# Is a sun-blocking orbit possible?

Is it possible for an orbital "sun shade satellite" to have a geosynchronous equatorial orbit such that it is in an approximately straight line between the Sun and a fixed point on the Earth's equator?

If it is possible, what diameter would the satellite need to have to block all direct sunlight? (I do realise that diffuse sky radiation prevents shadowing.) Obviously, it would need to be 32 arc minutes at apogee, but it may need to be even larger to account for necessary misalignment due to its orbit.

I have looked for online orbital simulators to try this, but was unable to find any that both included the direction of the sun and also allowed tracking a fixed point on the Earth's surface. I envision an elliptical orbit where the satellite has higher orbital velocity in the Earth's shadow to "catch up" and be ready for sunrise at the fixed equatorial point. Something like this:

Sorry for the crude illustration: Yellow shading is sunlight entering from the left. Black dotted line is satellite's orbit. Black vertical bar with white area to the right is the satellite and its shadow. Blue ball with white area to the right is the Earth and its shadow. Red dot is the fixed point on the equator.

• To get an idea how large the sun shade should be, you may calculate the 32 arc minutes at an orbit height of about 500 to 2000 km. The orbit may be calculated using Kepler's laws. – Uwe Jul 4 '18 at 21:23
• Do you mean always? Once a day? Twice a year? – Bob Jacobsen Jul 5 '18 at 2:26
• @BobJacobsen Most of every day's daytime. – Adám Jul 5 '18 at 5:24
• I'm wondering if it would be easier to accomplish this same task, but at the poles? – Magic Octopus Urn Jul 5 '18 at 12:26
• @MagicOctopusUrn That would be completely impossible. During the summer, the sun moves in an approximate circle in the sky, which would require the satellite to stay above the equator at all times. This is clearly impossible. – Adám Jul 5 '18 at 12:31

Is a sun-blocking orbit possible?

Yes!

Let's see what happens with the math.

For now let's use 2D and imagine the Earth's axis isn't tiled but points straight out of the plane.

Last time I checked (in the 1970's) the Earth's rotational period was about 23h 56m 4.09s or $$T =$$86164.09 sec. The rotational frequency $$\omega = 2 \pi / T$$ With an Equatorial radius $$R_E$$ of 6378.137 kilometers, that's an equatorial velocity of about $$\omega R_E =$$ 465.1 m/s.

Using the vis-viva equation (the only equation I can remember):

$$v^2 = GM_E \left( \frac{2}{r} - \frac{1}{a} \right)$$

where $$GM_E$$ is the standard gravitational parameter of the Earth, about 3.986E+14 m^3/s^2. Simplifying for a circular orbit and moving things around gives:

$$a = \frac{GM_E}{v^2}$$

which yields a semi-major axis of about 1.84 million kilometers. That's well past the L1 Lagrange point and the Earth's Hill sphere (about 1.5 million km, see this answer for how to calculate it) where the Sun's gravitational influence becomes as important as the Earth's.

So for such a slowly rotating Earth, a circular orbit is not going to work.

If Earth were rotating say perhaps twice as fast, the orbit would be smaller and there might be a chance it would work for a short span in the middle (an hour or so) for a short time (weeks or months) for a few days in the Summer or Winter, but perturbations from the Sun would cause an object in orbit so far from the Earth to drift off.

EDIT: But, what about an elliptical one as shown in the question? We can say that the periapsis is 400 km altitude minimum and re-solve. I know it can be done mathematically but I'll do it graphically:

With a periapsis of 6378+400 km, and an apoapsis of about 155,000 km, bingo, for the unrealistic 2D scenario, one can match the lateral speed of a point on the Earth, occasionally, from time to time.

• Seeing that it was out near L1 my thought was - hey lets just fly it as a solar sail and use the thrust to oscillate around L1. Then napkin calculated lateral motion of at least a moon diameter every 12 hours and went... maybe not. – GremlinWranger Jul 5 '18 at 7:46
• @GremlinWranger ya, building up ~460 m/s velocity in a fraction of a day without a stored reaction mass (propellant) of some kind is quite a challenge. – uhoh Jul 5 '18 at 7:50
• Furthermore, if we try to match the orbital period to 24hrs (ie. SMA ~42160km) as well as equatorial velocity at apogee, I calculate we have an apogee of ~82,400km and a perigee of ~1890km which is very much underground. – Jack Jul 5 '18 at 9:57
• Also note that we would want to match synodic day rather than sidereal, but that's an unimportant distinction and doesn't change the conclusion – Jack Jul 5 '18 at 9:59
• "Last time I checked (in the 1970's) the Earth's rotational period was about 23h 56m 4.09s" Impressive feat, remembering with hundredth second precision a figure you got 40 years ago (and presumably you didn't use much). :-) – Diego Sánchez Jul 5 '18 at 13:20

Hmm... Place a very dense mass at Sun-Earth L1 lagrangian point, (or thereabouts; such object is bound to change the actual location of our L1) and put a satellite in its orbit of 1 day period (synodic), minimally less than Earth radius of orbit radius and Earth axial tilt worth of inclination.

Half of the satellite's orbit would coincide with a point on Earth surface

With the satellite sized sufficiently to create actual umbra from there, it might work.

Let's try some numbers.

40,075 km (Earth equatorial circumference) -1% (see below why) = 39674.25km

That per synodic day is 459.2 m/s - that's the orbital velocity of our satellite.

Earth equatorial radius 6,371 km. The orbit will be 1% less (L1 is 0.99AU, so as distance is 1% shorter, so should be the radius) R=6307 km

$v^2 = GM/R$ for circular orbit.

Substituting, we need a body of $1.253×10^{23} kg$,or about 1.7 the mass of the Moon.

Now our satellite. It must cause a total eclipse of Sun, angular diameter 0.5 degree, so it must be 0.5 degree at 0.01AU. Using Lagrange Point Finder, and Angular Diameter Calculator, I obtain about 13,000km of diameter. Larger than Earth (diameter 12,742 km) - but should be hollow/thin not to exert significant mass on the central body.

It seems we'll need to make the satellite more massive, the "central body" less massive, and put them in a mutual orbit as a binary system with barycenter at L1. That way, at least, the solar pressure won't send it flying away as an enormous solar sail.

Is it possible for an orbital "sun shade satellite" to have a geosynchronous equatorial orbit such that it is in an approximately straight line between the Sun and a fixed point on the Earth's equator?

A satellite in GEO will only be directly between the Sun and it’s spot on the equator a couple times a century. But it comes close twice a year.

Viewed from the spot on the Earth under the satellite, the Sun moves north and south as the Earth goes around its orbit (remember seasons?).

The Sun is directly overhead the equator just at the two equinoxes: twice a year. If the equinox is at local noon, it’s overhead. But most equinoxes aren’t at local noon.

At the equinox, the Sun moving north or south by about one minute of arc per hour. That means it can be up 12’ north or south at its closest approach. Your sun shade size needs to take that into account or you won’t get total dark.

If it is possible, what diameter would the satellite need to have to block all direct sunlight? (I do realise that diffuse sky radiation prevents shadowing.) Obviously, it would need to be 32 arc minutes at apogee, but it may need to be even larger to account for necessary misalignment due to its orbit.

The other thing you’ll want to consider is how long you want the Sun to be covered. At its minimum size, the shadow just touches the Earth with its tip: it goes by instantly. A bigger shade makes a bigger shadow, which takes longer to go by. This is analogous to solar eclipses.

(Source)

• Regarding the last point, this image of ISS "eclipsing" the Sun should give you a bit of a hint on the sheer size-scale of such an endeavor. – SF. Jul 5 '18 at 8:48
• @SF. Based on that beautiful photo, it would have to be about 4 km i diameter. Large, but not impossible. – Adám Jul 5 '18 at 11:56
• @Adam: That's LEO, mere 400km, and atmosphere thick enough an object of this size and reasonably low weight would fall and burn up within days to weeks. And of course the eclipse would last about three seconds. – SF. Jul 5 '18 at 13:04