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Constant Acceleration / Deceleration Propulsion

  1. Nov 2, 2013 #1
    I'm making research for an article about constant acceleration / deceleration ship propulsion. Now, it's about 30 years since I was solving differential equations so there's no way I could solve, let alone devise, an appropriate differential equation. I'm not even sure an appropriate differential equation exists.

    What I'm interested in is the following.

    Let's suppose a ship leaves the earth and accelerates constantly at 1G. How much ship time will be required for the ship to reach, say, 0.99 c ? And conversely, What will be the ship velocity (relative to c) after, say, 10 years of the ship time (assuming it's accelerating at 1G starting at rest relative to the Earth).

    In general, given acceleration a and ship time interval t, what will be the speed relative to c at t0 + t. And conversely, given acceleration a and speed (relative to c), how much ship time is required to reach it.

    Basically, I'm interested in how much ship time is required to reach different destinations in cosmos if a ship were to use constant acceleration / deceleration propulsion (accelerating half a way and decelerating half a way).
     
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  3. Nov 2, 2013 #2
    I've noticed the description above is not correct although I think it's obvious what I'm after. The acceleration of course cannot be constant. What is constant is the weight of the cosmonauts (via F = m x a). The acceleration a actually decreases in time while m increases keeping the product (i.e. force) constant.
     
  4. Nov 2, 2013 #3

    Simon Bridge

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    http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]
     
    Last edited by a moderator: May 6, 2017
  5. Nov 2, 2013 #4
    Thanks for the reference but the conclusion doesn't seem right to me. The equations in that article are keeping acceleration constant but this would make cosmonauts heavier and heavier as their weight is given by the force they are experiencing and the force is F = ma which, by keeping the a constant, will become bigger and bigger as they approach the speed of light.

    It seems to me that if we want to keep the weight of the cosmonauts constant then the thrust of the rocket engines must also be constant. As we approach the speed of light the m will get bigger by the same amount a will get smaller thus keeping their weight constant.

    So I believe the calculation is much more complicated than the calculation in the referenced article. The times will be much bigger and the fuel consumption much smaller.
     
  6. Nov 2, 2013 #5

    stevendaryl

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    In Special Relativity, acceleration is frame-dependent. If to you, aboard the rocket, it feels like you are accelerating at rate 1 g, then to someone back on Earth where you started, the acceleration will be much less than 1 g.

    The equations in "The Relativistic Rocket" are about the case where it always feels like 1 g acceleration to those traveling in the rocket.
     
  7. Nov 2, 2013 #6

    Dale

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    The equations and conclusions are correct and directly answer the question you posed in the OP. The proper acceleration is constant, this is the "g-force" that the astronauts onboard the ship experience. The coordinate acceleration decreases as v -> c.
     
  8. Nov 2, 2013 #7

    Simon Bridge

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    @rajko:
    Intuition and what "seems right" tend to get unstuck when you are not used to relativity.

    It may be easier to think about it is the acceleration occurs in short steps, with a constant velocity between them.

    As far as the astronauts are concerned, there is nothing special about their weight, mass, or anything else they can measure once they have changed speed. Nothing has changed. There is no experiment they can do or test they can perform to show them that they have increased speed. [1] So they don't have to work any harder changing kinetic energy for the next increase in velocity than they had to for the last one.

    You may want to think about what a constant acceleration would mean though ... presumably the astronauts will never see their departure point retreat faster than light, no mater how long they accelerate for. This stuff is covered in the link.

    Remember - whenever you think about some value in relativity, you have to say which reference frame this is.

    The calculations were for a 1g acceleration, from the pov of the astronauts. [2] The people at mission control don't see a constant acceleration at all. Mission control still records even jumps in kinetic energy, but not in speed ... which is where the mass increase you keep talking about comes from.

    -----------------------

    [1] OK - maybe the water is sloshing in the captain's aquarium... but even then, that just says there was an abrupt change, not which direction.

    [2] First of all we need to be clear what we mean by continuous acceleration at 1g. The acceleration of the rocket must be measured at any given instant in a non-accelerating frame of reference travelling at the same instantaneous speed as the rocket (see relativity FAQ on accelerating clocks).
    ##\qquad ##-- http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html [Broken]
     
    Last edited by a moderator: May 6, 2017
  9. Nov 3, 2013 #8
    In relativity you need to constantly specify which reference frame the measurements are relative to. In this case when you say "The acceleration of course cannot be constant." then this is true if you mean as measured in the reference frame of mission control which is not accelerating. In the reference frame of the rocket the acceleration is constant. In the rocket, the observers can measure the acceleration by dropping a test particle a short distance. The result they obtain will be the same whatever, the velocity of the rocket relative to mission control is. They can also weigh a test mass inside the rocket and see that its weight does not change over time. The equation for force in relativity is F= γ3ma. Inside the rocket, the velocity of the test mass relative to the rocket is negligible so the gamma factor is γ=1. The acceleration a is constant and the weight (force) is constant so the mass must be constant. You should resist the temptation to think in terms of relativistic mass. For one thing the parallel relativistic mass is different to the transverse relativistic mass.

    You can also note that for momentum (P) the equation is P=γmv where the 'relativistic mass' increases by a factor of gamma and for force the 'relativistic mass' increases by a factor of gamma cubed. Having different forms of relativistic mass for momentum versus acceleration and for parallel versus transverse is unsatisfactory. Anyway, inside the rocket, the low relative velocities of the test mass over a short time interval means that the Newtonian equation F=ma holds in the infinitesimal limit and the force measured inside the rocket is constant over time.

    Now we can look at things as measured in the reference frame of mission control. If we call the force measured in this reference frame F', then the equation is F' = γ3ma'. Now in this reference frame the acceleration is constantly decreasing by a factor of gamma cubed as the velocity of the rocket increases. The end result is that F' = γ3m(a/γ3) = ma = F. In other words the force is constant in the mission control reference frame as well.

    To recap, according to measurements made inside the rocket, the acceleration is constant and the mass is constant so the force is constant

    In Newtonian physics, Kinetic Energy is defined as (1/2)mv2. This does not imply the mass is increasing by a factor of v2 or anything like that, it just how kinetic energy relates to velocity. Similarly, F = γ3ma does not imply that the mass is increasing by a factor of γ3. Now in the above quote, you are clearly talking about the acceleration as measured in the mission control reference frame and this does indeed reduce over time. If the initial acceleration is a0, then when the velocity of the rocket is v then the acceleration is a = a0(1-v2/c2)(3/2) = a03. This means the force measured in the mission control reference frame is F = m a0 which is constant over time because m and a0 are both constants.
     
    Last edited: Nov 3, 2013
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