scaredypants Posts: 32, Reputation: 1 Junior Member #1 Jan 20, 2009, 05:56 AM
Earths acceleration
What is the acceleration of the earths spin?
 tickle Posts: 23,800, Reputation: 2674 Expert #2 Jan 20, 2009, 06:20 AM

From wikipedia:

Standard gravity

Standard gravity, usually denoted by g0 or gn, is the nominal acceleration due to gravity at the Earth's surface at sea level. By definition it is equal to exactly 9.80665 m/s2 (approx. 32.174 ft/s2) (where the notation m/s2 means meters per second per second, and the notation ft/s2 means feet per second per second). This value was established by the 3rd CGPM (1901, CR 70).[1] [2]

The symbol g is sometimes also used for standard gravity, but g strictly means the local acceleration due to gravity, which varies depending on one's position on Earth (see Earth's gravity). The symbol g should not be confused with G, the gravitational constant, or g, the abbreviation for gram (which is not italicized). The g (sometimes written "gee") is also used as a unit of acceleration, with the value defined as above; see g-force.

The value of g0 defined above is a nominal midrange value on Earth, representing the acceleration of a body in free fall (in the absence of air resistance) at sea level at a geodetic latitude of about 45.5°. It is larger in magnitude than the average sea level acceleration on Earth, which is about 9.797 645 m/s2. Although the actual strength of gravity on Earth varies according to location, for weights and measures and many calculation purposes the standard gravity figure is used.

The SI unit of acceleration due to gravity (or, indeed, any acceleration), namely meters per square second, can also be written as newton per kilogram. The numeric value stays the same: gn = 9.80665 N/kg. This alternative representation can be understood by noting that the gravitational force acting on an object at the Earth's surface is proportional to the mass of the object: for each kilogram of mass, the Earth exerts a nominal force of 9.80665 newtons (though, as stated, the precise value varies depending on location).
 Capuchin Posts: 5,255, Reputation: 656 Uber Member #3 Jan 20, 2009, 07:41 AM

The deceleration causes the length of a year to lengthen by about 0.005 seconds every year. (with a year defined as 565.25 days)

Can you work out the deceleration in radians (or degrees) per second per second from this? (Or maybe you can find better units to make the number more intuitive?)
 tickle Posts: 23,800, Reputation: 2674 Expert #4 Jan 20, 2009, 09:58 AM

Hi, Capuchin, can you explain why it has nothing to do with the OPs question? I took the question to be related to the earth's eccelaration on its axis. Not so ? What am I missing in the OP's question. As if there is something to miss, it doesn't say much. I DID actually read and think about the answer before I accessed wikipedia.

You took it differently ?

Why did my answer rate a reddie, and especially from another expert ?
 Capuchin Posts: 5,255, Reputation: 656 Uber Member #5 Jan 20, 2009, 10:03 AM
Originally Posted by tickle
Hi, Capuchin, can you explain why it has nothing to do with the OPs question? I took the question to be related to the earth's eccelaration on its axis. Not so ? What am I missing in the OP's question. As if there is something to miss, it doesnt say much. I DID actually read and think about the answer before I accessed wikipedia.

You took it differently ?

Why did my answer rate a reddie, and especially from another expert ?
Yes, it is about the earth's rotation on it's axis... which has very little to do with gravitational acceleration of a body on the Earth?
 tickle Posts: 23,800, Reputation: 2674 Expert #6 Jan 20, 2009, 10:28 AM

My apologies then, and thanks for pointing that out, you could have done it more pleasantly. I am glad you made the correction.
 Capuchin Posts: 5,255, Reputation: 656 Uber Member #7 Jan 20, 2009, 10:29 AM
Originally Posted by tickle
My apologies then, and thanks for pointing that out, you could have done it more pleasantly. I am glad you made the correction.
? I think you put more weight into reddies than you should do. I use them because I believe that they make it more obvious to the OP that someone disagrees with that answer.
 ebaines Posts: 12,132, Reputation: 1307 Expert #8 Jan 22, 2009, 10:24 AM

Different people interpret things differently. Tickle read this as a question about the acceleration of falling objects due to earth's gravity. Capuchin read it as a question about the change of the earth's rotational velocity over time. I read this as a question about the centripetal acceleration of an object under constant rotational velocity: $a = r \omega ^2$. Three diffenent ways to answer one question!

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