1. How long would a day be if the earth were rotating sofast that the objects in

Question

1. How long would a day be if the earth were rotating sofast that the objects in the equator would be weightless? 2. Using one of the Keplers' laws, show that the ratioof the speeds of a planet at its near(perihelion) and far(apehelion) positions from the sun is equal to the inverse of theirdistances from the sun. 1. How long would a day be if the earth were rotating sofast that the objects in the equator would be weightless? 2. Using one of the Keplers' laws, show that the ratioof the speeds of a planet at its near(perihelion) and far(apehelion) positions from the sun is equal to the inverse of theirdistances from the sun.

Explanation / Answer

1. apply newton's second law to an object on the equator: sum of forces = ma the forces are the normal force of the earth acting up and weightacting down; these combine to produce a centrip force that actsdown to the center of the earth newton's second law is then N - mg = - mv^2/r or N=mg-mv^2/r an object will lose contact with the earth when N=0 or when g=v^2/r where r is the radius of the earth v=sqrt[gr]=sqrt[9.8ms/s/ x 6.4x10^6m] v=7.92x10^3 m/s to find the length of the day with this rotational speed, weuse time = distance/speed = 2pi r/speed time= 2x3.14x6.4x10^6m/7.92x10^3 m/s time = 5075s =1.41 hrs

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