h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

The footnote is incorrect. By substituting latitude in the Coriolis Equation (where the coriolis force is a product of the angular rotation of the earth, the mass of the object, the speed of the object and latitude), at the equator latitude is zero, therefore coriolis is zero, at the poles it is maximum latutude and therefore the coriolis effect is maximum at the poles. I am glad to see that the author has poo poo'd the idea of coriolis influencing the direction water flows out of a plughole. Once again, substituting mass in the equation, the multiplier is tiny, so the effect is therefore tiny, as the author says.

A formula for Kinetic Energy due to rotation around an axis = 0.5(I.w^2), joules. K.E. = 0.5(m r^2)w^2, j. If you take other values for the power of `r`, does the equation give other rotational units? For example, d = -1,centrifugal pressure = m.r^-1.w^-2, Pa? for d = 0, centrifugal surface tension = m.w^2, kg.s^-2 ? for d = 1, centrifugal force = m.r.w^2, newtons? and back to d = 2, K.E. = m.r^2.w^2, j. Just as strange, if you do the same with the angular momentum equation, L or h = I.w.r^(d-2) = m.w.r^d, kg.m^2.s^-1. Then for d=-1, dynamic viscosity? d=0, viscosity? d=1, momentum? and d=2, angular momentum. Please note w = 2 pie divided by time period, or w=(Gm/r)^0.5.

Oops, left out the r^3 in the omega equation previously, hence w=(G.m.r^-3)^0.5, rad/sec. Since Force=m.r.w^2=G.m.m.r^-2.

Is rotational pressure along the radial, 0.5(m.r^-1)w^2, Pa. The acceleration of line density, kg per meter per sec.sq.?

What if the speed of the missile being fired north from the equator were greater than or equal to the orbital speed of the Earth? Its in orbit! Therefore at speeds less than the earths orbital speed, the missile goes around a circle of radius, smaller than the complete Earth orbital radius. Orbital Velocity=(G.total Mass/a reducing radius)sq.rt.. circle radius = G M /V^2 ?

A body of Air slowly going northward from the equator, being attracted to the Earth by gravity, causes the spin radius to contract and speed of spin to increase, hence the body of air moves from west to east as well as from the equator northward.

As the air mass moves north, the curvature of the Earth reduces the length between the Earths axis of rotation and the increasingly northerly surface location of the Air mass. Angular rotation is mostly conserved (ignoring sources of friction), Hence as the Earths curvature reduces the length of rotation by the cosine of theta, the angle from the axis of rotation, between the equator and the new location of the air mass. Spin or angular rotation = l(total) = l(suns rotation) + l(orbit of Earth) + l(rotation of the Earth) + l(any other rotation).. Anyway l(Earth rotational momentum at the equator) = (I w) = (r x p) = (r x m v), which is conserved no matter where the air mass moves, what does change is the velocity of the air with reference to the equators rotational velocity, by the cosine of theta. Hence V(air speed west to east)= V(Earths rotational velocity at the equator)(1- cos(theta)). Angle Theta being the Earths Latitude.

F(Coriolis) = rate of change over time of the linear momentum (m v). Where velocity v depends on v(equator)(1- cos(theta)).

The obvious problem with my equation for the Coriolis Force, is that Air mass speed at the pole could reach the equators rotational speed. Coriolis Force is obviously damped considerably by other interactions?

Traveling North from the equator causes air & water to veer off to the East, so traveling south toward the equator causes air & water to veer off to the west, setting up vortexes, cool dense fluid chasing and undercutting the warm fluid. How can the Coriolis force be a convenient fiction, when it plays such a vital role in facilitating the weather?

Google searched for the Coriolis equation, top site answer was from wikipedia, good graphics but I would question some of the assumptions made. It`s use of the ohm symbol for angular momentum I find a bit confusing as on other other topics involving angular momentum it uses other letters( probably due to the lack of this symbol on the keyboard?). If you click on, metcheck.com, jet stream, the updated jet weather map displays wind arrows. The jet is sometimes seen to flow down around a low and back up again, as it does so, the wind speed of the jet increases, before slowing as it leaves the low behind? Is this an example of the Coriolis force in action?

You Tube has a lot about the Coriolis effect, although not much maths. My own attempt, v(west to east)=V(at equator)[1-cos(theta)],m/s, is incorrect. Using, sin(theta) = 1-cos(theta), and v(equator)=464 m/s, the equation for tangential velocity at latitude theta, is given by 464*sin(theta), m/s. An object in contact with the Earth at a higher latitude has less rotational speed and angular momentum(Earth rotation and angular momentum are altered by the objects change of latitude, although not by much). Coriolis effect is noticed when the moving object is not in contact with, or only weakly connected to the Earths surface environs(by lack of friction, viscosity, or rolling resistance). Thus the amount of Coriolis effect is in proportion to the amount of interaction with the frame of reference?

A car in motion will only exhibit the Coriolis effect, if its wheels loose traction with the road surface, and then only as a very small component of the angle of the skid? Since the velocity of the sliding car is much less than the rotational velocity of the earth.

Checked workings out, looks like I have confused V(west to east) with V(latitude), try again.. V(West to East)=V(equator)*sin(latitude). And V(latitude)=V(Equator)*cos(latitude), hope it`s right this time. Also please remember that the amount of Coriolis effect is dependent on how well the moving object couples to the rotating background.

Torque = m v^2 [ 1 + - k sin w t ]. Force = m v^2 r(t)^-1 [ 1 + - k sin w t ]. Where v is the initial velocity imparted to the moving object, r(t) is the distance traveled by the object moving at imparted velocity v during time interval t, the objects mass is m, the Coriolis effect is assumed to be of constant amplitude k, the background rotates with an angular frequency of w = 2 pie Hertz, giving w t, radians.

Some of the web sites that try to explain the Coriolis effect say that the inward force of gravity is counter balanced by an outward Centrifugal force. Well, if the rotation at the Equator is 464 m /s, and 464x464 = 215296, dividing by the Earths radius 6378 km = V^2 /R, 0.033756, m /s^2. about 100 times smaller than gravity 10 m /s^2. At higher latitudes the effective distance to the axis of rotation is reduced, but so is the angular velocity squared, causing the centrifugal acceleration to be even lower than acceleration due to gravity pulling toward the Earths c of g. Which I believe to be the case, unless that`s the reason for maximum speed limits in the Highway code?

The best web site found so far, is at," www.gmat.unsw.edu.au/snap/gps/clynch_pdf/intgrav.pdf", explains the 3D gravitational potential energy gradient.

Why do Wikipedia pages give the impression that the spherical Coriolis effect is the same as on a spinning disc, when clearly they are the result of quite different effects, although sharing the same name? The cause in the spherical case being due to changes in latitude, while on a rotating disc its due to changes in angular velocity of the object, as it moves radially inward or outward, although in both cases the Coriolis effect happens due to a lack of coupling (friction) between the rotating object and the object upon its surface.

As far as I can deduce, a projectiles tangential speed west to east = (GM/r+h)^0.5 -465 cos(latitude), m/s. V = 1120-465 = 10735 m/s at the equator. Or (2GM/r+h)^0.5 - 465 * cos(latitude), m/s if it is to escape from earth orbit?

Is there an explanation for why the energy of an object in orbit is 2pie^2 or roughly 20 times higher than the same objects centrifugal energy? Since, E(centrifugal) = m v^2, while E(rotational) = I w^2 /2 = m r^2 (4 pie^2)/ t^2 1/2 ~ 20 m r^2 /t^2, joules.