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Uniform Circular Motion

Uniform Circular Motion - gravitational attraction between...

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Uniform Circular Motion In chapter 4 we have seen that when a particle moves in a circle, it experiences an acceleration a, directed towards the center of the circle, with a magnitude equal to where v is the velocity of the particle, and r is the radius of the circle. The acceleration a is called the centripetal acceleration . To account for the centripetal acceleration, a centripetal force must be acting on this object. This force must be directed towards the center of the circle, and can be calculated from Newton's second law: An example of uniform circular motion is the motion of the moon around the earth. Suppose the period of this motion is T. What does this tell us about the distance r between the earth and the moon ? During one period, the moon covers a total distance equal to 2[pi]r. The velocity of the moon, v m , can be calculated: The corresponding centripetal force is Here we assumed that m m is the mass of the moon. The centripetal force is supplied by the
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Unformatted text preview: gravitational attraction between the earth and the moon. In Chapter 15 we will see that the strength of the gravitational interaction can be calculated as follows: where G is the gravitational constant and m e is the mass of the earth. For a constant circular motion, the gravitational force must provide the required centripetal force: The distance between the earth and the moon can therefore be calculated: The constant of gravity is known to be G = 6.67 x 10-11 m 3 /(s kg) and the mass of the earth is known to be m e = 5.98 x 10 24 kg. The measured period of the moon is 27.3 days (2.3 x 10 6 s). The distance between the moon and the earth can therefore be calculated: r = 3.82 x 10 8 m which agrees nicely with the distance obtained using other techniques (for example the measurement of the time it takes for light to travel from the earth to the moon and back). Figure 6.12. Forces acting on a car while rounding an unbanked curve....
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