# starship-design: Both right

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Steve writes:

>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.

Yes, that's right.  In the example you gave, the photons only had access
to 1kg (the energy), while the block had access to 2kg (1kg energy plus
1kg reaction mass) which made the comparison unfair.  You're right; for
a given amount of reaction mass AND energy you will get the most
momentum with photons.  But for a given amount of energy only (ignoring
the reaction mass) the best bet is massive objects.

But, given that the sail is somewhat useless once we start to
decelerate, we have reaction mass to spare, so this doesn't do anything
to harm my deceleration scheme.  You state:

>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.

The momentum of the photons is not ignored; it speeds up the ship, with
an absorbed momentum of E/c, where E is the "caught" energy.  I showed
earlier, though, how you could use this same amount of energy (plus part
of the sail itself as reaction mass) to slow down the ship by a momentum
equal to 3E/c.  The NET momentum loss (with no energy output) is only
2E/c, because you absorbed the momentum of the photons in the first
place.

As for the suggestion about making a heavy-element sail so there would
be more reaction mass to slow down the ship, don't forget we have to
speed this thing up before we can slow it down!  The lighter the better,
I think...

Ken

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