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The blade can be treated as a stack of layers going from the axis out to the edge of the blade. Each layer exerts an outward (centrifugal) force on the immediately adjacent, radially inward layer and an inward (centripetal) force on the immediately adjacent, radially outward layer.
Unlike the other two fictitious forces, the centrifugal force always points radially outward from the axis of rotation of the rotating frame, with magnitude , where is the component of the position vector perpendicular to , and unlike the Coriolis force in particular, it is independent of the motion of the particle in the rotating frame.
The fictitious centrifugal force in the co-rotating frame is mrΩ 2, radially outward. The velocity of the particle in the co-rotating frame also is radially outward, because dφ′/dt = 0. The fictitious Coriolis force therefore has a value −2m(dr/dt)Ω, pointed in the direction of increasing φ only.
In this case, the radiated field asymptotically takes the form of planewaves propagating radially outward (in the ^ direction) with ^ = and = ^ / where Z is the scalar impedance / of the surrounding medium.
As a result, air travels clockwise around high pressure in the Northern Hemisphere and anticlockwise in the Southern Hemisphere. Air around low-pressure rotates in the opposite direction, so that the Coriolis force is directed radially outward and nearly balances an inwardly radial pressure gradient. [38] [better source needed]
A centripetal force (from Latin centrum, "center" and petere, "to seek" [1]) is a force that makes a body follow a curved path.The direction of the centripetal force is always orthogonal to the motion of the body and towards the fixed point of the instantaneous center of curvature of the path.
In contrast, to second order in m, in the Schwarzschild vacuum the magnitude of the radially outward acceleration of a static observer is m r −2 + m^2 r −3; here too, the second term expresses the fact that Einstein gravity is slightly stronger "at corresponding points" than Nordström gravity. The tidal tensor measured by a static observer is
At the Earth's surface this becomes: = ^ where g is the downward 9.8 m/s 2 acceleration due to gravity, and ^ is a unit vector in the radially outward direction from the center of the gravitating body. Thus here an outward proper force of mg is needed to keep one from accelerating downward.