Gravity and centrifugal force

Originally posted by FabianFnas
If the entire planet of Earth stopped rotating in one sole second, but air and water still is free - I think every continent wolud be flooded by an enormous flood.

I think the wind is a minor problem. But, yes, I think we will have a *strong* wind for a while.

This has actually happened once if you read the bible. The sun stood still for 24 hours, ...[text shortened]... ating and then started again. The effect of this disaster is not documented by the bible though.

I wonder if anyone's connected the Biblical Flood with this sun stopping.

Originally posted by AThousandYoung
I wonder if anyone's connected the Biblical Flood with this sun stopping.

No, the Great Flood hapened before. This is far later. No flood reported.

Originally posted by FabianFnas
No, the Great Flood hapened before. This is far later. No flood reported.

So here is a question: The earth is spinning about 1600 Km/hr at the equator. How fast would it have to spin before it flew apart?

Originally posted by sonhouse
So here is a question: The earth is spinning about 1600 Km/hr at the equator. How fast would it have to spin before it flew apart?

For one to feel no gravity at the equator, (and being able to jump into space, as in dreams) the centrifugal force has to be equal to the force of gravity:

ma=mv^2/r | :m
a=v^2/r | *r
a*r=v^2

a=9.81 m/s^2
r=6 380 km

v=sqrt (9.81 m/s^2 * 6 380 000 m)=7 910 m/s

or 7.9 km/s. The length of day would be 1 hour and 24 minutes.

Originally posted by gambiitti
For one to feel no gravity at the equator, (and being able to jump into space, as in dreams) the centrifugal force has to be equal to the force of gravity:

ma=mv^2/r | :m
a=v^2/r | *r
a*r=v^2

a=9.81 m/s^2
r=6 380 km

v=sqrt (9.81 m/s^2 * 6 380 000 m)=7 910 m/s

or 7.9 km/s. The length of day would be 1 hour and 24 minutes.

When earth is just about (due to the calculations) to fall apart, the radious is no longer 6 380 km. The equator bulge is more predominant in this case than it is today. Does this change the calculated answer, conciderably or just marginally?

Originally posted by FabianFnas
When earth is just about (due to the calculations) to fall apart, the radious is no longer 6 380 km. The equator bulge is more predominant in this case than it is today. Does this change the calculated answer, conciderably or just marginally?

It gets messy because we don't know the mantle's strength to resist shape deformation; completely rigid, it will never bulge (only loose parts want to separate); completely liquid, we have a kind of negative feedback system: As r increases, g decreases (by the inverse square law) and the required v decreases.

Originally posted by gambiitti
It gets messy because we don't know the mantle's strength to resist shape deformation; completely rigid, it will never bulge (only loose parts want to separate); completely liquid, we have a kind of negative feedback system: As r increases, g decreases (by the inverse square law) and the required v decreases.

I don't think we can see the crust as anything but liquid. Any stress and it will break apart. This is the reason we have active tectonics. The crust is so thin compare to the size of the Earth. We can easily regard the Earth as an molten drop of iron and silicon with no alteration of the physical formulae.

(The Moon, on the other hand, is more solid. Does the Moon has an molten core at all? But now we're talking about the Earth...)

When the rotation increase the Earth will form the shape of an ellipsoid. But with a certain rotational speed the objects of the equator has an acceleration outwards exceeding that of gravitation at this distance from the center.

Now, with this in mind, how much will the answer from the calculation differ?

Originally posted by FabianFnas
I don't think we can see the crust as anything but liquid. Any stress and it will break apart. This is the reason we have active tectonics. The crust is so thin compare to the size of the Earth. We can easily regard the Earth as an molten drop of iron and silicon with no alteration of the physical formulae.

(The Moon, on the other hand, is more solid. m the center.

Now, with this in mind, how much will the answer from the calculation differ?

As you said, the crust is thin compared to the size of the Earth. If we regarded the crust as a liquid, the answer still could be "negligible". The crust would stay on the mantle until 7.9 km/s is reached.

Originally posted by gambiitti
As you said, the crust is thin compared to the size of the Earth. We can regard the crust as a liquid, and the answer is still "negligible". The crust will stay on the mantle until 7.9 km/s is reached.

... even when r is conciderable higher?

Originally posted by FabianFnas
... even when r is conciderable higher?

But would it be? You can't form a huge r with a thin crust, right?

Originally posted by gambiitti
But would it be? You can't form a huge r with a thin crust, right?

The crust has no more strength as the molten interior. It's too thin for that.

I can imagine a value of r being 50% up to 100% higher than now.