Like this, but floating in space.

enter image description here

I imagine that this would also be influenced by the distance of the object from the sun, and that there is some ideal distance for this object to be? (and obviously the distance of the Earth from the Sun as well)


1 Answer 1


The angular diameter of the sun is approximately 30 arcminutes (roughly .0087 radians), or half of a degree.

The formula for angular diameter of a round object in radians is: $$ \delta = 2arctang\frac{diameter}{(2)distance} $$

Given this, we should also think about position. Assuming we want to continually block the same spot on Earth we need to the orbital period to one year. (I'm not sure if there is a technical difference between orbiting the Earth once per year such that you're always between the Sun and Earth, and just plain orbiting the Sun)

According to this calculator, our sun blocker would need at a height of 2,152,050km to orbit Earth once per year. So, given that we can set up our formula with...

$$ .0087 = 2arctang\frac{diameter}{(2)2152050} $$

...and the needed diameter of an object should come out to be about 18,722km in diameter. Given that, Mr. Burns's terrestrial solution is far more practical, though certainly a nice orbital one is far more devious. Being evil isn't cheap.


This answer is a bit rough as it doesn't calculate for the 5% of the Earth's land mass stipulation. The answer which would meet this would depend on whether only an umbral shadow qualifies, or if a penumbral shadow would also suffice. I suspect probably the prior. The shadow would certainly reach the Earth, but probably not cover 5% of it, which is a shame considering its diameter is actually larger than Earth's. Regardless, it's not a very elegant solution.

  • $\begingroup$ Yikes. That's pretty big. (will accept after a couple more views) $\endgroup$
    – cgp
    May 20, 2014 at 2:33
  • $\begingroup$ Don't accept quite yet, I'd forgotten about the 5% which was in the title. I'm trying to work that out at the moment. $\endgroup$ May 20, 2014 at 2:39
  • 1
    $\begingroup$ @altCognito Unfortunately, I've run out of calculating steam for the night. I've made a small edit, but there is still plenty of room for a more complete answer. $\endgroup$ May 20, 2014 at 3:33
  • $\begingroup$ You would want the object to be at the earth-sun L1 point, about 1.5 E6 km away. This would need a bit of stationkeeping (it is an unstable point), but not much. That saves you a factor 2/7 in diameter, but the message doesn't change. $\endgroup$ May 22, 2014 at 21:53

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