Quote:
Originally Posted by jedaisoul
Quote:
Originally Posted by jerrygg38
We can get to the moon in 3.5 hours. We will ride the light beam at constant acceleration of G. Halfway to the moon we will reverse to a deceleration of -g. Thus except for a reversal of one minute, we will never be weightless. The photonic fuel cell is a very gentle method of space travel.
In Doppler Space Time 2000 we can get to Mars or Venus in a little over 2 days. Pluto requires 25 days. We can also get to Alpha Centauri in reasonable time.
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These claims seem unrealistic, but I admit I have not done the maths. How do you get from the Earth to the moon in 3.5 hours at only 1 G? |
Quote:
Originally Posted by jerrygg38
I just calculated these things in my Doppler Space Time book in 2000. Using S=0.5at^2 and V=at, and ignoring that we have to take off from the Earth at 2g initially until we clear the atmosophere, the trip to the moon will reach a maximum velocity of 137,400 miles per hour at the halfway point. The trip to Mars or Venus will reach a speed of 2.08 million miles per hour. ( I have not rechecked the numbers since the printing of the book).
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A quick check verifies the figures. Figuring half the distance to the moon using Jerry’s numbers:
Initial velocity =
= 0 m/s
Acceleration = a = 9.8
Time = t = 6300 sec.
Velocity = v =
= 61,740 m/s
this converts to 138,108 miles per hour which is near the 137,400 Jerry gives
= 194,481 km
Which is roughly half the distance to the moon. Decelerating the second half of the trip would double the distance in equal time, so this is correct. Mars and Venus checked out about the same. I think Pluto would be a bit less than 25 days.
As far as Alpha Centauri, you’d have to take relativistic effects into account. Using the
relativistic rocket equation for proper time of the passenger:
Half the distance to Alpha Centauri is 2.15 lightyears and acceleration (in light-year units) is 1.032 lightyears/years^2.
Rearranging and rewriting the equation above in a computer friendly format where c = 1 we get:
T=1/a*LN(((a*d)+1)+SQRT(((a*d)+1)^2-1))
where a = 1.032 ly/y^2 and d=2.15 ly:
T=1/1.032*LN(((1.032*2.15)+1)+SQRT(((1.032*2.15)+1)^2-1))
T=1.78 years
Doubling in order to decelerate the second leg of the trip makes our journey to the nearest star at 1g roughly a 3.56 year affair.
~modest
