The earth is rotating about its axis with a speed of 465 m/s toward east.
An object on the equator will rotate about the axis of earth with this speed of 465 m/s.
For an observer from the sun, the object is moving along a circle of radius 6871 km with a speed of 465 m/s.
We shall calculate the centripetal force or acceleration needed by this object to go round this circle.
v^2/r = 465² / 6871e3 = 0.031 m/s^2 .
The object will go round the earth if it is pulled toward the center with this small acceleration.
Imagine that the gravitational pull has some how reduced to this value of 0.031 m/s ^2 instead of the usual 9.8m/s^2.
The object is acted upon by this only gravitational force which is now equals 0.031m/s^2.
This being the centripetal fore, it will go round the circle. But note that it will not press down the ground beneath it.
However, in reality, this acceleration is 9.81 m/s^2. Hence it presses the ground by a force of
9.81- 0.031 = 9.779 m/s^2.
The ground as a reaction provides an upward force equal to 9.779.
The object is moving along the circle, since it has the necessary centripetal force.
Imagine what will happen all of a sudden the ground on which we are standing suddenly goes inside?
In fact when we are standing on earth, the net force acting on us is zero. The up thrust and the gravitational pull are equal and opposite.
However since we are moving along a circle, the up thrust will be less by that small amount and this reduced up thrust equals the reduced value of gravitational pull.
In 9.8m/s^2 acceleration an amount of 0.031 is used for circular motion.
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