Guess at slowing of Earth spin

Discussion in 'Physics & Math' started by Billy T, Mar 11, 2006.

?

How big is "SLIGHTLY" (in first post of Billy T if not here)

Poll closed Mar 25, 2006.
  1. Less than 1.0E-4Meters

    58.8%
  2. Less than 1.0E-2Meters

    5.9%
  3. Less than 1Meter

    17.6%
  4. Less than 100Meters

    5.9%
  5. Less than 1000Meters

    0 vote(s)
    0.0%
  6. More than 1000Meters

    11.8%
  1. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Messages:
    23,198
    True if the rocket were firing a machine gun or canon backwards, but not true if it is, as of course it is, a gas that is leaving the rear of the rocket. But again, before explaining this, let me note that neither the rocket nor any of the exhaust "go into orbit" Recall that to make this clear, impossible to miss understand, I said the fuel was aluminum dust and O2, which burn together to make Al2O3 dust that falls to the Earth surface,

    Now about you gas in space, even though it is contrary to the thread's assumptions:

    First note it will rapidly expand into a vacuum, and if in sunlight, surely ionize. Being very small particles, perhaps only atoms or molecules, these charged particles will orbit about the magnetic field lines. The solar wind pushes these field lines about quite strongly if you are far enough form Earth (a few Earth radii) to be speaking of "vacuum," instead of "residual atmosphere." (There is not clean line between these terms.)

    These particles will also feel solar radiation pressure. Recently a private group paid Russians to but up their "solar sailor" space craft, which was to go (slowly) far from Earth on this light flux, never achieving anything like "escape velocity" but, escaping. Unfortunately, the launce failed some way - I forget details.

    Because of solar particle and radiation fluxes plus the dynamics of Earth's magnetic field, especially the "tail" being whipped around by fluctuations in the solar wind (particle flux from) some of these gases do not need to coast away from Earth, with "escape velocity." They will be accelerated away by the sun. The sun will also keep them warm. In a thermal distribution of velocities (Some what doubtful that their collisional interactions can maintain this, so they really do not have any "temperature," but still will have some distribution of velocities.) a fraction will in fact have the escape velocity.

    In short summary:
    Because of these effects, and fact that even at geo-stationary distance the satellite is bound, (has significantly less than the escape velocity) I suspect that 99+% of all the gases that leave the micro thrustors used to fine position the satellite into its assigned geo-stationary position are:
    (1) With less than Earth escape velocity,
    yet
    (2) do escape from the Earth.
     
    Last edited by a moderator: Mar 25, 2006
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  3. DaleSpam TANSTAAFL Registered Senior Member

    Messages:
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    Well, here is how I worked it. Since the satellite is going to be in geosynchronous orbit we can consider the satellite/earth system to be a single rigid body with the satellite attached to the earth via a massless rod. We can then find the geosynchronous orbit radius as a function of the length of the massless rod.

    Starting with:
    T² = 4π/(GM) a³
    L = I ω
    ω = 2π/T
    I = Ie + m r²
    where T is the period, G is the universal gravitational constant, M is the mass of the earth, a is the geosynchronous orbit radius, L is the angular momentum, I is the moment of inertia of the satellite/earth, ω is the angular velocity, m is the satellite mass, r is the length of the rod, and Ie is the moment of inertia of the earth.

    Combining we get:
    √(a³) = √(GM)/L Ie + √(GM)/L m r²

    However, we are not actually interested in a but rather in changes in a. So differentiating both sides we get:
    3/2 √(a) da = √(GM)/L m 2r dr
    da = 4/3 mr/L √(GM/a) dr

    Now, integrating over r from re, the earth radius, to a we get:
    Δa = 2/3 m/L √(GM/a) (a²-re²)

    Which is about
    Δa = 6E-13 m

    Like superluminal this is also between the Angstrom length and Planck length, but closer to Angstroms than his result.

    -Dale
     
    Last edited: Mar 26, 2006
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  5. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

    Messages:
    23,198
    I'm pretty sure you have moment of inertia of the satellite wrong. Should it not be:
    m (R+r)^2
    ?, where R is the Earth radius. Perhaps your rod goes all the way to center of the Earth?, If yes, then you have no error here.

    also I am not clear why you want to integrate over r, or the limits you used {0 & r}? A few words about this would b helpful.

    Your general approach up to this integration part is probably like one I would have used, if I were not so lazy. One thing you should note, like I did, is the neat way DH's Taylor series tricks work, but his approach only gives the linearized equation delta t = K delta a, where K is the local "slope" of the t vs a graph (without needing to draw the graph) for the RELATIONSHIP between t and a. As it turns out this is all we need:
    turning this around:
    delta t = (3/2)*(86400/42164000)delat a = 0.0030672 delta a
    Thus for the bottom of my "a band" (1 cm) to be correct, I need the day to be lengthend by 0.30672 seconds. Clearly it is not as that is 112 second in a year.
    The E-4 to E-2 band bottom is 1.12 seconds per year, very very probably still to high, but certainly in the ball park of what is observed. Too bad it is mainly the moon, not NASA we must blame for this.

    The point of this is that if I had known the answer, I would have made the "biggest band" = "one meter or greater" and added more bands at the bottom, such as E-8 to E-6M, & E-6 to E-4M, then we would need some calculation of delat a, not just a Relationship between a & t.

    DH is not only clever with his Tayor series approach, he is also the beneficiary of my poor guessing the range for the poll. His method would not give the answer, if I had guessed better and then set up better set of choices.

    I was misslead by the fact (discussed earlier) that the angular momentum in the satellite is linear in the length of the moment arm (distance from center of Earth) but the velocity is not. So I thought: You must need a cm of difference to "store" B's angular momentum in as 42,000,000M changed by only a few cm is also "dam small."
     
    Last edited by a moderator: Mar 25, 2006
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  7. superluminal . Registered Senior Member

    Messages:
    10,717
    Here's how I got my result:

    L = Angular momentum
    I = moment of inertia
    w = angular velocity
    Ds = distance of satellite above the earth

    L = Iw

    For a solid sphere rotating about its own axis (the earth) I = 2/5 MeRe^2

    For a point mass rotating at some distance from the earth’s axis (the satellite on the surface) I = MsRe^2

    And for the satellite in orbit I = Ms(Re+Ds)^2

    Linit = Lfinal (conservation of angular momentum)

    so,

    Iinit x winit = Ifinal x wfinal

    I init = (2/5MeRe^2 + MsRe^2)

    Ifinal = (2/5MeRe^2 + Ms(Re+Ds)^2)

    wfinal/winit = Iinit / Ifinal

    This is a VERY VERY small change. I had to use a high precision calculator to get:

    0.99999999999999999998213...

    This means that the angular speed of the earth/satellite system decreases at a magnitude of
    10e-20. Since the geostationary altitude is related by the cube root of the square of the orbital period, it changes by something on the order of 10e-20 to 10e-21 also.

    An angstrom is 10-10meters. Planck length is 1.6e-35 meters

    I am quite dissapointed in my intuition.
     
  8. D H Some other guy Valued Senior Member

    Messages:
    2,257
    You can't do that. r is an alias for a -- they are one and the same variable. What you needed to do, instead, was solve your first equation for a.


    Edited to add,
    Those question marks should be square root symbols. The reply-to-post mechanism kindly changed DaleSpam's math for me.
     
    Last edited: Mar 26, 2006
  9. Tortise Registered Senior Member

    Messages:
    227
    Because the launch will most likely be in the day - (and if launched at night the earth would be blocking the sun - (there are just two narrow windows where this wouldn't be true)) the vast majority of the gas will be pushed away from the sun back into the earth's atmosphere.
    As the rocket is attaining orbital velocity, the gas is not, because it is going the oppisite direction, so the exaust gasses are not sharing the rockets growing independence of the earth and it's atmosphere - making it again much more likely to fall to earth. Add to that the fact that the vast majority of the fuel is expent durring the first part of launch (the first part of launch is when the rocket weighs the most and when the most thrust must be applied to get the rocket up to speed) and in the atmosphere. The remaining rocket/satellite system achieves somewhat of a ballistic trajectory by the time it is completely out of the atmosphere. Gas that excapes the earth's influence (as a fraction of the total amount of exaust) is very rare in a launch of a satellite.
     
    Last edited: Mar 26, 2006
  10. D H Some other guy Valued Senior Member

    Messages:
    2,257
    Assuming angular momentum is conserved, and ignoring the Earth's oblateness then this is quite easy using the first-order approximations,

    delta T/T = - delta L/L from L = 2*pi*I/T
    delta a/a = 2/3 delta T/T from a = k * T^(2/3)

    Thus
    delta a/a = -2/3 delta L/L

    All one has to do is solve for the pre-launch angular momentum, the pre-launch geosynchronous orbit radius, and the change in angular momentum.

    a = geosynchronous orbit radius = 4.2e7 m
    re = 6.4e6 m
    I = 8e37 kg-m^2
    T = 86400 s

    L = 2*pi*I/T = 2*pi*8e37kg-m^2/86400s = 6e33 kg-m^2/s
    delta L = -2*pi/T*m*(a^2-re^2) = 1000 kg*2*pi/86400*((4.2e7)^2-(6.4e6)^2)m^2 = -1.25e14
    delta a = 2/3*1.25e14/6e33*4.2e7 = 5.8e-13 m

    and similarly,

    delta T = - delta L/L T = 1.25e14/6e33*86400s = 1.8e-15 s

    Even with about ten thousand satellite launches since the onset of the space age, (average of 100 per year for 40 years), the cumulative effect is far from being observable.

    Does no one check my figures?
     
    Last edited: Mar 26, 2006
  11. DaleSpam TANSTAAFL Registered Senior Member

    Messages:
    1,723
    Yes, sorry I wasn't clear. All radii are measured from the center or axis of earth. Altitudes just confuse me.


    No, I integrated from re, the earth radius, to a, the geosynchronous orbital radius. The satellite starts out on the surface of the earth, re, and ends up at the geosynchronous radius, a. That is where the limits of integration came from.

    -Dale
     
  12. D H Some other guy Valued Senior Member

    Messages:
    2,257
    Dale,

    You didn't answer BillyT's question of why you chose that particular equation integrate, nor did you see my comment that that equation is invalid. You used the variable 'a' in one equation and 'r' in another, but they are clearly one and same variable (a is identically equal to r). dx/dx = 1, no matter what x(T) is.
     
  13. Pete It's not rocket surgery Moderator

    Messages:
    10,166

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    I think 1.3e13 kg.m^2.s^-1 is correct.

    Mass x Radius x Velocity = 4.2 x 3 x 10^12

    L x w should be exactly the same, I think...

    Mass x Radius^2 x Velocity/Radius = Mass x Radius x Velocity, right?
     
  14. superluminal . Registered Senior Member

    Messages:
    10,717
    Who cares.

    The current level of wine in my systenm argues that this is a point;less crock.
     
  15. D H Some other guy Valued Senior Member

    Messages:
    2,257
    Well, uh, yes. Back to the drawing board. I forgot a factor of 2*pi/86400.
     
  16. CANGAS Registered Senior Member

    Messages:
    1,612
    Amazingly agreeing with super', the current beer deficiency I am suffering, very soon to be remedied, enables me to clearly see that he has an extremely high probability of being correct.
     
  17. DaleSpam TANSTAAFL Registered Senior Member

    Messages:
    1,723
    They are not one and the same variable. r is the radius of the satellite (on the massless rod) and a is the geosynchronous orbital radius. Actually, a is a function of r. That is the whole point of the thread, and r varies from re to a. In any case, I did solve it for a, or at least √(a³)

    -Dale
     
  18. CANGAS Registered Senior Member

    Messages:
    1,612
    It is an hour before launch. The Enterprise is hovering on impulse somewhere beyond Mars orbit. Captain Kirk gives the order and Scotty instantaneously beams up the satelite. Nobody notices, so the launch sequence is completely executed.

    Immediately after the satelite vanishes POOF from the launch pad, does the Earth rotation change? Has the vanishing into thin hyperspace of the satelite and its angular momentum caused the remnant Earth rotation to change?
     
  19. Tortise Registered Senior Member

    Messages:
    227
    It depends on the nature of the teleportation. Some theories about how this could occur have to do with reconstructing an exact replaca from a quantum model. It just depends on how the teleportation works.
    If it is the purpose of the thought experiment to suppose that somehow it stops existing on earth and starts existing somewhere past mars, the answer is that the earth's rotation rate stays the same, but that since the satellite was part of the total inertia of earth, tidal forces ect. would slow the earth at a imperceptably faster rate.
     
    Last edited: Mar 26, 2006
  20. CANGAS Registered Senior Member

    Messages:
    1,612
    In my cool scenario, the transporter works by instantaneously moving the object into a hyperspace which has no measurable effect upon our normally sensible 4-space.

    Beyond quibbling, my scenario asks the question: would the hypothetical vanishing of the satelite and its mass and velocity and momentum make any difference at all to the rotation of the remnant of the Earth that was left behind?
     
  21. Tortise Registered Senior Member

    Messages:
    227
    If it is the purpose of the thought experiment to suppose that somehow it stops existing on earth and starts existing somewhere past mars, the answer is that the earth's rotation rate stays the same for that moment, but since the satellite was part of the total inertia of earth, tidal forces ect. would slow the earth at a imperceptably faster rate since the earth is already slowing due to tidal forces ect. This may not seem like much force, but over time it isn't insignificant. If a satellite on earth is going 1000. mph on the ground now, it will be going only 1/30 that speed in the very distant future (33 mph) this is not an insignificant amount of energy. 966 mph x aprox. 1000 lbs.
     
    Last edited: Mar 26, 2006
  22. CANGAS Registered Senior Member

    Messages:
    1,612
    It is indeed the gist of my godunkem that the instantaneous vanishing of the satelite would, in my humble (?) opinion, not change the rotation of the Earth at any amount measurable by the best state of the art instruments that money can buy.

    My reasoning being, since the prevanish angular momentum of the satelite is in perfect equillibrium with the prevanish angular momentum of the Earth/satelite system, the POOF disapearance of the satelite makes no measurable difference.

    I am basing my posts on the thread as originally stated, and the choices originally provided, not any willy nilly change of rules or choices in the middle of the game. My measurements of rotation would be made within the time era indigenous to the thread start conditions, not some time period far off and eventually happening.
     
    Last edited: Mar 26, 2006
  23. Pete It's not rocket surgery Moderator

    Messages:
    10,166
    I agree with CANGAS (and with super

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    ). A chunk of mass instantaneously disappearing from the Earth wouldn't affect the Earth's rotation at all.

    If the entire crust and mantle were teleported away, for example, the core would continue rotating at 1 rev per day.

    It's interesting to consider what would happen if the teleportation process conserves momentum, angular momentum, and energy, but I think that these could all be conserved at the other end, so to speak. Ie the Enterprise might perhaps be jolted as it "catches" the satellite, without affecting the Earth at all.
     

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