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

[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.

 

[Mister T]

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.