Is interstellar travel impossible due to kinetic energy magnitude?

[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.