Get Formula for Rolf's 24-Hour Mars Trip

[Reaptor]

Okay, first I wan´t to say that my english is not to good, so it will be a lot of wrong spellings in this text, hope you understand.

Okay, to the problem.

My masters (read: Teachers) wan´t me to write a story about the space.

The person in the story, Rolf, will travel to Mars for a base on the moon in just 24 hours.

I have made some calculitions, and if Mars is about 60 000 000 km away from the earth, he will have to travel at 2 500 000 km/h, or about 104 166 km/s.

As I have understand it, you will move slower trough time if you travel faster through space.

So, now I want to know how much time the travel will take in Rolfs perspective, if it will take 24 hours in the Earth's perspective.

I want the formula and an example of how you use the formula. (Sometimes, a demonstration of how the formula works will ave mcuh time. Atleast for poor me.)

 

[JesseM]

I have made some calculitions, and if Mars is about 60 000 000 km away from the earth, he will have to travel at 2 500 000 km/h, or about 104 166 km/s.

As I have understand it, you will move slower trough time if you travel faster through space.

So, now I want to know how much time the travel will take in Rolfs perspective, if it will take 24 hours in the Earth's perspective.

I want the formula and an example of how you use the formula. (Sometimes, a demonstration of how the formula works will ave mcuh time. Atleast for poor me.)

Depends if you want to assume he travels at constant velocity between Earth and mars, or if he accelerates and decelerates. If he goes at constant velocity, then you just multiply the time it takes from the Earth's perspective by $$\sqrt{1 - v^2/c^2}$$ to find the time it takes from his perspective (if he goes at 104 166 km/s = 0.347460 c, then the time will be about 0.938 days, or about 22.5 hours). If he accelerates at 1G for the first half of the trip and decelerates at 1G for the second half (which means it will no longer take a day to get there from Earth's perspective, I'm not sure what the exact time would be), you can use formulas from http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html page--in this case I think you'd want this formula:

T = (c/a)sh^-1(at/c) = (c/a) ch^-1 [ad/c^2 + 1]

Where T is the time it takes to get to the halfway point from the ship's perspective, t is the time it takes to get to the halfway point from the Earth's perspective, a is 1G and d is half the distance from Earth to Mars (then just double the time once you get the answer). Here "sh^-1" and "ch^-1" refer to the inverse hyperbolic sine and cosine functions, also written as sinh^-1 and cosh^-1.

On the other hand, if you still want to have him accelerate at a constant rate for the first half and then decelerate at the same rate for the second half, but you want to adjust the acceleration so the whole trip takes exactly a day from the Earth's perspective, you could use this formula:

t = (c/a) sh(aT/c) = sqrt[(d/c)^2 + 2d/a]

then if you solve for a, you get:

a = 2d/[t^2 - (d/c)^2]

So just plug in half a day for t, and half the distance from the Earth to Mars for d, and that'll give you the correct rate of acceleration, which you can then plug into the earlier equation to get the onboard time.

 

[Reaptor]

Thank you very much.

 

[JesseM]

On the other hand, if you still want to have him accelerate at a constant rate for the first half and then decelerate at the same rate for the second half, but you want to adjust the acceleration so the whole trip takes exactly a day from the Earth's perspective, you could use this formula:

t = (c/a) sh(aT/c) = sqrt[(d/c)^2 + 2d/a]

then if you solve for a, you get:

a = 2d/[t^2 - (d/c)^2]

So just plug in half a day for t, and half the distance from the Earth to Mars for d, and that'll give you the correct rate of acceleration, which you can then plug into the earlier equation to get the onboard time.

By the way, I checked the numbers here and the result I got was that he'd have spend the whole trip accelerating and decelerating at 32.15 meters/second^2, or about 3.3 G...could a person stand G-forces that high for an entire day? Also, if you have him spend most of the trip at constant velocity, he'll have to decelerate at a much higher rate when he gets close to Mars if he wants to stop there, maybe too high to survive depending on how brief the deceleration is. So this could be a plot hole in your story, maybe you want to have the trip last for two days instead of one.

 

[pervect]

For the constant acceleration case, you might want to check out the relativistic rocket equations at

http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html

This will give you a more realisitc answer. However, working out the problem in terms of constant velocity will probably give you a better insight into relativity. Since other posters have already done that, I'll work out the constant acceleration case a little bit in case it's interesting.

A note on notation: as per the web page, t and d are measured in the stationary frame (the Earth-Mars frame), and are time and distance, respectively, while T is measured in the accelerating rocket frame, and is
the elapsed "proper time" on the accelerating ship.

We are neglecting a few effects, including the motion of Mars (which is moving much slower than the rocket), and gravitational time dilation effects due to mars's position in the sun's gravity well. I don't believe these should be important.

The first equation of interest is the trip time t in the stationary frame in terms of the distance and the acceleration. The equation for this is

$$t = \sqrt{ \left( \frac{d}{c} \right) ^2 + \frac{2d}{a}}$$

Because we want to get to Mars and stop, we want to go 30 million kilometers in 12 hours, then go another 30 million kilometers in another 12 hours while we stop.

So we can re-write the above equation as

$$a = \frac{2d}{t^2-\left( \frac{d}{c}\right)^2 }$$

Using google calculator

http://www.google.com/search?hl=en&...((30+million+kilometers)/c)^2)++=&btnG=Search

we find that a = 32 m / s^2 as the previous poster has calculated for the non-relativistic case, which is rather high (about 3 g). It's probably survivable with proper acceleration couches, but uncomfortable - very uncomfortable if no breaks are taken for meals or other bodily functions.

We now want to calculate the ship time. Using the link, we find

$$T = \left( \frac{c}{a} \right) arcsinh \left( \frac{at}{c} \right)$$

Unfortunatly, google calculator doesn't seem to do arcsinh
Using another program I get, for the half-trip

Ship time = 43 199.84 seconds
Earth/mars time = 43 200 seconds (12 hours).

Double these for the whole trip. The net result is that only a small fraction of a second difference will occur.