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Effects of perpendicularly-approaching asteroid on Earth

  1. Nov 22, 2012 #1
    Assume the following hypothetical situation: an enormous body, 20,000 ly in diameter (I recognise that 'asteroids' per se tend to be smaller), approaches the Milky Way, perpendicular to its equatorial plane, and headed straight for planet Earth, or at least on a trajectory where it will definitely collide with the Earth at some point. Assume that the body (the 'evil space rock', from now on) is a perfectly spherical, homgenous body of density 3000 kg m^-3, that magically appeared 25,000 ly from the Milky Way (i.e. its gravitational waves have only just begun propagating towards the Earth). At what distance would it begin to have any effect on the Earth? At what distance would it upset the Earth's orbit enough to cause destructive climate change and a consequent extinction event? At what distance would it extract the Earth from orbit entirely and begin pulling it towards itself?

    And, as a side-query, what would happen if the evil space rock were to impact (or rather, get sucked into) Sagittarius A*? What if it was aimed there instead of at Earth?
     
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  3. Nov 22, 2012 #2

    mfb

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    A body with similar properties would be a black hole, and black holes have a fixed relation between size and density - with a size of 20000 light years, their density would be smaller.

    Your object cannot exist - at least not in general relativity, so it is pointless to ask about predictions of that theory.

    You are aware that 20000 light years is about the distance between us and the galactic center?
     
  4. Nov 22, 2012 #3
    Why can't it exist? HVC 127-41-330 is 20,000 ly across, but it consists 80% of dark matter and 20% of hydrogen. Is the evil space rock too massive, or what?

    At any rate, I'm actually looking more for answers on the classical physics (orbital dynamics, etc.) side of things. If we were to ignore the impossibilities of the objects existence, what would its effect be on the planet, in terms of basic kinematics and gravitation (F = -GMm/r^2, etc.)?
     
    Last edited: Nov 22, 2012
  5. Nov 23, 2012 #4

    BobG

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    You just stated it. There's no limit on how far away or how close the object has to be. Every star you see in the sky is tugging on you because of its gravitational pull. Likewise, your presence in the universe gravitationally affects every star you see in the sky (and even stars you can't see).

    The only limit is how weak of a force can you detect. For example, if you take the gravitational force between Jupiter and the Sun, and then add in the gravitational force between you and Jupiter, does your calculator display enough digits to display the difference?

    To figure out when it would pull the Earth out of its orbit, just plug in the masses involved and determine what distance is needed to equal the gravitational force between the Sun and the Earth. Although that wouldn't tell you when the Earth's orbit began to become horribly distorted. It will start perturbing the orbit immediately. You might arbitrarily choose some other boundary, such as when the Earth is becoming unacceptably cold for unacceptably long periods of time, etc.
     
    Last edited: Nov 23, 2012
  6. Nov 23, 2012 #5

    russ_watters

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    Yes, a rock is much denser than a cloud. I never would have considered calling a cloud an "object".
     
  7. Nov 23, 2012 #6
    OK, then what about a black hole absorbing a large mass, like the evil space rock? Would the emissions (if any) have any appreciable effect on the Earth?
     
  8. Nov 23, 2012 #7

    mfb

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    HVC 127-41-330 is a very good vacuum - just not as good as its environment.

    The gravity of other objects influence our solar system as soon as tidal forces become relevant - if everything is accelerated in the same way, it does not matter.

    Tidal accelerations are of the order of ##\frac{GM}{r^3}d## where d is the distance between objects in the solar system, r is the distance to the massive object and M is the mass of this object. This has to be compared with ##\frac{Gm}{d^2}## where m is the mass of our sun. Setting both equal gives ##\frac{M}{r^3} \approx \frac{m}{d^3}##. As an example, an object with 1000 solar masses would seriously disturb planetary orbits in a distance of ~10 times the orbital radius of planets, but does not matter (at least not for the orbits of planets) at ~100 times this distance.

    Based on this abstract, ~10^8 solar masses should be a good upper limit on the mass of HVC 127-41-330.
    This corresponds to a density of ##5\cdot 10^{-23}\frac{kg}{m^3}## (and 26 orders of magnitude below the value given in the first post). To get significant tidal effects, this mass would have to be in a distance of less than 1000 times the orbital radius. Using Neptune, this corresponds to about 0.5 light years.
    Obviously, a cloud of 20000 light years diameter cannot be 0.5 light years away. We would have to pass through the cloud and come close to some mass concentration inside (like a star or a black hole) to notice its gravitational influence on the solar system at all.
     
  9. Nov 27, 2012 #8
    your sense of scale is off..if it were like you explained in the first posts you could pick up electrons from atoms by holding a water melon next to your target atom.

    I dont think numbers are absolutely needed to "solve" this one...its rather observational and logic.
    An object so large cannot affect details on an object as small as our planet.
    It would pull on half the galaxy but not on our clouds and weather baloons.
     
  10. Nov 29, 2012 #9
    ...
    OK, I'm feeling reeeeaaaally embarrassed right now.
     
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