uniform circular motion The Earth turns once around its axis in 24.0 hours. We w
ID: 1328950 • Letter: U
Question
uniform circular motion
The Earth turns once around its axis in 24.0 hours. We will assume that it is perfectly spherical, with radius 6400 km. The mass of the Earth is taken to be 6.00 × 1024 kg. The gravitational acceleration on the surface is taken to be g = 9.80 m/s2.
a) What is the centripetal acceleration of a person at the equator? How big a fraction of the gravitational acceleration does this correspond to? Remember to draw a sketch of the situation.
b) How fast would the Earth have to turn for the centripetal acceleration to be exactly equal to gravity? Give the answer in revolutions per day. Remember to draw a sketch of the situation.
c) Define ? to be the latitude, so that ? = 0 corresponds to the equator, and ? = 90? is the North Pole. What is the centripetal acceleration of a person standing on the surface at a given value of ?? Remember to draw a sketch of the situation.
d) Taking into account that the centripetal acceleration is towards the ro- tation axis, and gravity is towards the centre of the Earth, how fast should the Earth now turn for gravity to be just enough to provide the necessary centripetal acceleration (as function of ?)? Remember to draw a sketch of the situation.
e) Using the force of gravity
Explanation / Answer
a. centripetal acclerationa = W^2R
here W = 2pIR/T = (2*3.14 * 6400000)/(24*60*60)
W = 465.186 rad./s
so a = 465.186* 465.186 * 6400000
a = 1.38 e 12 m/s^2
in terms of g, it is a = 1.41e11 g
---------------------------------------------------------
here g = GM/R^2 = RW^2
W^2 = GM/R^3
W^2 = 9.8/6400000
W = 0.00123 rad/sec
W = 0.00123 * 24*60*60/(2*3.14) = 17 rev per day
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