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How massive does a planet have to be to cut time in half?

  1. Jul 29, 2012 #1
    With respect to time in empty space, how massive does a planet have to be to affect the time to the point where the planet's time runs half the speed of empty space time?
     
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  3. Jul 29, 2012 #2

    Bill_K

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    The factor is 1 - rs/r where rs is the Scharzschild radius. So if you're twice the Schwarzschild radius away from the center, time will run at half the normal rate. It's not the mass of the planet that decides it, you just need to be sitting on an extremely dense object like a neutron star.
     
  4. Jul 29, 2012 #3

    phinds

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    goldk, in case Bill's statement is a bit confusing, let me put it another way. Although it IS (partially) the mass of the planet that matters, it is even MORE important how far you are away from the center of gravity (and will all of the planet still beneath you). That is, what matters is the strength of the gravitational well you are in, and you get strong gravity by having a very massive, but very small body.

    To actually get to 1/2 the time, I don't think a "planet" is possible. Bill's suggesting of a neutron star is more likely.
     
  5. Jul 29, 2012 #4

    pervect

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    You need to specify what the planet is made of to answer the question - or what the density is. I started off by assuming the planet had the density of water and that it was constant.

    Plugging the numbers into a constant density Schwarzschild solution and assuming the "planet" has a constant density of 1000 kg / m^3, I get the radius of the planet would be about 20 light minutes, i.e. about 347 million kilometers.

    I haven't calculated the central pressure, but I'm sure it'd be high enough to compress the water or any known form of matter into white dwarf star material (electron degenerate matter), making the constant density approximation I used pretty useless.

    The formula for the pressure was too complex for me to want to try and even type, much less evaluate. The good news is that it should be at least finite - a surface time dilation of 1/3 would be the maximum possible with a finite central pressure by Buchdal's theorem.

    The appropriate equations to solve for the constant density case are

    M = 4/3 pi * R^3 * rho (this comes from the defintion of m(r). It looks Newtonian, but that's a happy coincidence.)
    2GM/ c^2 = (3/4)*R (this comes from wanting the time dilation to be 1/2 at the surface)

    I used the following for a reference, which may or may not persist

    http://books.google.com/books?id=xm...q=constant density schwarzschild star&f=false
     
  6. Jul 29, 2012 #5
    Thank you all so much! Your answers are very enlightening! One more question......So is it at the Schwarzschild radius that time stops or the center?
     
  7. Jul 29, 2012 #6

    pervect

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    The time dilation factor becomes infinite at the Schwarzschild radius, which many people regard as "time stopping". However, that's probably not a good way of thinking about it - for example, you can fall through the Schwarzschild radius in a finite amount of time by your own wristwatch (proper time), which isn't terribly compatible with the common idea of "time stopping".
     
  8. Jul 30, 2012 #7
    Pervect: But from our perspective it does seem to stop for all practical purposes. I do see what you mean though.
     
  9. Jul 30, 2012 #8

    pervect

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    Suppose you were on a rocketship, accelerating at 1g. And you look down at the Earth through a telescope, monitoring a laser beam that has a precise set frequency (as seen on the Earth).

    About 1 year into your journey, the laser beam will redshift into extinction, essentially disappearing behind an event horizon created by your acceleration. This event horizon is formally very similar to that of a black hole, and is called the Rindler horizon.

    You might say that from the rocket ship's viewpoints "time stops" on the Earth at the instant it falls into the event horizon. The time dilation factor goes to zero. But people on the Earth will neither know nor care that they fell below the event horizon created by your rocketship's acceleration.

    There's a treatment of the Rindler horizon online at http://gregegan.customer.netspace.net.au/SCIENCE/Rindler/RindlerHorizon.html

    it may or may not be too advanced (I dont know your background). For a textbook treatment, I'd recommend MTW (but it requires tensors).

    Understanding the Rindler event horizon is a big help in understanding the rather similar black hole event horizons.
     
  10. Jul 31, 2012 #9
    It is a bit perplexing to me... I read most of the link.... I'm not qualified in the equations anymore.... Calculus 3 was a blur in college :redface: however, From what I understand about SR is that time outside the spaceship (from the perspective of the person accelerating in the spaceship ) goes by quicker than normal. I understand the redshift example, but it seems as.... if the time went by quicker on the outside (from the perspective of the person in the spaceship)..... then the laser beam frequency on earth would seem quicker, shifting more and more to blue.
     
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