Is interstellar travel impossible due to kinetic energy magnitude?

[BruceW]

Hi, Bruce,

No, I'm not advocating the pessimistic position of the Joint Propulsion people, and in fact, I'm pretty upset about it. I really want my own starship, and how am I going to get one if the propulsion people have already given up?!

I admit to trying to "hide" my feelings since what I really want is unbiased information, but perhaps I went too far advocating the propulsion expert's position, which -- if the posts I've seen so far both here on PF and elsewhere represent the trend -- is a majority opinion. However, "Poppycock!" is not a reasoned counterargument, so I need bullets. Pass the ammo, please.

Chris

haha, awesome. Well, they were definitely talking about the near future. No-one can say anything for certain about technology in the far future. I think the bad news is that neither of us will be getting a starship anytime in the near future.

edit: to anyone who says that interstellar travel will never be possible (even in the far future), just say "Any sufficiently advanced technology is indistinguishable from magic" (Arthur C. Clarke).

 

[BitWiz]

Hi, Nugatory,

...
Yes, because no matter which observer you use to base the kinetic energy calculations on, you will get the same answer for the fuel burn and amount of energy that has to be generated by the rocket's propulsion system to send the rocket on its journey. ...

If I use a precise quantity of energy to accelerate an ideal rocket, ignoring gravity et al, then it will attain a final velocity that precisely corresponds to its measured KE from its zero velocity origin. Is this correct?

If an observer in the rocket integrates proper, measured acceleration with proper time, will he come up with the same final velocity?

Thanks,

Chris

 

[bob012345]

When we say "the frame of <something>" or "its frame" where "it" is some thing, that's just a convenient and somewhat sloppy shorthand for the more precise "a frame in which <something> has velocity zero".

Thus, the kinetic energy of the rocket is perfectly well defined in both the Earth's frame and the rocket's frame, just as the kinetic energy of the Earth is defined in both frames. All objects can always be described in any and all frames, and the dynamical quantities such as momentum and kinetic energy are defined for all objects no matter which frame you choose to do the arithmetic in. The numerical values of some of these (momentum and kinetic energy, for example) may be frame dependent, but they are not undefined.

The kinetic energy in the Earth's frame is well defined and makes sense but the real question is does that accurately represent the energy actually expended from the ships point of view. I argued this with a friend who says yes, it does. I'm still struggling with the concept.

A simple example. Consider firing a bullet from a gun with velocity v. Call its kinetic energy K. Now, consider a big plane flying at the same v. Fire the gun in the direction of motion inside the plane. The kinetic energy wrt the planes frame is K. But the bullet is going at 2v wrt the ground if velocities add linearly as they do so the kinetic energy as computed from the ground is 4K. It started with K wrt the ground, being on board the plane moving at v yet the chemical energy released was the same in both cases. So the question is how did the bullet end up with 4K of kinetic energy wrt the ground frame when only K of additional chemical energy was added to its initial energy of K wrt the ground? I'm stumped.

 

[jbriggs444]

So the question is how did the bullet end end up with 4K of kinetic energy wrt the ground frame when only K of additional chemical energy was added to its initial energy of K wrt the ground? I'm stumped.

What about the recoil of the gun? If you choose to use a frame where the gun is moving, the energy subtracted from it by recoil is non-negligible and accounts for the discrepancy you see.

 

[bob012345]

What about the recoil of the gun? If you choose to use a frame where the gun is moving, the energy subtracted from it by recoil is non-negligible and accounts for the discrepancy you see.

The recoil would make the relative velocity wrt the ground frame less, not more. I could devise an equivalent frame such as another planet moving past us such that the recoil effects would be the same as in the ground frame. Basically, I think we can ignore the recoils. Treat it as an idealized problem please.

 

[jbriggs444]

The recoil would make the relative velocity wrt the ground frame less, not more. I could devise an equivalent frame such as another planet moving past us such that the recoil effects would be the same as in the ground frame. Basically, I think we can ignore the recoils. Treat it as an idealized problem please.

This is incorrect. Recoil is the source of the discrepancy. The recoil of the gun makes the gun's energy less, not more. That's what balances the books with the extra energy that shows up in the bullet.

 

[bob012345]

This is incorrect. Recoil is the source of the discrepancy. The recoil of the gun makes the gun's energy less, not more. That's what balances the books with the extra energy that shows up in the bullet.

I don't get that especially if the recoil is not going to change the velocity of the plane much since it has an almost infinite mass by comparison. Also, I showed a scenario where the recoil is the same in both frames.

The point not clear in you answer is what is the actual energy it takes to make the bullet go at basically 2v wrt the ground. Kinetic energies don't subtract. Thanks.

 

[jbriggs444]

I don't get that especially if the recoil is not going to change the velocity of the plane much since it has an almost infinite mass by comparison.

Infinite mass times an infinitesimal change in velocity gives a finite change in energy. The answer I have given twice now is still correct. Do the math.

 

[bob012345]

Infinite mass times an infinitesimal change in velocity gives a finite change in energy. The answer I have given twice now is still correct. Do the math.

I think you are saying that the whole plane slowed down due to recoil just enough to 'balance' the energy so the total amount of energy of the system of plane, gun and bullet is only increased by the the amount of energy the gun powder released. The bullet is going about 2v and has about 4K but borrowed some of it from the planes vast reserves.

 

[jbriggs444]

I think you are saying that the whole plane slowed down due to recoil just enough to 'balance' the energy so the total amount of energy of the system of plane, gun and bullet is only increased by the the amount of energy the gun powder released. The bullet is going about 2v and has about 4K but borrowed some of it from the planes vast reserves.

Yes, that is correct.

 

[Nugatory]

I don't get that especially if the recoil is not going to change the velocity of the plane much since it has an almost infinite mass by comparison.

Work through the math - the recoil doesn't change the velocity by much, but it changes it by enough. I really cannot stress enough how important it is that you actually work through the math - until and unless you've worked through the math you won't really understand how the physics works here.

Also, I showed a scenario where the recoil is the same in both frames.

It will be clearer if you work through the math yourself. The change in the plane's speed and hence the momentum change from the recoil is the same in both frames. The energy change is not the same and cannot be. Let's say that the plane is moving at speed $u$ before the gun is fired, and the recoil changes its speed by an amount $\Delta{v}$. The kinetic energy before was $E_0=m(u^2)/2$, the kinetic energy afterwards is $$E_1=m(u-\Delta{v})^2/2=m(u^2-2u\Delta{v}+\Delta{v}^2)/2\approx{m(u^2-2u\Delta{v}})/2=E_0-2u\Delta{v}$$
(The approximation is allowed because $\Delta{v}$ is small compared with $u$ - if you work through the math you will see that for reasonable assumptions about the mass of the plane and of the bullet $\Delta{v}^2$ will be a one part in a billion rounding error).
So the change in kinetic energy is $E_1-E_0=-2u\Delta{v}$, and that's going to be different for different values of $u$ - it is frame-dependent.

Note also that the frame in which the plane is at rest is the only one in which the change in the plane's kinetic energy doesn't matter. It doesn't matter because in that frame $u$ is zero so $2u\Delta{v}$ is also zero.

Did I mention that you should work through the math yourself? No? I'm sorry, that was an oversight. I certainly meant to say it.

[Edit: enough people have looked at this post already that I'm not going to edit it... But if you check my algebra you'll see that I lost a factor of 1/2 and the $\Delta{E}$ is $-u\Delta{v}$ not $-2u\Delta{v}$. This doesn't change the basic point of the calculation though]

 

[sophiecentaur]

Forget the recoil. It has a vanishingly small effect on the energy of the plane/gun. Extra energy (in the Earth frame of reference was supplied when the gun, bullet and plane were accelerated by the chemical energy in the fuel at takeoff.
The force times distance that provided the ke of the bullet on the plane PLUS the force times distance during the firing of the gun is the total ke of the bullet.
The sneaky part is that the distance traveled by the bullet during firing includes the forward distance traveled by the gun and the plane during the firing. (Earth frame) in the given example, that distance is as great as the barrel length.

 

[jbriggs444]

Forget the recoil. It has a vanishingly small effect on the energy of the plane/gun.

Energy is a frame-relative concept. Do the math using first a frame in which the plane starts at rest, next a frame in which the bullet ends at rest and finally a frame in which a convenient cosmic ray is at rest. You will find that the recoil is quite important.

Edit: That is, quite important if you want to account for where the energy in the gunpowder went.

 

[sophiecentaur]

The "four times energy" in the original question is in the Earth reference frame. That is sufficient to deal with.
The sharing of energy between bullet and gun depends on relative masses and tends to 100% for the bullet as the bullet mass approaches zero. That is the simplest case so I chose to go with that. The ke added during the firing is three times the ke of the unfurled bullet. Integral under the vt graph.

 

[jbriggs444]

The "four times energy" in the original question is in the Earth reference frame. That is sufficient to deal with.
The sharing of energy between bullet and gun depends on relative masses and tends to 100% for the bullet as the bullet mass approaches zero.

Utter rubbish.

The energy expended by gunpowder in a plane flying at a velocity of v which results in a bullet flying at velocity 2v is NOT distributed 100% to the bullet. It is distributed 300% to the bullet and -200% to the gun [in the frame in which the velocities are as given].

 

[bob012345]

Work through the math - the recoil doesn't change the velocity by much, but it changes it by enough. I really cannot stress enough how important it is that you actually work through the math - until and unless you've worked through the math you won't really understand how the physics works here.

It will be clearer if you work through the math yourself. The change in the plane's speed and hence the momentum change from the recoil is the same in both frames. The energy change is not the same and cannot be. Let's say that the plane is moving at speed $u$ before the gun is fired, and the recoil changes its speed by an amount $\Delta{v}$. The kinetic energy before was $E_0=m(u^2)/2$, the kinetic energy afterwards is $$E_1=m(u-\Delta{v})^2/2=m(u^2-2u\Delta{v}+\Delta{v}^2)/2\approx{m(u^2-2u\Delta{v}})/2=E_0-2u\Delta{v}$$
(The approximation is allowed because $\Delta{v}$ is small compared with $u$ - if you work through the math you will see that for reasonable assumptions about the mass of the plane and of the bullet $\Delta{v}^2$ will be a one part in a billion rounding error).
So the change in kinetic energy is $E_1-E_0=-2u\Delta{v}$, and that's going to be different for different values of $u$ - it is frame-dependent.

Note also that the frame in which the plane is at rest is the only one in which the change in the plane's kinetic energy doesn't matter. It doesn't matter because in that frame $u$ is zero so $2u\Delta{v}$ is also zero.

Did I mention that you should work through the math yourself? No? I'm sorry, that was an oversight. I certainly meant to say it.

Thanks, I did the math, though probably not as formally as you would. While I agree what you and jbriggs are saying, the example I picked does not convey the dilemma fully. I'll try and think of another.

 

[Nugatory]

Thanks, I did the math, though probably not as formally as you would. While I agree what you and jbriggs are saying, the example I picked does not convey the dilemma fully. I'll try and think of another.

There's much more than what I posted, that was just a sort of hint to get you started. So if you haven't done it as formally as I did... you haven't done it at all. A few more hints to get you started:
0) Is choosing a value for $u$ equivalent to choosing a frame? That's not a trick question, it's more establishing a solid anchorage.
1) How do you determine the $\Delta{v}$ for the plane and for the bullet as a function of their masses?
2) Is $\Delta{v}$ for the plane and the bullet independent of $u$? If so, why? If not, what is the relationship between it and $u$?
3) Choose two frames. In both of them, use conservation of energy and momentum to calculate the final speeds of the plane and the bullet. Then demonstrate that if the amount of energy released by the propellant charge is the same in both frames (as you have rightly insisted it must be) and equal to the total increase in kinetic energy then these speeds are related by the Galilean transform, meaning that the speeds in one frame all differ from the speeds in the other frame by the same constant amount.

the example I picked does not convey the dilemma fully. I'll try and think of another.

Don't do that yet, you'll be wasting your time. On the one hand, until you have correctly and completely analyzed this problem you won't be prepared to take on a more complicated example. On the other hand, once you have this problem down cold you'll find that the apparent dilemma disappears even in the more complicated examples.

 

[bob012345]

There's much more than what I posted, that was just a sort of hint to get you started. So if you haven't done it as formally as I did... you haven't done it at all. A few more hints to get you started:
0) Is choosing a value for $u$ equivalent to choosing a frame? That's not a trick question, it's more establishing a solid anchorage.
1) How do you determine the $\Delta{v}$ for the plane and for the bullet as a function of their masses?
2) Is $\Delta{v}$ for the plane and the bullet independent of $u$? If so, why? If not, what is the relationship between it and $u$?
3) Choose two frames. In both of them, use conservation of energy and momentum to calculate the final speeds of the plane and the bullet. Then demonstrate that if the amount of energy released by the propellant charge is the same in both frames (as you have rightly insisted it must be) and equal to the total increase in kinetic energy then these speeds are related by the Galilean transform, meaning that the speeds in one frame all differ from the speeds in the other frame by the same constant amount.

Don't do that yet, you'll be wasting your time. On the one hand, until you have correctly and completely analyzed this problem you won't be prepared to take on a more complicated example. On the other hand, once you have this problem down cold you'll find that the apparent dilemma disappears even in the more complicated examples.

Thanks.
0) I think yes.
1) I use the center of mass frame of plane and bullet.
2) I think so.
3) I'll do that.

 

[Nugatory]

The sneaky part is that the distance traveled by the bullet during firing includes the forward distance traveled by the gun and the plane during the firing. (Earth frame) in the given example, that distance is as great as the barrel length.

You are right that that provides an alternative way of calculating the kinetic energy that the bullet acquires (to very good approximation and exactly in the limit as the ratio of the bullet's mass to the plane's mass approaches zero). However, you aren't helping bob012345 with his dilemma because the frame-dependent kinetic energy imparted to the bullet will not in general be equal to the frame-independent chemical energy released by the explosion - and that's the essence of his dilemma.

You can save conservation of energy by saying that the airplane has kinetic energy (it does, in every frame except the one in which it is at rest, and that's the frame where the bullet's kinetic energy is equal to the energy released by the explosion so there's no discrepancy to explain) and some of that kinetic energy is transferred to the bullet. However, that implies that the plane is losing kinetic energy, which means that it is losing speed... Hence jbrigg's and my insistence that the change in the plane's speed, although small, is essential to completely and correctly analyzing the problem. (Another hint that it is essential is that momentum is not conserved when you ignore it).

 

[sophiecentaur]

1) I use the center of mass frame of plane and bullet.

That's your choice and it will tell you the KE of the bullet relative to that CM. But how is that relevant when dealing with any collision the bullet might have with anything else? The choice of a frame of reference is surely best made with reference to the situation.

Utter rubbish.

The energy expended by gunpowder in a plane flying at a velocity of v which results in a bullet flying at velocity 2v is NOT distributed 100% to the bullet. It is distributed 300% to the bullet and -200% to the gun [in the frame in which the velocities are as given].

That cannot be right because the relative masses are what governs the share of the energy. Your "utter rubbish" statement clearly didn't consider the small matter of Momentum. The velocity of the gun / plane can end up as near zero as you choose, relative to the CM. Even with an isolated gun, the ratio of masses will be up to 100:1 so that will imply that the bullet will get the lion's share of the chemical energy.

 

[Nugatory]

Even with an isolated gun, the ratio of masses will be up to 100:1 so that will imply that the bullet will get the lion's share of the chemical energy.

The bullet will get the lion's share of the chemical energy, but depending on your choice of frame that can be an arbitrarily small fraction of the increase in the bullet's kinetic energy. As I pointed out above, the increase in the bullet's kinetic energy is $u\Delta{v}$; that quantity can be arbitrarily large compared with the chemical energy.

We're digressing from the original question to such an extent that the thread should be closed. As always, PM me or any other mentor if you need it reopened to add t the discussion.