Thread: "No gravity" spin on Earth? View Single Post
 Emeritus Sci Advisor PF Gold P: 2,352 Another way of looking at it is that the velocity an object has to have at the equator to be "weightless" is the same as the velocity it has to have to orbit the Earth at an altitude of sea-level. So if you take the formula for gravitational force: $$F_{g}= \frac{GMm}{r^{2}}$$ and the formula for centripetal Force: $$F_{c}= \frac{mv^2}{r}$$ then equate them: $$\frac{GMm}{r^{2}}=\frac{mv^2}{r}$$ solving for v will give you the needed velocity. Divide the circumference of the Earth by this velocity, and you have your answer.