4 ms·

It's approximately only a problem of density. For large objects, you can neglect tensile strength, so "How fast until it's torn apart" becomes "how fast would i

It's approximately only a problem of density. For large objects, you can neglect tensile strength, so "How fast until it's torn apart" becomes "how fast would it have to spin for an object on the equator to be in orbit?" F_c = m_1 r ω^2 = F_g = G m_1 m_2 / r^2 cancel out the orbiting mass (m_1), rearrange it a bit ω^2 = G (m_2 / r^3) where you'll note that term in parenthesis is directly proportional to the density. So the answer would probably be ~a 2.2 hour day.

mrb 2y ago
"how fast would it have to spin for an object on the equator to be in orbit" I think this is correct, but then it can be simply calculated knowing the escape velocity on Earth is 11.2 km/sec and the circumference is 40e3 km: 40e3 / 11.2 = 3571 seconds Therefore it's the Earth span at one revolution per hour, objects at the equator would escape gravity... I'm not sure where is your error in your formula that leads you to a different response.