Newsgroups: sci.space.shuttle
Path: utzoo!henry
From: henry@zoo.toronto.edu (Henry Spencer)
Subject: Re: Propellant velocity
Message-ID: <1991May27.195120.16369@zoo.toronto.edu>
Date: Mon, 27 May 1991 19:51:20 GMT
References: <1991May25.215849.15606@zoo.toronto.edu> <12463@uwm.edu> <1991May27.022456.2921@agate.berkeley.edu> <12468@uwm.edu>
Organization: U of Toronto Zoology

In article <12468@uwm.edu> markh@csd4.csd.uwm.edu (Mark William Hopkins) writes:
>With 50 km/s fuel there's no practical way to get much above a 50 km/s change
>of velocity.

Not true; with 3-4 km/s fuels we routinely achieve orbit (8+ km/s) and
escape (11+ km/s).  A more precise statement is that achieving more
than 2-3 times the exhaust velocity is difficult unless thrust can be
very low, and going beyond 4-5x quickly becomes impossible.

>Incidentally, 50 km/sec exhaust velocity would STILL enable you to reach the
>moon in 4 hours, ignoring the effects of having to pull against Earth's
>gravity, if you could accelerate at 1G to 50 km/sec and decelerate likewise.
>You'd get to Mars even with this fuel in a few weeks, and not a whole year.

You don't seem to understand that there are basic problems with this happy
scenario *other than* the exhaust velocity involved.  You do not accelerate
at 1G with a 50 km/s engine!  The high-exhaust-velocity engines are mostly
eletrical rockets.  To accelerate a ship of (say) 100 tons at 1G with a
50 km/s exhaust velocity requires 25 gigawatts of power.  There is simply
no foreseeable near-future power source that can pack that kind of electrical
output into 100 tons.

For the interested, the derivation of that number... by conservation of
momentum, taking the time derivative we have

	mass_flow * exhaust_velocity = ship_mass * acceleration

(This ignores the loss of ship mass in the exhaust; assume for the moment
that that is negligible.)  The power of the exhaust is the time derivative
of its kinetic energy, which in the ship frame of reference -- the one that
matters in this case -- is simply

	power = 0.5 * mass_flow * exhaust_velocity^2

Substituting the first into the second, we get

	power = 0.5 * ship_mass * acceleration * exhaust_velocity

This assumes, of course, zero losses.  Ha ha.  Most power sources have more
like 50-75% losses, never mind the efficiency of the engines themselves.
Getting rid of 50+ GW of waste heat from a 100T ship is not going to be fun.

Note, I'm not saying that respectable acceleration at an exhaust velocity
of 50 km/s is impossible, but you *won't* do it with electrical rockets,
and it won't be easy even with not-here-yet technology like fusion rockets.
-- 
"We're thinking about upgrading from    | Henry Spencer @ U of Toronto Zoology
SunOS 4.1.1 to SunOS 3.5."              |  henry@zoo.toronto.edu  utzoo!henry