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Orbital Light Speed

  1. Apr 18, 2008 #1
    Hi,

    Is it possible for a man made object to remain in a circular solar orbit at speeds approaching light speed without a coriolis force destroying the object or disrupting the orbit.

    If so, what distance from the sun would be required and what would the orbital period be?


    KodeK
    reel8.co.uk
     
    Last edited: Apr 18, 2008
  2. jcsd
  3. Apr 18, 2008 #2
    No it´s not possible as you easily see from newtonian mechanics when you try to calculate what radius such an orbit would have try it :)
     
  4. Apr 18, 2008 #3
    Thanks Mr Brown.

    But it's been years since I did anything but read physics and I'm afraid that I wouldn't know where to start with the calculation. Is the same true for fractions of light speed say a quarter or a half?

    KodeK
    reel8.co.uk
     
  5. Apr 18, 2008 #4
    Well it´s pretty easy to show what the maximum speed is:
    You know on any orbit that the gravitational force inwards to the sun has to equal the radial force in order to keep the object on the orbit:
    m... Mass of the object
    M ... Mass of the sun


    So:

    [tex]G \frac{m \cdot M}{r^2}=\frac{m \cdot v^2}{r} [/tex]
    gives -> [tex]v=\sqrt{\frac{G\cdot M}{r}} [/tex], which shows that the possible speeds are pretty severly limited by the Radius and Mass of the sun :)
     
  6. Apr 18, 2008 #5

    HallsofIvy

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    Kodek, it took me a moment to see the obvious! The faster the orbit, the closer to the sun you must be (not farther as I was momentarily thinking :redface:). The appropriate distance for a speed anywhere near c would be well within the sun. That's what Mr. Brown is saying.
     
  7. Apr 18, 2008 #6

    tiny-tim

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    Welcome to PF!

    Hi KodeK! Welcome to PF! :smile:

    For the mass of the sun, the circular orbit at the speed of light would be at a radius of about 3 miles.

    The sun's radius is about 43,000 miles … so if orbital velocity is inversely proportional to square-root of radius, then the maximum orbital velocity round the sun would be about c/120. :smile:
     
  8. Apr 18, 2008 #7

    D H

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    Tim, two errors in your post. One is kinda big.

    The smaller error: For a point mass with the same mass as the sun, a circular orbit at the speed of light would have a radius of less than a mile. Calculation

    The bigger error: You are off by a factor of ten on the Sun's radius. Orbital speed (ignoring relativistic effects) at the surface of the Sun is c/686, not c/120. Calculation
     
    Last edited: Apr 18, 2008
  9. Apr 18, 2008 #8

    tiny-tim

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    oops!

    :redface: oops! of course I am! :redface:

    (i was thinking of the number of seconds in a day, and dividing by 2!)
     
    Last edited: Apr 18, 2008
  10. Apr 18, 2008 #9
    Hi Kodek,

    The non intuitive part is that you "know" if a rocket is in a stable orbit is accelerated it will spiral outwards to a new larger orbital radius. What happens is that conservation of angular momentum (L = mvr) slows down the boosted rocket as it spirals outwards.

    To give a semi-numerical example, say a rocket is in a stable orbit [tex]GMm/R^2 = mv^2/R[/itex] with velocity v at radius R. Rapidly boosting the rocket to 2v will cause it to spiral outwards to a new orbital radius of 4R. Conservation on angular momentum means the velocity will have dropped from the initial boosted velocity of 2v to a final velocity of v/2. The force of gravity at this new radius will be [itex]GMm/(4R)^2 = (GMm/R^2)/16[/itex] and the centripetal force [itex]= m (v/2)^2/(4R) = (mv^2/r)/16[/itex]

    Another interesting question (that is possibly even less intuitive) is what will happen to the orbital radius of the moon as tidal friction slows it down?
     
  11. Apr 18, 2008 #10
    Coriolis force is not present in case of circular motion. It occurs when an objects moves in radial direction. Actually it is not even a real force, but a pseudo force, which must be added only in rotating coordinate system.
    Tidal forces are the reason why an object in very close orbit could be ripped apart. Tidal forces are the consequence of nonuniform gravitational field: the force of gravity decreases with distance, so the near end of a satelite will experience greater force than the far end. The average gravitational force will not deform the satelite because the satelite is in free fall motion, but the diferences in that force will cause deformation.
     
  12. Apr 18, 2008 #11
    To kev:

    A rocket that is suddenly accelerated to a higher velocity with not go into higher circular orbit. It will move in eliptic orbit. The equation which sets radial acceleration equal to gravity acceleration is not true in this case.
    And I don't know how you calculated that 4 times greater radious R follows from 2 times greater velocity. If a rocket moves in a circular orbit and double it's velocity, it will move infinitely far, since second cosmic velocity is sqrt(2)*first cosmic velocity.
     
    Last edited: Apr 18, 2008
  13. Apr 18, 2008 #12
    You are right that a single boost wil put the rocket into an elliptical orbit. During a typical Hohmann transfer there is a boost at the lower circular lower to put the rocket into an ellliptical orbit followed by a second boost at the point of the elliptical orbit furthest from Earth (the apogee) to stabilise the rocket into a new circular orbit.

    Not sure about the second part of your post. (Should have known rocket science wouldn't be easy :P) I can see that stable circular orbit velocity is [itex] \sqrt{GM/R}[/itex] and escape velocity is [itex] \sqrt{2GM/R}[/itex] suggesting increasing the velocity by [itex]\sqrt{2}[/itex] would take the satellite to infinity but I have to ask if the escape velocity applies to a vertical (radial) velocity rather than a tangential velocity?

    P.S. What is going to happen to the moon?
     
  14. Apr 18, 2008 #13

    D H

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    Hohmann burns are always tangential. A tangential burn changes the semi-major axis. A small radial burn changes the eccentricity but has little effect on the semi-major axis.
     
  15. Apr 18, 2008 #14
    Very interesting. Of course I should have known that it wasn't a coriolis force I was thinking about, as I said It's been a while since I left the domain of armchair physics.

    I think it was the chemist Kary Mullis who I first read suggesting that the moon is falling.

    Well, the reason I've been trying to get my head around this problem is because I've been trying to design a practical theoretical time capsule. A man made object which accelerates in order to change its relative time reference, like the twins paradox, without leaving the bounds of the solar system. A time traveling artificial comet if you like. Is it perhaps possible by using the sun and Jupiter as gravity slingshots for alternating acceleration and braking? There is no real reason why it would have to approach full light speed. Or am I just scratching around in the dark like a hack Sci-fi author?

    KodeK
    reel8.co.uk
     
  16. Apr 18, 2008 #15

    D H

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    The Moon has been receding from the Earth since the initial formation of the Moon and continues to recede from the Earth to this day. It is not falling toward it. The final state of the Moon is the subject of some debate, but that is a long, long, long time from now.
     
  17. Apr 18, 2008 #16

    [EDIT] OK, just did a quick brush up on orbital mechanics (overdue, I know :)

    Assuming initial velocity (v=1) in the low orbit (r=1) and the required final orbital radius is 4 then the required initial transfer velocity is 1.26 (a boost of +0.26) and the final transfer velocity at the apogee is 0.63 requiring a further reverse boost of -0.13 to stabilise the satellite to its final circular orbital velocity of 0.5

    You are right that a transfer velocity of [itex]\sqrt{2}[/itex] would take the satellite to infinity and does not require the escape velocity to be radially directed.
     
  18. Apr 18, 2008 #17

    D H

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    Suppose a vehicle is orbiting circularly at a velocity=1 in some handy set of units. The escape velocity is [itex]\sqrt 2[/itex]. Suppose now we add a velocity of [itex]\sqrt 2 -1 [/itex] in some direction. If the added velocity is directed along the pre-delta V circular orbit velocity vector, the vehicle's post-delta V velocity will be [itex]\sqrt 2[/itex], i.e., escape velocity. If the added velocity is directed radially (inward or outward, it doesn't matter because the radial direction is normal to the circular orbit velocity vector), the vehicle's post-delta V velocity will be [itex]\sqrt {4-2\sqrt2} \approx 1.08[/itex], quite a bit less than escape velocity. In fact, a delta V of magnitude [itex]\sqrt 2 -1 [/itex] in any direction but directly aligned with the pre-delta V circular orbit velocity vector the magnitude of the resultant velocity vector will be less than escape velocity.
     
  19. Apr 18, 2008 #18
    So what would the maximum instantaneous velocity of an eliptical orbit be, if the object was eventually to return to the solar system? I assume the maximum velocity would be limited at the perihelion to about 1.444 of the velocity an a object in circular orbit would have if its radius was equal to the perihelion radius of the eliptical orbit. Is the maximum velocity of an eliptical orbit at the periapsis?
     
  20. Apr 19, 2008 #19
    Just had a look at this wiki page on the subject: http://en.wikipedia.org/wiki/Tidal_acceleration Apparently the Earth spins faster than the Moon accelerating the moon. Enventually the moon will become "tidally locked" to the Earth and will stay above one point on the Earth appearing "stationary" to an observer on the Earth. Where will it stop? Lunar roulette ...place your bets now! :P

    As the moon accelerates, it will gradually move outwards to a larger orbit. The Earth spin rate is gradually slowed down a little bit by the same process. Presumably the final angular velocity of the Moon will be greater than it is now and a little less than the current angular velocity of the Earth. This makes me wonder if a very gradual acceleration of an orbiting object (say over hundreds or thousands of orbits) would result in a final larger orbit that is essentially circular rather than the elliptical orbit that results from a rapid boost?
     
  21. Apr 19, 2008 #20
    The final tidally-locked orbit, if there are no remaining wobbles in the Earth or Moon, has to be circular. Of course there are really two circular orbits, one for each object.

    But this is getting away from relativity. To get back on track, consider Wheeler's geon, an object that orbits itself at the speed of light. The mass of the orbiting light holds the object together without the need for a central mass.
     
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