10,000 KM is inside the Roche Limit for a moon-sized object.
Considering what the moon does at 240,000 miles to create tides, I would imagine at the very least the same mass at 10,000 miles would at very least release every pent-up slippage between crustal plates. (Like the “big one” much of the West Coast is just waiting for, or anything in the Ring of Fire.) I would think that taking out a huge number of satellites and wrecking the orbits of many more would be the lowest importance of disruption. (Plus most of those satellites don’t have the ability to make a HUGE orbital correction.)
If the moon at 240,000 miles can raise tides within 6 hours (twice a day, 6 hours up, 6 hours down), I shudder to think what something can do even in 1.5 hours with 1/24 the distance so 576 times the tidal force and gravitational pull. I’m going to make a wild guess of tides in the neighbourhood of 100 feet in the area closest to the approaching mass; and some residual sloshing for a few days after that. .
Remember, tide is differential gravity, so pull on the near side of earth is from 12,000 miles from the center of its gravity (2000 miles diameter), and the other side of earth is 20,000 miles from its center of gravity.
The Roche limit for earth is about 20,000 miles, but the object will not be around long to break up - although some breaking up from Earth’s pull is a distinct possibility. This is why I think tectonic distortion is a risk for earth too.
11,470 miles.
Tidal force is inverse cube of distance. So it would be 13,824x greater.
So, might see a smidge of coastal flooding in areas.
Plus side, if you are quick enough, there will be some seafloor to explore that hasn’t been exposed to air in millions of years.
Kurzgesagt did a related video, on what would happen if the Moon started spiraling into the Earth (with the caveat that such a thing is impossible).
Doh! Radius not diameter… Thanks.
I wonder how much velocity figures into the Roche equation? As the object zips by with a sharp angle change in a hyperbolic orbit, will the outer part be flung off due to centrifugal force?
I’d think that the object would have enough structural integrity to hold together while it is passing only slightly within the Roche limit for a brief period.
OTOH, it would mean that any loose rock or debris (or astronauts) on the near side would float off of the surface and eventually come raining down on the Earth.
No, they’d have the same velocity - they’d just end up on a tighter (more acute angle) hyperbolic orbit than the rest of the body. Roche breakup of a hyperbolic body would simple change it to a shotgun scatter of differently angled hyperbolic orbits- assuming its gravity didn’t pull it all back together again after it was beyond earth’s influence.
I agree that blink comparison would find it.
North pole or south pole wouldn’t make a difference - the parallax is due to Earth’s orbit (worst case for detection would be coming in in Earth’s orbital plane, because the motion would be back and forth)
I’m not sure that’s the right way to think about tides. The moon has created a distortion in the Earth’s (and the oceans’) shape - and that distortion doesn’t move (much) relative to the Moon. The Earth just rotates “underneath” the permanent tide. A transient effect from a rapidly moving object would really have to move water and rock - and I think it wouldn’t do that very effectively (though I might be wrong).