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- Engineering
- Under the condition of weightlessness due to rotation of earth at equator time period of earth will be
.
Answers (1)
If you do the math (equating to
and using M(Mass of the earth) =
and
), it follows that the size of the object is entirely irrelevant and that only the density ρ of the sphere enters into the equation for f, the number of revolutions per unit time:
For ρ=5.5×10^3 kilogram per cubic meter (the density of planet earth) it follows that f=0.197×10^(−3) revolutions per second, corresponding to a revolution period of 50705070seconds (1 hour and 24 minutes).
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