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

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

<p>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. <em>(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)</em></p>

<p>According to <a href="http://www.calctool.org/CALC/phys/astronomy/earth_orbit" rel="nofollow">this calculator</a>, our sun blocker would need at a height of <strong>2,152,050km</strong> to orbit Earth once per year. So, given that we can set up our formula with...</p>

<p>$$
.0087 = 2arctang\frac{diameter}{(2)2152050}
$$</p>

<p>...and the needed diameter of an object should come out to be about <strong>18,722km in diameter</strong>. 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.</p>

<h3>Edit:</h3>

<p>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 <a href="http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit2/eclipses.html" rel="nofollow">umbral shadow qualifies, or if a penumbral shadow</a> would also suffice. I suspect probably the prior. The shadow <em>would</em> certainly reach the Earth, but probably not cover 5% of it, which is a shame considering its diameter is actually <em>larger</em> than Earth's. Regardless, it's not a very elegant solution.</p>

CC BY-SA 3.0
2014-05-20T02:33:07.013Z
Yikes. That's pretty big. (will accept after a couple more views)
cgp
CC BY-SA 3.0
2014-05-20T02:39:38.297Z
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.
Mitch Goshorn
CC BY-SA 3.0
2014-05-20T03:33:50.007Z
@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.
Mitch Goshorn
CC BY-SA 3.0
2014-05-22T21:53:46.053Z
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.
Ross Millikan
CC BY-SA 3.0
2014-05-20T02:05:46.543Z
2014-05-20T03:32:58.683Z
2014-05-20T03:32:58.683Z
Mitch Goshorn
<p>Like this, but floating in space.</p>

<p><img src="https://i.sstatic.net/aK6ds.png" alt="enter image description here"></p>

<p>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)</p>

<the-sun>
How big would a disc have to be to blot out the sun covering 5% of the land mass of earth?
cgp
# How big would a disc have to be to blot out the sun covering 5% of the land mass of earth?

(Asked by: username_0 on 2014-05-19)

Like this, but floating in space.

![enter image description here](https://i.sstatic.net/aK6ds.png)

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)

**Tags:** the-sun

---

## 1 answers

### Answer by username_1 on 2014-05-20. Score: 3 | Accepted answer

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](http://www.calctool.org/CALC/phys/astronomy/earth_orbit), 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.

### Edit:

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](http://www.astronomy.ohio-state.edu/~pogge/Ast161/Unit2/eclipses.html) 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.

> **Comments:**
> **username_0**: Yikes. That's pretty big. (will accept after a couple more views)  
> **username_1**: 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.  
> **username_1**: @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.  
> **username_2**: 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.  

---

