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If the earth is rotating at some $465~\text{m}/\text{s}$ at the equator and that's really fast.

  1. Shouldn't we in that case be in orbit with the earth just not fast enough?
  2. How fast do we need to move?
  3. If the earth stopped rotating (gradually), shouldn't gravity increase significantly?
  4. Do satellites need extra speed when they are released or does it just depend on their inertia from the earth's rotation?
  5. If the satellite decelerated against the earth direction of rotation will it fall?
  6. If the earth rotated extremely fast, shouldn't the earth's crust along with everything else get ejected away by the centrifugal force?

I'm sorry if the questions cover many matters but I preferred to ask them all at once instead of asking each separately.

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I once calculated that you need to go $7906~\text{m}/\text{s}$ to become weightless. – Keep these mind Apr 11 '13 at 19:05
@Gugg with or without the inertia you already have? – Force Apr 11 '13 at 19:07
If you like this question you may also enjoy reading this and this Phys.SE posts. – Qmechanic Apr 11 '13 at 19:16
@Force Relative to the earth (and along its surface). – Keep these mind Apr 11 '13 at 19:19
Force, we prefer that you not put multiple unrelated questions in the same post. The questions you're asking here are somewhat related, so perhaps they could all stay together, but do keep in mind for the future that the general principle is one question per post. – David Z Apr 11 '13 at 19:37
up vote 4 down vote accepted

The acceleration of uniform circular motion is a very basic computation that we do for first year students. $$ a = \frac{v^2}{r} $$ which for someone standing on the Earth's equator comes to $$ a_\text{equator} \approx \frac{\left(465\text{ m/s}\right)^2}{6400\text{ km}} = 0.03\text{ m/s}^2$$ or less than 1% of g.

That is a measurable quantity, but not very significant. Indeed fluxuations of local $g$ at that level can (and do) occur simple due to local deposits of heavy ore. Mining and oil companies use precise gravitation maps in surveys for exactly this kind of reason.

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I'm quite familiar with the equation F=mv^2/r but I guess wasn't brave enough to use it. I was anticipating other factors when dealing with astronomy. Thanks for reminding me though.. – Force Apr 12 '13 at 0:30

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