How Does 1J of Energy Propel a 1kg Mass in Space?

[guss]

Let's say we have a spaceship with a 1kg mass, and 1J stored in a battery or something. I am going to explore two things we can do with this setup to propel the craft. The first one is simply shooting the mass itself out the back.

If we shoot the mass out the back by using that 1J of energy, our craft would gain 1J in the opposite direction, right?

What if we converted our 1kg mass to light energy, and shot that out the back with a laser, though? For simplicity, let's say we only shoot a single photon containing all our energy. From energy-mass equivalence, the total energy of the photon would be $c^{2}+1$ joules.

This is where I get confused. I'm not sure if the total energy the ship would gain in the opposite direction of the photon should be $c^{2}+1$ joules or just 1 joule. I'm thinking it could be the former because we are sending $c^{2}+1$ joules in the opposite direction. I'm also, however, thinking it could be the latter because the relativistic mass of the photon would be $\frac{c^{2}+1}{c^{2}}$, and because the total energy gained should match up with the first case scenario where we just shoot the 1kg mass out.

Can someone help? Thanks!

 

[Drakkith]

If you have 1 joule of available energy to propel the ship, then you will never be able to have the ship gain more than 1 joule of energy in velocity. Since we have no way of converting 1 kg of matter into energy, nor use 100% of that energy, we can't really talk about it. Instead just have a different scenario with about 1x10^22 joules of energy instead of 1, but no 1 kg mass to use. The photon's maximum energy would be half of 1x10^22 joules, as that is the measure of energy. For a photon you can determine its energy using E=(hc)/wavelength, where h is planks constant and c is the speed of light.
I THINK the momentum the ship would gain would be the other half of 1x10^22, but I'm not certain.

 

[K^2]

If we shoot the mass out the back by using that 1J of energy, our craft would gain 1J in the opposite direction, right?

No. Only a fraction of this energy will become kinetic energy of the craft. The rest will become the kinetic energy of the mass you expelled. And if mass you expelled is much smaller than mass of the craft, almost all of this 1J will become energy of the expelled mass, and very little will become energy of the craft.

The quantity that is conserved the way you think is momentum. If you impart 1Ns of momentum to 1kg mass, your craft will get 1Ns of momentum in the opposite direction. So long as nothing moves at relativistic speeds, the kinetic energy is related to momentum as E=p²/(2m).

When you use light to propel yourself, momentum is still conserved, but energy-momentum relation for light is completely different: E=pc. If you are trying to gain a huge amount of momentum, enough for near-speed-of-light travel, this relation is favorable, because it is linear in p. However, for typical spaceship velocities, the speed of light multiplier makes it a poor choice. Ejecting a mass ends up energetically favorable.

 

[guss]

No. Only a fraction of this energy will become kinetic energy of the craft. The rest will become the kinetic energy of the mass you expelled. And if mass you expelled is much smaller than mass of the craft, almost all of this 1J will become energy of the expelled mass, and very little will become energy of the craft.

The quantity that is conserved the way you think is momentum. If you impart 1Ns of momentum to 1kg mass, your craft will get 1Ns of momentum in the opposite direction. So long as nothing moves at relativistic speeds, the kinetic energy is related to momentum as E=p²/(2m).

When you use light to propel yourself, momentum is still conserved, but energy-momentum relation for light is completely different: E=pc. If you are trying to gain a huge amount of momentum, enough for near-speed-of-light travel, this relation is favorable, because it is linear in p. However, for typical spaceship velocities, the speed of light multiplier makes it a poor choice. Ejecting a mass ends up energetically favorable.

Thanks, the first things you explained seem obvious now, I made some really stupid mistakes. But do you mean "almost all of this 1J will add to the velocity of the expelled mass, and very little will add to the velocity of the craft"? Because I think both impulses would be equal but opposite, it's just the spacecraft would probably be more massive so it would take more energy to accelerate.

I also still don't see how simply ejecting the mass could lead to more energy. Let's say we have a stationary 1000kg ship.

In the mass ejection scenario:
We have a 1kg mass and 1J. When the mass is ejected, half the energy goes to the ship, half to the mass. So the new velocity of the ship can be found with E = (mv^2)/2, replacing m with 1000kg and E with .5J. Solving for v I got .032 m/s for the velocity of the ship.

In the light propulsion scenario:
We have (c^2+1)J. I am confused on what to do next, though. Does half the (c^2+1)J contribute to the energy of the ship, and half to the photon? I don't know. If I assume it does, the energy added to the ship will be ((c^2+1)/2)J. Using E = (mv^2)/2, and replacing E with ((c^2+1)/2)J and m with 1000kg, I got a huge number.

So, what did I do wrong? I still don't understand how ejecting the mass could add more energy to the spacecraft than the photon. Maybe the problem lies with me working with energy and not momentum, I don't know.

 

[guss]

Still looking for an answer if anyone has one.

 

[JeffKoch]

When the mass is ejected, half the energy goes to the ship, half to the mass..

No. The ship has the same momentum as the ejected mass, since in the ship's frame of reference the initial momentum is zero - actually to be precise, since momentum is a vector quantity, the momentums after ejection have equal magnitudes but opposite directions, so the vector sum is always zero since it started as zero. Therefore the ejected mass has 1000x the velocity as the ship, and you can work out the ratios of the two energies - they clearly are far from equal.