Transiting the Sun

[KrazyHorse]

If you could give me some sciencey sounding words as to why it could potentially fall apart, that would be rad.

Coming up with sciencey words to explain why it wouldn't is the more difficult task...

[BeBMan]

Just repolarise something to hold everything together...

[Lorizael]

Coming up with sciencey words to explain why it wouldn't is the more difficult task...

I don't know if this helps, but I haven't given a complete description of the object. In fact, it's not just one object, but a gigantic collection of smaller satellite-ish objects that, taken as a whole, look like the contact lens of a star.

[KrazyHorse]

Then you need something to holds them in place. They aren't following free orbits.

[Lorizael]

I've been going on the assumption that they'd probably have attitude thrusters or something along those lines.

[Perfection]

My calculations are showing that in terms of the percent of time the sun is completely obscured, the orbital period doesn't matter, it seems to depend completely on Earth's orbit and the solar radius. In any year, no matter what the period (so long as it's within the orbit of Earth ) of the body is, you will spend on average about 13 hours completely obscured. Since there are 5 transits a year under your scenario each one is about 2 hours 35 minutes.

The total transit time (that is the time the sun is partially obscured), does depend on the orbital radius (the closer the object is, the more time spent partially obscured). In this case the time is 11 Hours 10 minutes each transit.

Note that I've made some small angle approximations here, the results might be off by a small fraction.

[Lorizael]

Whoa. Thanks. Results being off by a small fraction is just fine, given that I have no idea how to do this calculation myself, you had no reason to do it for me, and I only need ballpark figures.

[KrazyHorse]

Those results are absolute nonsense on their face. Don't use them.

[Lorizael]

Cool!

[MikeH]

Just make up a number and have the tiny number of science nerds who work it out write about how wrong you were on the internet.

[Lorizael]

Okay, my awful attempt at mathiness. At 45,000,000 km from the Sun, the object's orbit has a circumference of 280,000,000 km. It completes this orbit in 60 days, giving it an orbital speed of 54 km/s. The orbital speed of the Earth is 29 km/s, so the object's umbra will move across the Earth's surface at 25 km/s. Because the object is 1,391,000 km across, it will take that umbra 15.5 hours to pass completely over the Earth. How wrong am I?

[The Mad Monk]

Are you allowing for the umbra moving much faster than 54 km/s? Remember, the umbra will have the same angular velocity, but it will be tracking along Earth's orbit, not the object's.

edit: okay, it's obvious you're not. The umbra's "year" is about 1/6th that of Earth's so it's running about six times faster, or 174 km/s.

[The Mad Monk]

.

[Lorizael]

Right. So, closer to 147 km/s across Earth's surface, meaning it will last 2 hours and 37 minutes. Which is Perfection's result (coincidentally?). Which KH says is crap.

[Perfection]

Those results are absolute nonsense on their face. Don't use them.

A cool property which I didn't realize and which explains why Lorizael's math works (as well as mine) is that since the sun and the object have the same radius, the umbra will always be the same radius, (that radius is the radius of the sun). If the object was larger or smaller, you'd need to use a different calculation.

This also confirms my finding that the percent of the time obscured does not vary with the radius of the object's orbit, the umbra is just as big no matter how close or far the object is.