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5 votes

Could a 200m diameter asteroid be put into a graveyard orbit and not be noticed by people on the ground?

Others have made good suggestions about ways to minimize the albedo to make it less visible, but it's hard to completely remove it. I think that what you could do is instead of trying to hide it ...
AndyD273's user avatar
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3 votes

Could a 200m diameter asteroid be put into a graveyard orbit and not be noticed by people on the ground?

Put it into the L2 Lagrange point of the Earth-Moon-system. A station at L2 would need some fuel for stationkeeping, but it would be hidden behind Luna.
o.m.'s user avatar
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2 votes

Could a 200m diameter asteroid be put into a graveyard orbit and not be noticed by people on the ground?

Frame challenge: what is Anvil's shape? If it was long and thin, it would be almost invisible when viewed end-on. If it was in an equatorial orbit, pointed towards Earth's centre, then someone in the ...
Richard Kirk's user avatar
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16 votes

Could a 200m diameter asteroid be put into a graveyard orbit and not be noticed by people on the ground?

Forget about telescopes for a minute. Anvil's first worry is sunlight visibility to the naked eye. The sun will light them up. Based on this table, an 0.2 km asteroid at 1 AU would have a visual ...
causative's user avatar
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12 votes

Could a 200m diameter asteroid be put into a graveyard orbit and not be noticed by people on the ground?

Radar exists, and has existed for three quarters of the century. A 200 meters wide target on an equatorial orbit will be detected very very quickly. If the thing is not on an equatorial orbit it might ...
AlexP's user avatar
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8 votes
Accepted

Calculating Living Area on a Concentric Shellworld

Considering that probably you do not need a twenty-digit precision, you can calculate with volume. What you need: Calculate the volume of a 15000km sphere Substract the volume of a 3000km sphere ...
Gray Sheep's user avatar
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14 votes

Calculating Living Area on a Concentric Shellworld

1.183*10ยนยน square km The total area is $4\pi \sum_0^{100}(3000+120n)^2$. We ask Wolfram Alpha (leaving out the 4$\pi$): Now we just have to multiply the result by 4$\pi$: 4$\pi$ * 9.417.240.000 = 1....
Klaus Æ. Mogensen's user avatar

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