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Physics, Astronomy & Cosmology / Re: How do the oceans ignore centrifugal force?
« on: 11/02/2020 16:34:16 »
To put some math to this question, I'll use the centrifugal force equation:
Fc = (mv2)/r, where
"Fc" is the force in newtons,
"m" is the mass in kilograms,
"v" is the local rotational velocity in meters per second, and
"r" is the distance from the center of rotation in meters.
For a 70 kilogram human standing along the equator, the resulting centrifugal force acting on them is:
Fc = (mv2)/r
Fc = ((70 kg)(461 m/s)2)/6,334,080 m
Fc = 2.3486 newtons
For comparison, gravity pulls on a 70 kilogram human with a force of 70 kg x 9.807 m/s2 = 686.49 newtons. The centrifugal force is only 0.34% of the gravitational force at the equator. The human (or the ocean) is in no danger of being flung off of the Earth.
Fc = (mv2)/r, where
"Fc" is the force in newtons,
"m" is the mass in kilograms,
"v" is the local rotational velocity in meters per second, and
"r" is the distance from the center of rotation in meters.
For a 70 kilogram human standing along the equator, the resulting centrifugal force acting on them is:
Fc = (mv2)/r
Fc = ((70 kg)(461 m/s)2)/6,334,080 m
Fc = 2.3486 newtons
For comparison, gravity pulls on a 70 kilogram human with a force of 70 kg x 9.807 m/s2 = 686.49 newtons. The centrifugal force is only 0.34% of the gravitational force at the equator. The human (or the ocean) is in no danger of being flung off of the Earth.
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