Vector drive - faster than light?

android34
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If a spaceship is traveling at 90% the speed of light on the X axis. And it is also traveling 90% the speed of light on the Y axis. What is its velocity?
 
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1.27c. But your supposition doesn't describe the behavior of a physical spaceship.
 
The solution to the apparent paradox is to model how it could get to those speeds. The ultimate answer is: it couldn't.

The reason is that speeds near C are not summed. They are the result of the formula 1/(v1 = 1-(v0^2/c^2)^.5).
 
Morgan (l982) estimated that the exhaust velocity of a pion- relecting matter/antimatter rocket could be in excess of 0.9c. If the spaceship is a sphere, and one antimatter rocket is aimed in the X axis, and one antimatter at 90 degrees to X 1/4 the way across the sphere is aimed in the Y axis, and both push to 0.9c, is what you say still true?
 
android34 said:
Morgan (l982) estimated that ...

You need a better and more complete reference than that!

Zz.
 
http://www.aiaa.org/Participate/Uploads/2003-4676.pdf

Burried in there is 0.9c
 
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android34 said:
Morgan (l982) estimated that the exhaust velocity of a pion- relecting matter/antimatter rocket could be in excess of 0.9c. If the spaceship is a sphere, and one antimatter rocket is aimed in the X axis, and one antimatter at 90 degrees to X 1/4 the way across the sphere is aimed in the Y axis, and both push to 0.9c, is what you say still true?
Hi android34, welcome to PF

This is actually not very difficult to approximate using the http://en.wikipedia.org/wiki/Four-momentum" .

We will start with a rocket with 1.0 kg payload/vehicle and 2.0 kg fuel (3.0 kg total). We will consider the entire payload of fuel to be burnt instantaneously with 100% efficiency at an exhaust velocity of 0.9 (using units where c=1). We will consider two cases, the first where all of the exhaust goes in the -x direction and the second where half goes 45º above the -x axis and half goes 45º below. We will simply use the conservation of four-momentum before and after the burn to determine the speed of the rocket.

Case 1:

3.0 (1,0,0,0) = 1.0 (γ,γv,0,0) + m (2.29,-2.06,0,0)
eliminating m and solving for v
v = 0.78

Case 2:

3.0 (1,0,0,0) = 1.0 (γ,γv,0,0) + m (2.29,-1.46,1.46,0) + m (2.29,-1.46,-1.46,0)
eliminating m and solving for v
v = 0.71

So you will go faster by sending your exhaust off in the opposite direction you want to go instead of splitting it up on two orthogonal axes.
 
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android34 said:
and both push to 0.9c, is what you say still true?
Propulsion systems do not "push to [a velocity]"; they provide (a force which results in) an acceleration.

If you accelerate for time X to reach v=.9c, and then accelerate for time X again, your final velocity will only be about .99c.

Yes, it is still true.
 
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  • #10
ZapperZ said:
You need a better and more complete reference than that!

Zz.
The choice of propulsion is largely academic, so lack of a reference is not a show-stopper. Any propulsion that can (theoretically) accelerate a craft to .9c will do. There are a few well-known ones that are unlikely to be contested.
 

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