{2}(Mv_r^2 + 2mv_rv_e + v_e^2\frac{m^2}{M}) + \frac{1}{2}(mv_r^2 - 2mv_rv_e + mv_e^2)$$

Now the $+2mv_rv_e$ cancels with the $-2mv_rv_e$ so the result becomes $$KE_f = \frac{1}{2}(Mv_r^2 + v_e^2\frac{m^2}{M}) + \frac{1}{2}(mv_r^2 + mv_e^2)$$

Remember that we are after the delta between final and initial KE. So subtract initial KE (see formula above) yielding $$KE_f - KE_i = \frac{1}{2}\frac{m^2}{M}v_e^2 + \frac{1}{2}mv_e^2$$

The rocket's initial velocity does not enter into the result. Galilean relativity is upheld.

One way of finding the the bonus rocket energy is to look at that $2mv_rv_e$ term that appears in the left hand sum. Recall that sum represents the final rocket KE. The greater the rocket velocity, the greater that term in the sum becomes.

But there is no free lunch. The $-2mv_rv_e$ term in right handsum (the exhaust energy) exactly cancels it out.

 

[Dale]

I seem to be ignoring the KE of the exhaust gases because I am not fully understanding the relation. Yes, I have seen it mentioned and I know the math works out.

The problem is that you know the math works out only because others have told you so, you have not worked it out yourself. Nobody actually learns physics without working problems, that is why professional classes require so much homework. In the process of doing the homework you learn tremendously. Such practical use of the laws of physics is essential for understanding the laws of physics. Let me help you by proposing a simplified problem for you to work to understand the physical principles involved:

Consider a 1000 kg “rocket” that, instead of firing a continuous stream of exhaust, fires a 1 kg bowling ball using a bowling ball gun (details not essential). The gun exerts 1000 N of force on the ball (and by Newton’s 3rd the ball exerts an equal and opposite force on the rocket/gun). The barrel of the gun is long enough so that the force lasts for 1 s. Calculate:

1) the initial KE of the rocket
2) the initial KE of the ball
3) the total initial KE
4) the final KE of the above
5) the change in KE of the above
6) the change in velocity of the rocket

Assume that the initial velocity is 0, then repeat the calculations for an initial velocity of 500 m/s, 1000 m/s, and 10000 m/s.

What trend do you notice?

the more you understand, the easier it is to explain in the most simple terms.

This isn’t the issue. The terms already used are simple. The explanations you have received are clear. Understanding comes primarily through mental effort on the learner’s part. You need to work this out on your own to grasp it fully.

can you do a energy inventory for me so I can put this one to bed?

This has already been done for you 3 times now. You need to work through it yourself.

Please assess your learning style for things like your favorite hobby, etc? Are you a “hands on” or “learn by doing” type or do you learn best by reading explanations or theory about your hobby? From this thread I am guessing that you are a “hands on” learner in general. Many people are, and problems like the one I posed above are the physics version of hands on learning.

 

[Herman Trivilino]

im trying to explain the squared factor of KE, based on energy that would be required to accelerate the same from a higher velocity vs the lower velocity

The best way to do proceed would be to show him a couple examples. In each example you can have the same change in velocity, but the velocities themselves are very far apart. Like a change from 10 to 20 in one example, and in the other example from 80 to 90. Once you can get him to grasp this, then you follow the advice you've been given, like when @Dale recommended using the work-energy principle to explain.

 

[sophiecentaur]

can you do a energy inventory for me so I can put this one to bed?

That is too much like spoon-feeding, imo. I posted an alternative view (post #35) which perhaps you didn't look at because it was not precisely in your terms. It shows in a very simple way, that the Power needed (Energy per unit of time) goes up as the speed increases. Some effort may be needed on your part to link between that and the rocket question but the basic explanation is all there, without the added complication of a rocket motor.
To "put this one to bed" you may need more personal effort and to be a bit flexible in what you are prepared to accept.