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At the equator, the velocity of Earth's surface is about 465 metres per second (1,674 km/h; 1,040 mph). The amount of centripetal force required to cause an object to move along a circular path with a radius of 6378 kilometres (the Earth's equatorial radius), at 465 m/s, is about 0.034 newtons per kilogram of mass. For a 10,000-gram internal ...
Figure 3: (Left) Ball in a circular motion – rope provides centripetal force to keep the ball in a circle (Right) Rope is cut and the ball continues in a straight line with the velocity at the time of cutting the rope, in accord with Newton's law of inertia, because centripetal force is no longer there.
In Newtonian mechanics, gravity provides the centripetal force causing astronomical orbits. One common example involving centripetal force is the case in which a body moves with uniform speed along a circular path. The centripetal force is directed at right angles to the motion and also along the radius towards the centre of the circular path.
Animation depicting evolution of a Cornu spiral with the tangential circle with the same radius of curvature as at its tip, also known as an osculating circle.. To travel along a circular path, an object needs to be subject to a centripetal acceleration (for example: the Moon circles around the Earth because of gravity; a car turns its front wheels inward to generate a centripetal force).
The speed (or the magnitude of velocity) relative to the centre of mass is constant: [1]: 30 = = where: , is the gravitational constant, is the mass of both orbiting bodies (+), although in common practice, if the greater mass is significantly larger, the lesser mass is often neglected, with minimal change in the result.
The only requirement is that the central force exactly equals the centripetal force, which determines the required angular velocity for a given circular radius. Non-central forces (i.e., those that depend on the angular variables as well as the radius) are ignored here, since they do not produce circular orbits in general.
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There are many useful features of the effective potential, such as . To find the radius of a circular orbit, simply minimize the effective potential with respect to , or equivalently set the net force to zero and then solve for : = After solving for , plug this back into to find the maximum value of the effective potential .