It depends on its density. If it is very dense then yes, it could, so the question is, how dense would it have to be.
The formula for the Roche limit for a liquid satellite - how close it can get before it gets torn apart by tidal forces, is L = 2.44 R (D/d)^(1/3)
(Where R, D are the radius and density of the Earth and d is density of the satellite and a is its orbital radius). There, the more dense the satellite is, the higher d is, then the lower L is, meaning that it can orbit more closely to the Earth.
So solving for d, to achieve a Roche limit of L, we have
d = L*(a/(2.44*R))^3
There R = radius of the Earth = 6,371 km
a = radius of geostationary orbit = 35,786 km
D = density of the Earth = 5.51 g/cm³
so
d = 5.51 * (35,786/(2.44*6,371))^3 = 67.22 g/cm³
Platinum is 21.4 g/cm³ at 20C.
Mercury is 13.594 g/cm³
Nickel 8.9 g/cm³
Iron 7.87 g/cm³
where Nickel / Iron is most likely, possibly with traces of other metals like gold, platinum etc.
So, I think it would be a struggle. Of course denser at the center but still.
For smaller asteroids, then they are held together by the cohesive forces. That's why we as humans can survive. Obviously we live just fine well within the Roche limit of our planet, even though we are largely made of water. If we were just held together by gravity, as human sized satellites orbiting the Earth we'd be pulled apart by tidal effects, but the other forces easily overcome the tidal effects of gravity.
There I'm assuming you mean the present day geostationary orbit.
If the Earth rotated once a lunar month for instance, then the Moon would be in geostationary orbit :). So a slower rotating planet can easily have Moons in geostationary orbit.
Like Pluto also - Charon is in geostationary orbit around Pluto as both are tidally locked to each other.
It is the only example in our solar system of a planet with a moon in geostationary (or rather pluto stationary) orbit.
The formula for the Roche limit for a liquid satellite - how close it can get before it gets torn apart by tidal forces, is L = 2.44 R (D/d)^(1/3)
(Where R, D are the radius and density of the Earth and d is density of the satellite and a is its orbital radius). There, the more dense the satellite is, the higher d is, then the lower L is, meaning that it can orbit more closely to the Earth.
So solving for d, to achieve a Roche limit of L, we have
d = L*(a/(2.44*R))^3
There R = radius of the Earth = 6,371 km
a = radius of geostationary orbit = 35,786 km
D = density of the Earth = 5.51 g/cm³
so
d = 5.51 * (35,786/(2.44*6,371))^3 = 67.22 g/cm³
Platinum is 21.4 g/cm³ at 20C.
Mercury is 13.594 g/cm³
Nickel 8.9 g/cm³
Iron 7.87 g/cm³
where Nickel / Iron is most likely, possibly with traces of other metals like gold, platinum etc.
Largest discovered meteorite fragment at Meteor Crater
So, I think it would be a struggle. Of course denser at the center but still.
For smaller asteroids, then they are held together by the cohesive forces. That's why we as humans can survive. Obviously we live just fine well within the Roche limit of our planet, even though we are largely made of water. If we were just held together by gravity, as human sized satellites orbiting the Earth we'd be pulled apart by tidal effects, but the other forces easily overcome the tidal effects of gravity.
There I'm assuming you mean the present day geostationary orbit.
If the Earth rotated once a lunar month for instance, then the Moon would be in geostationary orbit :). So a slower rotating planet can easily have Moons in geostationary orbit.
Like Pluto also - Charon is in geostationary orbit around Pluto as both are tidally locked to each other.
It is the only example in our solar system of a planet with a moon in geostationary (or rather pluto stationary) orbit.
About the Author
Robert Walker
Writer of articles on Mars and Space issues - Software Developer of Tune Smithy, Bounce Metronome etc.Studied at Wolfson College, Oxford
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