A plane shoots a bullet -- energy at different frame?

[lemd]

Let say a plane flights at 1km/s, shoots a 2kg projectile at 1km/s, so to the Earth, the projectile flights at 2km/s with 2*2*2/2 = 4MJ.

The same gun when fixed on Earth, shoots the same 2kg projectile at 1km/s, achieves 2*1*1/2 = 1MJ

Assume that the energy comes from burning propellant or electrical energy (railgun), how can the gun on the plane does more work than the same gun fixed on Earth? Is the additional energy from the plane?

How can energy transfer from plane to the projectile? I.e. energy is transferred by force, and the pressure that pushes the projectile is the same in both cases. How can the projectile fired from a plane has higher energy?

 

[Orodruin]

Energy is not frame invariant and neither is work done. In order for the energy conservation laws to hold, you must describe everything in the same frame. However, energy is going to be conserved in both cases as long as you also include the recoil energy of whatever you fixed the gun to.

 

[PeroK]

Let say a plane flights at 1km/s, shoots a 2kg projectile at 1km/s, so to the Earth, the projectile flights at 2km/s with 2*2*2/2 = 4MJ.

The same gun when fixed on Earth, shoots the same 2kg projectile at 1km/s, achieves 2*1*1/2 = 1MJ

Assume that the energy comes from burning propellant or electrical energy (railgun), how can the gun on the plane does more work than the same gun fixed on Earth? Is the additional energy from the plane?

How can energy transfer from plane to the projectile? I.e. energy is transferred by force, and the pressure that pushes the projectile is the same in both cases. How can the projectile fired from a plane has higher energy?

Because of conservation of momentum, the plane will lose speed. The faster the plane is flying, the more energy this represents. The plane is so large that it is difficult to see. But, if the gun is mounted on the plane and is free to recoil, you can see what is happening more easily:

On the ground, the bullet travels at v and the gun recoils at u, say. That represents a total gain of kinetic energy from the firing mechanism.

On the plane, the gun recoils at u and this represents a loss of KE in the gun. If you add up the loss of KE in the gun and the energy from the firing mechanism this equals the gain in KE of the bullet.

 

[DEvens]

Indeed, how can it happen that way?

The answer is, you are forgetting something about conservation laws. Energy is not the only thing conserved. So also is momentum. And so you need to extend your calculation a little. The plane experiences re-coil. So it has to change its energy.

So make things simple. Make the plane 200 kg. So it will get a recoil of 1/100 the forward speed of the bullet.

So the starting kinetic energy of the whole system is (remember the whole system starts at v)

1/2 (M + m) v^2

Here v can be 0 if the plane starts stationary relative to us, or 1000 m/s if it starts at 1 km/s. Or even negative if we start out running ahead of the plane.

And the bullet winds up going v + 1km/s, but the plane winds up going v - 10 m/s

So the energy after is

1/2 M (v-10)^2 + 1/2 m (v+1000)^2

And if you use the fact that M = 100 m, where m=2, you find that the change in total kinetic energy turns out to be

1/2 m (10,000 + 1,000,000)

Notice that v has disappeared. That is, the change in total kinetic energy is the same no matter how fast the plane starts out when it shoots the bullet.

So for homework:
Do the general case where the mass of the bullet and the mass of the plane can be anything, and the speed the bullet is fired can be anything. Show that the change in kinetic energy does not depend on the starting velocity of the plane.

And for extra advanced credit:
Do the relativistic case where the plane is replaced by a particle moving at some fraction of the speed of light, and it breaks up into two decay products with smaller total rest mass. Do the simple case where the decay products move forward and backward relative to the motion of the initial particle. Show that total energy is conserved, as well as total momentum. Use all of he correct special relativistic formulas.

 

[russ_watters]

Let say a plane flights at 1km/s, shoots a 2kg projectile at 1km/s, so to the Earth, the projectile flights at 2km/s with 2*2*2/2 = 4MJ.

The same gun when fixed on Earth, shoots the same 2kg projectile at 1km/s, achieves 2*1*1/2 = 1MJ

Assume that the energy comes from burning propellant or electrical energy (railgun), how can the gun on the plane does more work than the same gun fixed on Earth? Is the additional energy from the plane?

How can energy transfer from plane to the projectile? I.e. energy is transferred by force, and the pressure that pushes the projectile is the same in both cases. How can the projectile fired from a plane has higher energy?

Frankly, I think you understand it just fine except for the very last bit: that force for the extra kinetic energy came from the plane's engines when it took off.

 

[DEvens]

Frankly, I think you understand it just fine except for the very last bit: that force for the extra kinetic energy came from the plane's engines when it took off.

Let's be careful. The engines gave the bullet 1 MJ on takeoff. But the firing of the gun mechanism in flight moves it from 1km/s to 2 km/s, so it gives it 3 MJ to total to 4MJ. Meaning the engines did not give the bullet the extra kinetic energy. The engines gave it only 1 MJ. Firing it gave it 3 MJ when fired in flight compared to 1 MJ when fired on the ground.

To understand this you have to do the recoil analysis presented up-thread. That's the only way you can conserve energy.

 

[A.T.]

that force for the extra kinetic energy came from the plane's engines when it took off.

It has nothing to with the plane's engines, just with reference frames. When you view the plane standing on the ground from a reference frame where the plane moves, it's bullets will also have more energy than in the frame of the plane. But the engines were never on.