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The sun Qn

  1. Apr 20, 2005 #1

    So, we see the sun everyday, right?

    So let's say you are standing on one point of the earth looking at the sun rise... At where must you point a laser cannon in order to hit the sun dead centre? and why? :confused:
  2. jcsd
  3. Apr 20, 2005 #2


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    I won't do the calculation, but several effects come to mind:

    1) Time delay - It takes about 8.5 minutes for light to reach the sun from here, so you'd have to take the sun's motion during that time into account.
    2) Atmospheric refraction - The sun's actual position is shifted from the apparent one by the bending of its light in the atmosphere. Although the laser beam would be bent as well, this bending is wavelength-dependent, so the shift wouldn't be the same. Also, the earth would be at a different point in its orbit when the laser's light was bent, so the path would be slightly different.
    3) Aberration - The sun's apparent position is shifted due to the earth's motion. This is the same effect that has to be considered when measuring stellar parallaxes.
    4) GR effects - The light beam will follow a geodesic in spacetime, not a straight path in space. The difference should be very small, but non-zero.

    In other words, you couldn't just point the beam at the center. I think that would get you pretty close, though.
  4. Apr 23, 2005 #3
    ok...thx spaceTiger
    but is there a way to find the exact spot eg:how much cm/metres to the centre of the sun?
  5. Apr 24, 2005 #4


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    Yes there is; for starters you need an accurate ephemeris, your own position on the Earth (accurately), the wavelength of the laser (in the laser cannon, and probably its intensity), all the right equations, and a means of doing the calculations.

    In addition to the effects SpaceTiger mentioned, you may also need to consider how the path of the light will be affected traversing the Sun' corona, chromosphere, and upper photosphere. There's also the question of what "the sun['s] dead centre" actually is!
  6. Apr 25, 2005 #5


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    You could theoretically perform such a calculation, but I wouldn't recommend it unless you had a really good reason. You could define an exact center of mass (in the classical limit), but again, this would be really messy.
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