[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]

*To*: starship-design@lists.uoregon.edu*Subject*: starship-design: Momentum*From*: Steve VanDevender <stevev@efn.org>*Date*: Fri, 25 Jul 1997 15:40:15 -0700*In-Reply-To*: <199707242232.PAA16586@watt>*References*: <199707242232.PAA16586@watt>*Reply-To*: Steve VanDevender <stevev@efn.org>*Sender*: owner-starship-design

Ken Wharton writes: > Steve, > > I'll restate the formula I just derived: > > P = [(E^2 / c^2) + 2 E M]^0.5 > > If you disagree with the way I derived this, let me know. Keep in mind > that E is kinetic energy, not total energy (total energy includes rest > mass). If you don't like the looks of it, check it out in the limits of > M->0 and M>>E/c^2; you'll see you get the momentum/energy relationship > of zero-mass and sub-relativistic objects respectively. On the surface, your equation is correct enough. I think the problem I'm having is that in a different context I derived rather convincingly that photons are the best possible reaction product, in the sense that you get the most momentum per unit of fuel if you can convert the fuel completely to photons, which is somewhat different than what you're saying. I'm going to have to go over my notes when I get home to figure out what we're differing on in more detail. Until then, I'll leave you with the question: When your spaceship picks up the energ E from some photons, where are you putting the momentum of the photons? It can't be ignored.

**References**:**starship-design: Momentum***From:*wharton@physics.ucla.edu (Ken Wharton)

- Prev by Date:
**starship-design: Propulsion ideas** - Next by Date:
**starship-design: Both right** - Prev by thread:
**starship-design: Momentum** - Next by thread:
**starship-design: Propulsion ideas** - Index(es):