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If a planet is rotating fast enough, no object can remain on the surface at the...

If a planet is rotating fast enough, no object can remain on the surface at the equator and is flung off due to the planet's rotation at the equator. Suppose a planet is spinning that fast and the length a day on that planet happens to the length of a day on Earth. Another random planet has a mass which is a factor of 3.08 times larger that of first planet and a radius which a factor of 3.17 times larger than that of the first planet. What does the length of the day need to be on the planet in terms an Earth day if this second planet is spinning fast enough such that everything is flung off the equator as well?

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change in the value of aviration accelertin Since objec eguateo it mean 2 0 2 Similary fr s e cmd blane Taking rahio 2.

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