Suppose I would be time travelling back in the past, so I exist now two times. And then I get an arrangement that one of us is going to work, and the other one is staying at home. Most certainly we would get into a fight, because both of us would want to saty at home. Strange, eh?
It'd be great entertainment lol
Not really to do with quantum physics, but I thought you guys would like to see what I've been conversing with my brother, who is a physicist, on complexities of travelling to Mars with current or forseeable tech:
This is a simplified trajectory analysis of travel between Earth and Mars. In order to keep things simple, I have ignored the orbital mechanics aspects of the solar system. So, for instance, I am ignoring the gravitational pull of the sun, initial velocity conditions due to the travelerís initial start in the Earthís reference system, any velocity needed to match Marís velocity upon arrival, any drag and/or resistance due to solar wind, light pressure from the Sun, or changes in inertia due to the ejection of propellant. Some of these issues may be important. I could in fact attempt to perform the gravitational field part of the analysis, and the answer would be somewhat different than the one Iím about to yield, but I think the analysis I am performing will get the point across, and the analysis is far less difficult to execute. The analysis I am going to perform is basically straight-line motion, between Earth and Mars.
So, letís start by summarizing the only assumptions needed for this analysis:
The distance to Mars from Earth is 100 million km, per Wikipedia
D1 = 100 ∙ 109 m
D1/2 = 50 ∙ 109 m
The desired time to get to Mars is 45 days, per PhysOrg
t1 = 3.888 ∙ 106 s
t1/2 = 1.944 ∙ 106 s
We are going to engage in constant acceleration motion, with a direction of acceleration reversal halfway there, and no initial velocity.
This means that for the first half of the trip, our rocket is firing in the direction of Earth, to speed us up, and in the second half of the trip it is firing towards Mars, to slow us back down.
Now, letís derive the equation of motion for this system.
1. a = (d2x / dt2)
So, integrating to solve for x:
2. x = x(0) + v(0)t + 0.5at2 (valid for constant acceleration)
Now, assuming that we have no initial velocity, as we stated, and that the start position is labelled as zero, then we can solve for the acceleration:
3. a = 2x/t2
So, letís make some substitutions, and solve for the acceleration:
ē x = D1/2
ē t = t1/2
a = (2)(50∙109) / (1.944∙106)2 = 0.0265
So, weíve got an acceleration of 0.0265 m/s2. If you compare that against the standard surface-of-Earth gravitational acceleration g, which is 9.81 m/s2, then you might say that the acceleration of the spacecraft would be 0.0270 g.
So, with this low level of acceleration, the travelers in our spacecraft would not really notice too much the acceleration level at all.
As an afterthought, the moment at which the spacecraft velocity is highest is going to be the moment of turn-around. What is that velocity? Well, using the same type of analysis as above, it can be shown that the maximum velocity is given by vmax = a∙ t1/2, which can be solved to be 51440 m/s, which is roughly 185185 km/hr.
Tell your brother about Orbiter Space Simulator if he's interested. It basically simulates orbital mechanics with gravity inetia etc. It does not take into account solar wind etc however. But it's interesting to try and get to Mars
http://orbit.medphys.ucl.ac.uk/ - thats the website.