So, SpaceBillboard was a bust, great. so forget about mylar sheets–I want to harness an asteroid, fly it back to Earth, and park it in an orbit that will pass over New York City once a day. I don’t care about it being anything more than visible as an asteroid or chunk of rock, just that it’s more than a bright light wizzing by–I’m going to recoup my startup costs by selling naming rights (“Oh look, there’s the SDMBasteroid overhead!”), so it needs to be memorable but not tacky like a Billboard.
Bonus question–if it’s large enough to be seen, would an average person be able to detect changes in gravity or anything else when it passed over? Scales change when it’s directly overhead? Pigeons dropping like flies? Flies flying like pigeons?
First off, the size in the sky is related to two things (ok, three things):
Actual size of the thingie.
How CLOSE the thingie is.
How BRIGHT the thingie is.
At International Space Station height, they’re about 270 miles up, and MOSTLY stay in orbit naturally. You start getting any *closer *(especially if you’re heavier) and you’re going to be essentially using rocket fuel to “fly” in really thin atmosphere while your rock tries to plunge dramatically to earth. If you’re much heavier, then you’ve got to be further out to stay up.
Your best bet is to find an asteroid made of sponge, paint it white, and set it out about ISS height going at a good whack. You may get more of a visual blurry streak than a leisurely sphere heading across, but you would certainly be able to see it clearly. Just hope it doesn’t run into anything.
You could get a very rough estimate by doing the geometry. Assume a jet flying at 35,000 feet is just slightly bigger than a point source and it is, say, 250 feet long (like a 747).
So, at roughly 7 miles up, 250 feet subtends an arc angle of “a”. So you’ve got 2 lines: one from you to the front of the jet, and one from you to back of the jet, with an arc angle of “a”, and you know at 7 miles up, the distance is 250 feet. Now just choose any height you want, and you can find out the size by calculating the distance between the 2 lines at that height.
Ok, I’m sure it is glaringly obvious that I don’t really know my geometry that well. And I can’t really give you any formulas that will let you do what I’m describing, but I think the idea is sound for back-of-the-envelope calculations. Others will be around soon, I’m sure, to correct this.
For a rough estimate, you can assume it needs to appear to be the size of the moon to be easily viewable/readable.
The moon’s angular size from earth is about half a degree. So if you park your giant space object to be visible (readable) at the same orbital height as the ISS (about 270 miles):
No, your orbit does not depend on your mass since heavy objects and light objects both fall at the same speed, barring friction from the atmosphere. An orbit is an object falling toward the planet, but having enough velocity sideways that it keeps missing. This means that your sideways vector has to match your downwards vector. And the closer you are to the planet, the more gravity pulls you down, and so the faster your sideways vector must be. This is why objects very far from Earth have very long orbits–the Moon at 380,000 km away takes 30 days to orbit the Earth, while geostationary satellites at 36,000 km take 1 day, while the ISS at 400 km takes 92 minutes. Any object of any mass at the same height would have the same orbital period.
This website explaining what Phobos and Deimos would look like from the surface of Mars might be helpful. Deimos appears as just a bright star, but Phobos appears about 1/3 the diameter of the Moon. Phobos is 11 km in diameter and 6000 km up and orbits Mars every 7 hours. But since Mars is much smaller than Earth an object of comparable size in Earth orbit at the same height would have to orbit much faster. http://earthsky.org/space/phobos-and-deimos
So Phobos is about 14 miles across, orbiting about 4000 miles high, showing up about 1/3 the size of the moon. I can move a much smaller rock (about 2.36 miles per Dorjän) to just 270 miles up and it’ll be the same size as the moon.
Is it safe to assume that at that size/distance there will be no noticeable affects (to the lay person, not sensitive measuring devices)? Would there be if I could get a 14-mile object into a steady 2.36-mile orbit (how would that scale viewing angle? I can’t quite follow Dorjän’s calculations–when I try and find SIN(.5) I get 0.049979, not 0.0087).
And if I wanted to affect tides (even a little), just how large would this two-plus mile orbiting rock need to be?
To simply be “more than a point of light”, it could get by with a much smaller angular size than the moon - a couple of minutes of angle should suffice.
And if you do consider friction from the atmosphere, then heavier objects actually have the advantage. Inertia scales with mass, but atmospheric resistance scales with the cross-sectional area. So drag will impose a greater acceleration on a lighter object.
I think you accidentally asked the wrong question, since “how large is a two-mile object” is rather like “who’s buried in Grant’s tomb?”. You also need to specify just how much effect on the tides counts as “even a little”.
For gravitational effects, the apparent size of the object divided by it’s density will tell you the tidal effects. So an object that looks to be the same size as the moon, with the same density as the moon, will cause tides of the same intensity as the moon. Note that the sun has the same apparent size as the moon, but since it is much less dense than the moon it causes much smaller tides.
Note that since the sun is essentially stationary, and the moon takes a month to orbit the earth, the major cause of the changes in the tides is the earth’s rotation. But a very close object is going to be changing position in the sky very differently. An object in geostationary orbit wouldn’t move at all, it would cause permanent tidal bulges, changing sea levels permanently lower in some places and higher in others by up to several meters.
An object orbiting closer or farther is going to cause tidal changes every orbit, adding another variable to our current lunar and solar tides. We’d have some even higher high tides, some even lower low tides, and lots of chaotic medium tides.
But gravitationally you wouldn’t notice any difference. You don’t measure your weight as higher or lower based on the position of the moon or sun, do you?
The size of the object is irrelevant; you need mass.
Hmmm… how does gravitational lensing around a black hole affect the answers so far? Can you put a smaller object on the far side of a black hole and make it seem larger from the ground?
Hmm. Trying to parse what he asked, he’s thinking that a 2 mile object orbiting at 270 miles up would be the same apparent size as the moon. Then he imagines putting a 14 miles object in the same orbit as the 2 mile object.
I think he’s confused and still imagines that objects of different sizes would have to orbit at different heights. But that’s not the case, if I’m at the International Space Station and throw away a screwdriver, the screwdriver and the space station both have almost identical orbits even though the screwdriver is a tiny fraction of the size of the space station.
So any size of object can orbit at any height, the size and orbital height have nothing to do with each other. The only thing that matters is that lower orbits need to be faster and faster, while higher orbits need to be slower and slower. And this is simply because the speed at which an object falls towards earth increases as you get closer to earth and decreases as you get farther from earth, and an object in a circular orbit is always falling toward earth at a constant speed but always missing.
If instead he means an object orbiting the earth at 2 miles up, that is impossible since it would be well within Earth’s atmosphere and would be slowed by the air and crash. Even if the Earth didn’t have an atmosphere there are lots of mountains that are higher than 2 miles, and plenty of mass concentrations that would make the orbit chaotic enough that eventually the object would smack into something.
Also note that of course you could detect differences in weight with very sophisticated instruments based on the position of the sun and moon, just that it’s only a few parts in a million or so, and ordinary vibrations and nearby mass concentrations will effect the measurement a lot more than the lunar or solar positions. Remember as a rule of thumb that an object’s gravitational influence is about the same based on its apparent size multiplied by its density. So a nearby rock that looks to be about the same size as the full moon will exert about the same gravitational attraction on you as the moon does. Now notice that there are a lot of objects near you. As long as those objects don’t move relative to your instrument then you won’t notice a change, and you might be able to pick out the daily solar and lunar signals among the noise. Just don’t sneeze. Or move.
Here’s a related question: What’s the minimum size at which a point of light is more than just a point of light?
I’m not being facetious. The ancients saw the planets with the naked eye. On the one hand, they were mere points of light. Yet they were undeniably larger than the other stars. So what it the cutoff point that you’re looking for?
My suggestion is that if you have a sufficiently noticable SHAPE, then you can manage with a much smaller size. Consider an equilateral triangle, for example. “I saw the Basteroid last night! Not really any bigger than the other stars, but there’s no mistaking the triangle shape!”
Or an irregular oblong might work just as well, if it were 3 to 4 times longer than it was wide, for example.
You can visually tell the difference between a planet and a star due to the planet showing a visible disk. It’s a very subtle difference, but it is detectable to most people. It probably would be more noticeable for a distinctive shape, though, like Keeve said.
If it worked that way, the stone paperweight on my desk would be exerting tidal effects more powerful than those of the moon.
The disks of the planets aren’t large enough to be distinguished from points of light by direct naked-eye observation, but they are large enough to prevent the entire image from being displaced by random refraction caused by atmospheric turbulence. Or, to give you the [noparse]TL:DR[/noparse] version: stars are points of light that twinkle, planets are points of light that don’t twinkle.
Yup, it is, at least on you. The tidal effects from either the Moon or your paperweight are so small as to be undetectable (at least, without extremely precise equipment) on such a small size scale. It’s only when you get to a size scale comparable to a planet that you can detect such small tidal effects, and of course, by the time you get as far from your paperweight as the center of the Earth, its angular size is negligibly small.
