# B I need an equation for relativistic exhaust velocity (Ve)

1. Jul 6, 2017

### BitWiz

Does anyone have an equation that -- given energy applied to a mass over a distance -- will give me an (ideal) final velocity of the mass?

If I direct a gigajoule to accelerate a gram over a distance of a meter, I obviously get into real trouble with c using conventional equations.

I've found equations that apply Lorentz -- but they seem to do it as an adjustment after a raw, conventional velocity has been obtained, using this final, conventional velocity to plug into Lorentz. That can't work in this case. I'm looking for a function Ve( joules, mass, distance ) that works for all non-negative parameter values, and takes relativity into account. Can you help?

Thank you very much!

2. Jul 6, 2017

### Staff: Mentor

Take the ratio of total energy to rest energy: that gives you the relativistic $\gamma$ factor. Then invert the formula for $\gamma$ in terms of $v$ to get $v$.

3. Jul 6, 2017

### Staff: Mentor

You should check your math first. Hint: what is the rest energy of one gram of mass, in gigajoules?

4. Jul 6, 2017

### Staff: Mentor

The distance is actually irrelevant to the final velocity. It is only relevant to determining the acceleration required to achieve that velocity. The final velocity only depends on the total energy applied.

5. Jul 6, 2017

c2 / 1000 ?

6. Jul 6, 2017

### Staff: Mentor

If you calculate this numerically and convert to gigajoules, yes. What do you get?

7. Jul 6, 2017

### BitWiz

So, v = v' * rest_mass / (Rest_mass + Jo
~898756 Gj ?

8. Jul 6, 2017

### Staff: Mentor

No. Try starting from what I said in post #2.

I think you're too high by an order of magnitude.

9. Jul 6, 2017

### BitWiz

Whoops.
If a joule = 1 (kg) / c2, then 1kg = c2j, and a gram is 1000th of that?

If c is ~ 3e8, then c2 is 9e16j for a kg. For a gram, 9e13j. For a gigajoule, 9e4?

10. Jul 6, 2017

### BitWiz

You're saying that gamma is 1 + ( mass-equivalent-of-energy-added / rest mass ) ?

Thanks!

11. Jul 6, 2017

### Staff: Mentor

That's one way of putting it, yes. Or you could just do everything in energy units; it works out the same either way.