`On Wed, 2 Dec 1998, Roeland Th. Jansen wrote:> On Wed, Dec 02, 1998 at 03:25:24PM -0000, Simon Kenyon wrote:> > On 02-Dec-98 Alex Buell wrote:> > > What is the first thing that ceases when you spin down a planet? That's> > > right, gravity. Everything not tied down would just fly out into space,> > > and that includes the atmosphere. Simple. > > i don't see a smiley, but i have to assume that that was a little joke! right?> well, the problem is that... it's true. so don't use ower management on the> earth spin...No. The problem is that it's _not_ true. Sorry but I can not resist.Gravity between two bodies is a function like this (Universal GravityLaw, Newton, formulated long ago):F=f(m1, m2, x)where:F	forcem1, m2	masses of the two bodies involvedx	distance of the two bodiesOn the surface of earth we have:         (M * m)F = K * ---------           r^2whereM is the mass of earthm is the mass of body on earth surfacer is the radius of earthWe know what is the number g (gravity acceleration):g = 9.81 m/s^2But confronting g with the Law on the surface, we can write:g = K * M/(r^2), where K, M and r are constantsso that we obtain the well known:F = m * gthat is the force exercised by the earth on a body on her surface.Now, this force is _not_ dependent of earth rotation.The rotation simply balances part of the F force giving, for vectorialaddition, the well known "1 gravity" on the surface. This one is simple todemonstrate:         /|\          | Force (centripetal, vector) exercised by earth rotation:          | Fc = m * w^2 * r          | where          | w = earth spinning in 1/sec          | r = earth radius          | m = mass of body on her surface          *         [|] <----- body         / \        ----- [earth surface] ----> earth spinning (w)          |          | Gravity force calculated with Newton's law:          | F = m * g          | where          | m = mass of body          | g = gravity acceleration          |         \|/Now you can compose the two vectors into the total force exercised by theearth:Ft = F - Fc = m * g - m * w^2 * rWe can now, for example calculate what is the w that is necessary to fly:To fly the Ft force _must_ be 0 (no forces acting on the body) so that:m * g = m * w^2 * rg is a constant (oh yes ... here I must note that it is not really a                 constant, but for this approximate calculation we can                 assume it to be equal to 9.81 m/sec^2)r is a constant equal approximatively 6378 Km = 6.378 * 10^6 mso that:w(fly) = sqrt(g/r) = 1.24 * 10^-3 sec^-1so that the earth should make 1 spin in:w(fly) = 1.24 * 10^-3 [1/sec] * 3600 [sec/h] = 4.464 h^-1For a body on her surface to fly, the earth must spin at more than 4 spinsin an hour (or, if you prefer, 0.22 hours for one single spin) !This result contradict your statement that if earth spins down all thebody on her surface will fly (it is exactly the countrary ... if theearth spins down the force acting on bodies is maximum).Note that the centripetal force is not so high, and it varies with theposition on the surface of earth you are calculating that vector.For example at the poles the centripetal force is zero because you are onthe axis of rotation of earth (approximatively) while on the equator ismaximum (still approximatively, all the calculations done here madeconsidering the body at the equator). Of course making these statements Ihave considered the earth an omogeneus body: g varies with densityvariation of earth too.Note that this was a very approximate calculus, in any way you look atit. Anyway the result, knowing its approximation, is a quite welldescription of the reality, given that the considered body is stationarywith respect of a reference system blocked on the earth, because if it isin movement you must account for many other forces that change a lot theresultant Ft).Ciao,	Riccardo.-To unsubscribe from this list: send the line "unsubscribe linux-kernel" inthe body of a message to majordomo@vger.rutgers.eduPlease read the FAQ at http://www.tux.org/lkml/`