# starship-design: Momentum

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>Why do you people keep saying this?  Photons are the _best_ way to turn
>energy into momentum.  What's the momentum of 1 kg of mass converted
>into photons vs. the momentum of 1 kg of mass converted to energy and
>used to accelerate another 1 kg of mass?

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.

If you take 1kg of mass and convert it to energy you get (1kg x c^2) =
10^17 Joules.  The momentum of E = 10^17 Joules of photons is simply
E/c, or P = 3 10^8 kg m/s.

However, if you accelerate a 1kg object with the 10^17 Joules instead,
you can use the above equation to find that

P = [(10^17)^2 / c^2 + 2 10^17 1kg]^0.5

P = [3 10^17]^0.5 kg m/s

P = 5.5 10^8 kg m/s (I've rounded here, but the answer should be exactly
the square root of three greater than the photon case.)

In this case the ratio between Mc^2 and E was 1; as this ratio gets
bigger (either the mass is larger or the energy drops) the difference
from the photon case gets even larger.  I suggested a ratio of 4:1 as a
way to get three times the momentum as the incident beamed energy.

put a 100 watt bulb in a mirrored box on wheels, with a hole cut in one
side.  The box wouldn't go anywhere, despite the fact that 100Joules per
second was streaming out in photons.  The momentum per second would only
be 100Joules/(3 10^8 m/s) = 3 10^-7 kg m/s.  But if instead you put 100
Joules into a 2kg weight every second, they'll be flying out at 10m/s,
one a second, with a very sizable momentum indeed (namely 20 kg m/s per
second).  The huge mass/energy ratio causes quite a difference in
momentum.

Ken

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