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About the Starship Titanic and joules/accelerations.

  1. Jan 31, 2014 #1
    Upon discovering the Starship Titanic while doing some research about time travel, I was wondering what joules would be necessary to accelerate even faster to attempt a more severe time dilation: Namely, a 50,000kg ship to 99.999999999%C. Am I correct in my calculations that this requires app. 1x10^30 joules?

    If not, what are the joules required to do so? And what would the peak wattage be if I am accelerating said ship over 1 year?

  2. jcsd
  3. Jan 31, 2014 #2


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    Staff: Mentor

    Show us the details of how you worked it out, and someone will be glad to tell you whether your basic method is correct and whether you carried out the arithmetic properly.
  4. Jan 31, 2014 #3

    Filip Larsen

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    Gold Member

    The relativistic kinetic energy [1] of an object with (rest-) mass m is Ek = (γ-1)mc2, where γ is the Lorentz factor [2], so you should be able to estimate the minimum amount of energy you need to accelerate that ship of yours and from there the average power needed if this has to take a year. If the energy is stored in a fuel and brought along the ship you may want to take a peek at the relativistic rocket [3] to see what high speeds means for the fuel mass to payload mass ratio.

    [1] http://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy_of_rigid_bodies
    [2] http://en.wikipedia.org/wiki/Lorentz_factor
    [3] http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]
    Last edited by a moderator: May 6, 2017
  5. Feb 1, 2014 #4
    Thanks for the links, they were very helpful. I should have added the caveat that my figure was an approximation. In it, I had extrapolated from the link at the bottom and (despite nagging doubts) saw a relatively linear relationship from 99.9-99.9999%C, which resulted in my guess of ~1x10^30. I had initially feared the actual, more accurate math would be much more complicated, but I had already actually seen the Lorentz transformation formula regarding time dilation (which is why I wanted my ship to go that specific velocity). Just didn't know that the formula could be used this way.

  6. Feb 1, 2014 #5


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    How is your rightmost column calculated? Assuming an instant acceleration to the final speed? If so, your astronauts will not age at all, they will die smashed to pulp on the very first day of the trip.
  7. Feb 1, 2014 #6
    I'm not sure if it accounted for acceleration, probably was just calculated to illustrate the amount of time dilation. The reason I used a year for my acceleration was because that's approximately how long it would take to accelerate to my velocity at 1G.
  8. Feb 1, 2014 #7
    1G for a year will only get you about 0.77c. You answer is only true if you neglect relativistic effects. Reaching 0.99999999999c will take much longer longer than a year at 1 G.
  9. Feb 1, 2014 #8
    I assume that you're doing this as a relativistic rocketry problem. I'll call the rapidity of the rocket θ. This is related to its velocity by [itex]v=ctanh\left(\theta \right)[/itex]. Lets say the ship has an initial mass including fuel of [itex]m_{i}[/itex] and a final mass [itex]m[/itex]. It will have a ship frame exaust port exaust velocity of [itex]v_{ex}[/itex] which relates to the specific impulse [itex]I_{sp}[/itex] of the fuel by [itex]v_{ex}=gI_{sp}[/itex].
    A change in the rapidity of the ship will then be given by
    [itex]\Delta \theta =\frac{v_{ex}}{c}ln\left(\frac{m_{i}}{m}\right)[/itex]
    You find the change in the rapidity from this formula and calculate the final velocity from the rapidity according to the first equation. In the case of zero initial velocity you can compose these as
    [itex]v=ctanh\left(\frac{v_{ex}}{c}ln\left(\frac{m_{i}}{m}\right) \right)[/itex]
    I like this version as its easy to crunch numbers on a scientific calculator, but if you want you can use log properties and the hyperbolic trig version of Euler's identities to write this without the hyperbolic trig function like at a site someone else linked.
    In the case of an ideal matter anti-matter rocket [itex]v_{ex}=c[/itex].
    As for the question of power, this final velocity - mass ratio relation for the rocket is independent of the flow rate at which you choose to burn it off. Burn it off how you like. However, it is common to analyse the case that you are burning off the fuel at a time dependent rate such that it accelerates at a constant "proper acceleration". This is where the occupant feel like they undergo a constant "g-force" or in otherwords, if one stood on a weight scale it would read a constant weight for the cosmonaught for the whole burn. If the proper acceleration is [itex]\alpha [/itex] then the ship time rate of the mass burn relates to the proper acceleration generally by
    [itex]\alpha =\frac{v_{ex}}{m}\frac{dm}{dt'}[/itex], and for constant proper acceleration this integrates to result in
    [itex]\alpha \frac{\Delta t'}{c} =\frac{v_{ex}}{c}ln\left(\frac{m_{i}}{m}\right)[/itex]
    You can then replace this into the equation for the velocity to get the velocity as a function of ship time for constant proper acceleration. If you want more information about this, just let me know.
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