[Spaceships] Interstellar Magsail Deceleration

munin · 07-23-2010, 10:11 PM

I'm hoping someone who has actually used calculus in the last two decades can check my thinking here.

The Dirac-class Exploration Cruiser (p. SS5:11) says that it can use its magsail to decelerate from 23% light-speed. I was curious to see how much distance it required to perform this deceleration.

GURPS Spaceships says to use a ship's average distance from the sun to determine the acceleration provided by a magsail (p. 39). This didn't seem to make sense to me when you're coming in from an interstellar trip at some non-negligible fraction of light-speed -- do you use the mid-point distance between the two stars where the magsail's acceleration will be negligible? Three-quarters the distance as half of the deceleration phase, where the magsail's acceleration is still negligible? Neither seemed right, so I decided to try real math instead of the explicit simplification.

Using an energy balance equation similar to a derivation of escape velocity (conceptually you can think of this as a magsail-equipped spacecraft starting at the sun's surface, seeing what kinetic energy it will have after a specific distance of acceleration, and what velocity that produces -- and then reversing it to find the distance needed for a deceleration from a specific velocity):

1/2 * m * v^2 = INTEGRATE (a * m * (u/x)^2 * dx) FROM r0 TO r

where: m is the spacecraft's mass (which quickly cancels out of the equation); a = 0.001 * 9.81 m/s^2 (0.001G for a magsail); u = 1.50e11 m (1 AU); r0 = 7.00e8 m (our sun's radius); and r is the resulting distance from the sun's center and v is the resulting velocity. All units are meters and seconds.

Solving for r in terms of v, I get:

r = (2 * a * u^2 * r0)/(2 * a * u^2 - v^2 * r0)

Obviously, there are some values of v which will produce a negative result for r, which wouldn't make sense. Like the escape velocity derivation, v converges on a finite value even as r goes to infinity -- in this case, when v = (2 * a * u^2 / r0)^1/2 = 7.94e5 m/s = 494 mps (miles per second).

So it seems like, realistically, the most a single magsail could reduce your velocity by is 494 mps (0.27% light-speed), even if decelerating from infinity all the way down to the sun's surface (though actually 98% of the deceleration is done in the last 0.1 AU). You'd need 14 magsails just to get down from ramscoop velocity (1,800 mps) and effectively 7,500 magsails to decelerate from 0.23c (magsails to mps is non-linear).

Is this correct, or have I screwed up my math, or simply taken the wrong approach?

07-23-2010, 10:11 PM	#1
munin Join Date: Aug 2007 Location: Vermont, USA	[Spaceships] Interstellar Magsail Deceleration I'm hoping someone who has actually used calculus in the last two decades can check my thinking here. The Dirac-class Exploration Cruiser (p. SS5:11) says that it can use its magsail to decelerate from 23% light-speed. I was curious to see how much distance it required to perform this deceleration. GURPS Spaceships says to use a ship's average distance from the sun to determine the acceleration provided by a magsail (p. 39). This didn't seem to make sense to me when you're coming in from an interstellar trip at some non-negligible fraction of light-speed -- do you use the mid-point distance between the two stars where the magsail's acceleration will be negligible? Three-quarters the distance as half of the deceleration phase, where the magsail's acceleration is still negligible? Neither seemed right, so I decided to try real math instead of the explicit simplification. Using an energy balance equation similar to a derivation of escape velocity (conceptually you can think of this as a magsail-equipped spacecraft starting at the sun's surface, seeing what kinetic energy it will have after a specific distance of acceleration, and what velocity that produces -- and then reversing it to find the distance needed for a deceleration from a specific velocity): 1/2 * m * v^2 = INTEGRATE (a * m * (u/x)^2 * dx) FROM r0 TO r where: m is the spacecraft's mass (which quickly cancels out of the equation); a = 0.001 * 9.81 m/s^2 (0.001G for a magsail); u = 1.50e11 m (1 AU); r0 = 7.00e8 m (our sun's radius); and r is the resulting distance from the sun's center and v is the resulting velocity. All units are meters and seconds. Solving for r in terms of v, I get: r = (2 * a * u^2 * r0)/(2 * a * u^2 - v^2 * r0) Obviously, there are some values of v which will produce a negative result for r, which wouldn't make sense. Like the escape velocity derivation, v converges on a finite value even as r goes to infinity -- in this case, when v = (2 * a * u^2 / r0)^1/2 = 7.94e5 m/s = 494 mps (miles per second). So it seems like, realistically, the most a single magsail could reduce your velocity by is 494 mps (0.27% light-speed), even if decelerating from infinity all the way down to the sun's surface (though actually 98% of the deceleration is done in the last 0.1 AU). You'd need 14 magsails just to get down from ramscoop velocity (1,800 mps) and effectively 7,500 magsails to decelerate from 0.23c (magsails to mps is non-linear). Is this correct, or have I screwed up my math, or simply taken the wrong approach?