Velocity after a certain acceleration.

I am reading a sci fi story by Michail McCollum called 'the Sails of Tau Ceti' and I am checking his math. At one point in the story, he has the intrepid hero's going out in a spacecraft and getting to "12 times the distance from pluto to the sun in 6 weeks". However, a couple pages on, he says the spacecraft reaches 4 percent of C in the same time. I think different. Anyone want to try the math to see how accurate that statement is? The craft is accelerating at a steady rate, except for short periods to jettison fuel tanks, so for all intents and purposes, it can be called a steady state acceleration.
So what do you say the craft is doing in terms of G's of accel, and what percent of C does it reach in that time. They have to get to about one light month away from Earth for the plot so after 6 weeks of accel, they coast then six weeks of decel. They spend 510 days coasting at top speed which gets them to 650 light hours from Earth according to McCollum. Will they get that far in 510 days after accel for the figures mentioned?

Originally posted by sonhouse
I am reading a sci fi story by Michail McCollum called 'the Sails of Tau Ceti' and I am checking his math. At one point in the story, he has the intrepid hero's going out in a spacecraft and getting to "12 times the distance from pluto to the sun in 6 weeks". However, a couple pages on, he says the spacecraft reaches 4 percent of C in the same time. I think o McCollum. Will they get that far in 510 days after accel for the figures mentioned?39.48

The average distance from Pluto to the Sun is 39.48 AU. An AU is 149,597,870,691 meters. Assuming he's talking about 12 times the average distance of Pluto from the Sun, the distance is 1795174448292 meters. To keep it simple, let's use 1.8 x 10^12 m.

Let's imagine the spacecraft moved the entire distance at 0.04c,or 1.2 x10^7 m/s. At this speed, it would take 150,000 seconds, or about 6 days.

So, at that speed it's easily possible. I guess we have to do the hard math then 🙁

We know initial speed (0 m/s) and we know final speed at the halfway point (1.2 x 10^7 m/s). We know this equation works:

x = xi + 1/2(v+vi)t

Since initial speed and displacement are zero,

x = 1/2vt

1.8 x 10^12m = 1/2*(1.2*10^7 m/s)(t)

Now if we solve for t and get anything more than 3 weeks we know it doesn't work. I'm getting bored now though so I'll stop here.

http://www.windows.ucar.edu/tour/link=/pluto/statistics.html
http://en.wikipedia.org/wiki/Astronomical_unit

EDIT - Oops, didn't realize it was a 12 week trip with a coasting period at top speed. Oh well...

Originally posted by AThousandYoung
The average distance from Pluto to the Sun is 39.48 AU. An AU is 149,597,870,691 meters. Assuming he's talking about 12 times the average distance of Pluto from the Sun, the distance is 1795174448292 meters. To keep it simple, let's use 1.8 x 10^12 m.

Let's imagine the spacecraft moved the entire distance at 0.04c,or 1.2 x10^7 m/s. At this spe s, didn't realize it was a 12 week trip with a coasting period at top speed. Oh well...

Not 12 weeks coasting, 510 days coasting, 17 months. I just wondered if anyone else would come up with my #'s, which was top speed about 2% of C and something like 248 light hours out from the sun, not 650 as the final distance. If you confirm my calcs, I am going to see if I can contact McCollum and tell him, ask him how he arrived at his numbers.
No big deal, just got some slack time at work🙂 We have ten computers we have to deal with in our optics testing room and they are all down, we have a sister company in Sweden and they do the software and the link is broke, regular internet is fine but the swedish connection sucks right now so have some boredom time to kill🙂