thrown off

How fast would the earth have to spin, so that stuff
started flying off?

--
Rich

On 20/06/2013 10:28 PM, wrote:
How fast would the earth have to spin, so that stuff
started flying off?


Given the Earth's radius is 6367.5 km, a 1 kg mass, being held down by 1
N of gravitational force would need to balance out that 1 N of gravity
with an equal and opposite 1 N centripetal force, at a minimum.
Therefore the Earth would need to rotate at an incredible speed of
rotation speed of 2.523 km/s = 5645 mph = 9084 km/h at the equator!

The Earth's actual current rotational speed is only 0.4651013 km/s =
1040.4 mph = 1674.365 km/h. This results in a miniscule centripetal
force of only 0.03397 N, against the Earth's gravitational force of 1 N
on a 1 kg mass.

Yousuf Khan

Yousuf Khan wrote:
On 20/06/2013 10:28 PM, wrote:
How fast would the earth have to spin, so that stuff
started flying off?


Given the Earth's radius is 6367.5 km, a 1 kg mass, being held down
by 1 N of gravitational force would need to balance out that 1 N of
gravity with an equal and opposite 1 N centripetal force, at a
minimum. Therefore the Earth would need to rotate at an incredible
speed of rotation speed of 2.523 km/s = 5645 mph = 9084 km/h at the
equator!
The Earth's actual current rotational speed is only 0.4651013 km/s =
1040.4 mph = 1674.365 km/h. This results in a miniscule centripetal
force of only 0.03397 N, against the Earth's gravitational force of 1
N on a 1 kg mass.

Yousuf Khan

You are forgetting the Earth's gravitational acceleration in the equation!
Earth's gravitational force on a 1 kg mass is 9.8 N. g = 9.8 m/s^2.

Hence for balance, mg = m v^2/r or, circular velocity v for balance of
centripetal acceleration and gravitational attraction is v = sqrt (rg) =
7900 m/s or 7.9 km/s. For a circumference of 40000 km, the rotation period
would need to be 1.406 hours (5063 sec).

The reduction in weight at the equator vs the poles for today's 23h56m
rotation period (neglecting obliquity of the Earth) is about 0.0034 or 0.34
%.

--
Mike Dworetsky

(Remove pants sp*mbl*ck to reply)

On Thu, 20 Jun 2013 19:28:24 -0700 (PDT),
wrote:

How fast would the earth have to spin, so that stuff
started flying off?


I have a question. How much would a 200 pound person weigh if the
earth did not spin

wrote:
On Thu, 20 Jun 2013 19:28:24 -0700 (PDT),
wrote:

How fast would the earth have to spin, so that stuff
started flying off?


I have a question. How much would a 200 pound person weigh if the
earth did not spin

A 200-lb polar explorer at 90 deg latitude would weigh the same whether or
not the Earth was spinning.

At the equator the weight (measured by a spring scale) would be 0.68 kg less
than at the pole, with the Earth's current rate of spin. But the person's
mass would still be the same.

--
Mike Dworetsky

(Remove pants sp*mbl*ck to reply)

Mike Dworetsky wrote:
Yousuf Khan wrote:
On 20/06/2013 10:28 PM, wrote:
How fast would the earth have to spin, so that stuff
started flying off?


Given the Earth's radius is 6367.5 km, a 1 kg mass, being held down
by 1 N of gravitational force would need to balance out that 1 N of
gravity with an equal and opposite 1 N centripetal force, at a
minimum. Therefore the Earth would need to rotate at an incredible
speed of rotation speed of 2.523 km/s = 5645 mph = 9084 km/h at the
equator!
The Earth's actual current rotational speed is only 0.4651013 km/s =
1040.4 mph = 1674.365 km/h. This results in a miniscule centripetal
force of only 0.03397 N, against the Earth's gravitational force of 1
N on a 1 kg mass.

Yousuf Khan

You are forgetting the Earth's gravitational acceleration in the
equation! Earth's gravitational force on a 1 kg mass is 9.8 N. g =
9.8 m/s^2.
Hence for balance, mg = m v^2/r or, circular velocity v for balance of
centripetal acceleration and gravitational attraction is v = sqrt (rg) =
7900 m/s or 7.9 km/s. For a circumference of 40000 km, the
rotation period would need to be 1.406 hours (5063 sec).

The reduction in weight at the equator vs the poles for today's 23h56m
rotation period (neglecting obliquity of the Earth) is about 0.0034
or 0.34 %.

Thinking about it for a moment, the required speed would be the same as that
needed for reaching orbital velocity at altitude zero. If there were no
atmosphere that would work, so stuff would start "flying off", but only at
the equator.

--
Mike Dworetsky

(Remove pants sp*mbl*ck to reply)

Dear r_dela...:

On Thursday, June 20, 2013 7:28:24 PM UTC-7, wrote:
How fast would the earth have to spin, so that stuff
started flying off?

Others have given you an answer based on the Earth being a solid body with its current shape.

The Earth is like a creme filled chocolate, and if you even tried to double its speed, it would lobe up, and spin off another Moon. Even a 50% increase would be pretty disastrous.

David A. Smith

dlzc wrote:
Dear r_dela...:

On Thursday, June 20, 2013 7:28:24 PM UTC-7,
wrote:
How fast would the earth have to spin, so that stuff
started flying off?

Others have given you an answer based on the Earth being a solid body
with its current shape.

The Earth is like a creme filled chocolate, and if you even tried to
double its speed, it would lobe up, and spin off another Moon. Even
a 50% increase would be pretty disastrous.

David A. Smith

If by double its speed, you mean an equatorial velocity of 0.93 km/s, it
would have only a very small effect (slightly more oblate) and in fact the
Earth probably had such a speed a very long time ago. The idea that the
Earth "spun off" the Moon is very old and nowadays pretty well discredited.

--
Mike Dworetsky

(Remove pants sp*mbl*ck to reply)

On 21/06/2013 10:11 AM, dlzc wrote:
Dear r_dela...:

On Thursday, June 20, 2013 7:28:24 PM UTC-7,
wrote:
How fast would the earth have to spin, so that stuff started flying
off?

Others have given you an answer based on the Earth being a solid body
with its current shape.

The Earth is like a creme filled chocolate, and if you even tried to
double its speed, it would lobe up, and spin off another Moon. Even
a 50% increase would be pretty disastrous.

Ummm, creme-filled chocolate Moon.

Yousuf Khan

Dear Mike Dworetsky:

On Friday, June 21, 2013 8:04:33 AM UTC-7, Mike Dworetsky wrote:
dlzc wrote:

Dear r_dela...:

On Thursday, June 20, 2013 7:28:24 PM UTC-7,

wrote:

How fast would the earth have to spin, so that stuff
started flying off?

Others have given you an answer based on the Earth
being a solid body with its current shape.

The Earth is like a creme filled chocolate, and if
you even tried to double its speed, it would lobe
up, and spin off another Moon. Even a 50% increase
would be pretty disastrous.

If by double its speed, you mean an equatorial
velocity of 0.93 km/s, it would have only a very
small effect (slightly more oblate) and in fact the
Earth probably had such a speed a very long time ago.

We had 16 hour days about 2.2 billion years ago (tidal rhytmites). This was when the Moon was much closer. Today, the Moon affects a 4m high lump in the Earth's crust. Spin it faster, and I think that lump will not have time to form, which will increase drag on the plates themselves. It will be really easy for the system to become unstable...

The idea that the Earth "spun off" the Moon is
very old and nowadays pretty well discredited.

I think it is inherent to a Theia collision, post collision, allowing for sorting of light elements into a lobe of sharper curvature... prior to the masses separating. Spinning off uncaused, yes I agree with you. Spinning off because the mergence had too much angular momentum, makes sense to me.

David A. Smith