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RE: starship-design: Deceleration scheme


>> It then will take  (A*6.87E4)/(2*0.00423*A) = 8.1E6 seconds to stop the sail
>> This is about 100 days.
>What, you think light pressure from a star just "magically" cuts off at some 
>distance from the star?

No, but by then it is neglectable:
  Light intensity at 6.8E16 m (45 lightdays): 6.9E-9 Watt/m^2
  Light intensity at 5E9 m                  : 1.3E6 Watt/m^2
That's about 14 magnitudes less (thus neglectable)

>Time for a refresher course:

<Lots of refreshing removed>

>If the combined sail, cable, and payload mass is M and the circular sail's 
>radius is Rs, the acceleration of the sail outward from the Sun is:
>     (1 + k) (6.3 x 10^17) Rs^2 
>as = ------------------------- m/sec^2
>                2M r^2

So far OK, then you continue with:

>The final interstellar departure velocity, V@, in meters/sec of a solar 
>sail starting with velocity, Vo at closest approach to the Sun we have 
>to be:
>      V@ = [[O.5(l + k) (1.26 x 10^18 Rs^2 - 2.66 x 10^28M]/Mro + Vo^2]^0.5
>where ro is the closest approach distance from the center of the Sun.

Neat formula, but how did you get from "as" to "V@"?
And where did this "l" come from?

I haven't seen any relation between the sail size and the sail mass. That
relation is the most important one when you are talking about sails.
This alone makes your formula quite doubtful.

True, if you can make it light enough per square mile, then you indeed can
reach relativitic velocities. But I think there aren't materials that work
under the circumstances we need.

In the rest of your letter where you show that for a sail WITH ship it
indeed isn't possible to slow down from relativistic velocities.

This isn't true either (ASSUMING you can make a sail without a ship that can
slow down from relativitic speeds).
You simply would need to make the sail larger so that the "total mass" per
"area of sail" would go down to similar numbers as for a sail without ship.