If the rotation of a planet of radius 6.91 × 106 m and free-fall acceleration 7.97 m/s2 increased to the point?
If the rotation of a planet of radius 6.91 × 106 m and free-fall acceleration 7.97 m/s2 increased to the point that the centripetal acceleration was equal to the gravitational acceleration at the equator, what would be the tangential speed of a person standing at the equator?
Answer in units of m/s
How long would a day be in this case?
Answer in units of h
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g = V²/R → V = √[R*g] = 7421.1 m/s
T = 2πR/V = 5850.5 sec = 1.625 hr
i could call for help, it is the sole factor, sorry. Bears run quicker than human beings, and that they climb quicker too, to not point out I even have short legs. do not think of i will outrun a undergo.