How fast would the Earth have to spin to lose the atmosphere?

And it made me curious. Is there a certain limit of a planet's spin speed, where it becomes unfit to wear an atmosphere?

Based on the calculation in the question:

The velocity of the Earth at its surface can be calculated by noting it takes $$1 \text{ day} = 86400 \text{ s}$$ to rotate a full $$2\pi$$ radians, so

$$v=\frac{2\pi R_\text{Earth}}{T_\text{day}}\sim\frac{2\pi\times6400\times10^3\text{ m}}{86400\text{ s}}\sim470\text{ m s}^{-1}$$

The centripetal acceleration is then

$$a=\frac{v^2}{R_\text{Earth}}\sim\frac{(470\text{ m s}^{-1})^2}{6400\times10^3\text{ m}}\sim0.035\text{ m s}^{-2}\ll9.8\text{ m s}^{-2}$$

So basically, the Earth would need to spin 300 times faster?

Because 0.035m/s^2 *300 ~ 9.8 m/s^2?

This is where gravity would equal the centrifugal force?

Where did I make a mistake?

Question:

1. What is the spin limit, where Earth would not be able to keep the atmosphere?
• Why do you think you made a mistake? If the planet is splinning fast enough to overcome gravity everything will go, not just the atmosphere - there's no other glue holding it all together. Atmosphere can be stripped from planets due to incident radiation from the local star, this is a big factor. – JMLCarter Oct 14 '18 at 23:22
• You don't need to totally cancel $g$. Consider: the Moon has a surface gravity around $g/6$, but its atmosphere is negligible. – PM 2Ring Oct 14 '18 at 23:45
• @JMLCarter Gravity isn't the only glue. There's also chemical bonding (which is electromagnetic). A small rock is held together by its chemical bonds, its self-gravity is tiny. But certainly, if the Earth were made to spin fast enough to cancel gravity at the equator, Bad Things would happen. – PM 2Ring Oct 14 '18 at 23:52
• Right - "little" other glue. – JMLCarter Oct 15 '18 at 0:24