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Principle of relativity hypothetical question

  1. May 5, 2013 #1
    Ok, so I have had a lot of spare time and started pondering the things of the universe. Mainly problems looking at time dilation, mass dilation, length dilation and gravity. So one day I came up with a question that I have been unable to answer and I wonder if it is answerable or if there are too many unknowns.

    The Situation:
    Let's say you have a jetpack out in space. Both you and the jetpack have a mass of 100kg together. This jetpack allows you to approach the speed of light. You are located 10,000,000,000,000 km's away from the earth. So you decide to use this jetpack to approach the speed of light in a circle with diameter of 100km's. Your aim is to relocate the earth to your position using the gravitational force created from your mass increase as you approach the speed of light. You approach the speed of light for what appears to be 20 minutes from your perspective. In relation to the speed of light ( c ) how fast were you travelling?

    Extra notes:
    - Neither you or the earth are affected by any other gravitational force.
    - Your jetpack never runs out of fuel.
    - You don't need to eat, drink, breathe or go to the toilet.
    - Your circle is not affected by centrivical force.
    - Scale between distance and time could contradict eachother
    - If i have missed any other crucial exceptions please do mention them but then carry on as if they were irelevant
     
    Last edited: May 5, 2013
  2. jcsd
  3. May 5, 2013 #2

    mfb

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    2016 Award

    Staff: Mentor

    • To increase the gravitational force, you have to increase the total energy content of the system. The jetpack cannot do this, not even in principle.
    • As your energy content in your local rest frame does not increase, you have the same mass independent of the velocity
    Without additional information, this is impossible to answer. What prevents us from having an arbitrary value (<1 of course) here?

    Where can I buy this?
     
    Last edited: May 5, 2013
  4. May 5, 2013 #3

    • So what your saying is, is that you can not create a gravitational force by increasing your mass through accelerating to the speed of light as your mass from your inertial frame of reference stays the same? So therefore in the formula for gravity ( g=Gm/r) you would input Mo into the formula instead of Mv? ( from the mass dilation formula Mv=Mo/squareroot of1-(v2/c2) )So your force of gravity is still as insignificant as it ever was?

      Are you absolutely sure? Like I wouldnt be surprised if we were right that there isn't enough information here as that was my conclusion but ive been wondering if mabye it is answerable through manipulating enough formulas, especially ones i havnt so far considered.

      Haha I wonder that myself ;) if you ever find this out can ya let me know ;)
     
  5. May 5, 2013 #4

    Dale

    Staff: Mentor

    This is a very common misconception. Gravity in GR is not as simple as taking Newtonian gravity and replacing rest mass with relativistic mass. The source of gravity in GR is the full stress-energy tensor. It includes things like total energy (relativistic mass), but it also includes things like momentum flux and pressure. A relativistically moving mass not only has a high total energy, but it also has a large momentum. Its gravity becomes complicated.
     
  6. May 5, 2013 #5
    Oh ok, im somewhat getting what your saying but could you further elaborate please? What do you mean by gravity in GR? Also what do you mean by newtonian gravity? Also could you have some formulas that include these variables?
     
  7. May 5, 2013 #6

    Dale

    Staff: Mentor

    GR = general relativity. It is the theory of gravity which is consistent with relativity. The formula you were using is Newton's formula, and it is not compatible with relativity. So you cannot ask a question about relativistic effects using that formula. They contradict each other.

    The primary formula for gravity in GR is known as the Einstein Field Equations, or EFE. You can read about it here: http://en.wikipedia.org/wiki/Einstein_field_equations

    The primary formula for gravity in Newton's theory is known as Newton's law of universal gravitation. You can read about it here: http://en.wikipedia.org/wiki/Newton's_law_of_universal_gravitation
     
  8. May 5, 2013 #7

    pervect

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    Also note that if your jetpack were 100% efficient, using the equations from http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken], you can find your theoretical maximum possible velocity.

    Wiki has some similar equations.
    http://en.wikipedia.org/w/index.php?title=Relativistic_rocket&oldid=542512156

    You basically specified that the fuel/mass ratio was 1:1 as that was the mass ratio of you to your jetpack. The theoretically most efficient possible rocket has an exhaust velocity of c, the speed of light. This gives a maximum velocity of 3/5 c, and a gamma factor of 5/4.

    This is the highest velocity you'll ever get from a self-contained jetpack which has fuel that weighs as much as you do and satisfies the conservation laws.
     
    Last edited by a moderator: May 6, 2017
  9. May 5, 2013 #8
    No. That’s not what was meant. If you have a rocket pack (jets don’t work in space) then its ejecting mass in order to accelerate you. The faster you go the more mass lost from your rocket pack. The increase in relativistic mass from accelerating will have less energy than when you started off from at rest.

    You assumed that your jetpack allows you to approach the speed of light which is an incorrect assumption.

    Your equations are incorrect.. I’ve never calculated the gravitational field due to a moving point object so I don’t know the exact formula for the force exerted in a body. However I have calculated the gravitational force exerted on a moving test body due to a stationary body, the spacetime/gravitational field of which is described by the Schwarzchild metric. The force is

    F = GMm[1 + (v/c)2 ]/r2

    where M = proper mass of gravitating body, m = m0/sqrt(1 + 2phi/c2 - (v/c)2 ) is the passive gravitational mass = inertial mass of the particle.

    Note: The m I use here is the same m that Einstein uses on page 102 of his text The Meaning of Relativity which he refer to as inert mass which I believe is his way of saying backk then what we now call inertial mass. Einstein wrote The inert mass is proportional to 1 + phi, and therefore increases when ponderable masses approach the test body.

    Yes.
     
    Last edited: May 5, 2013
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