When an object is on the surface fo the Earth, it experiences a centrifugal force that depends on the latitude of the object on Earth. If the Earth were not spinning, the force on the object would have been mg. However, the object experiences an additional centrifugal force due to spinning of the Earth.
This centrifugal force is given by `momega^(2)R.`.
`OP_(z) cos lamda = (PZ)/(OP) = (R.)/(R)`
`R. = R = cos lamda`
where `lamda` is the latitude. The component of centrifugal acceleration experienced by the object in the direction opposite to g is
`a_(PQ) = omega^(2)R cos lamda = omega^(2)R cos^(2) lamda`
Since `R. = R cos lamda`
Therefore, `g.= g - omega^(2) R cos^(2) lamda`
From the above expression, we can infer that at equator, `lamda = 0, g. = g - omega^(2)R`. The acceleration due to gravity is minimum. At poles `lamda = 90, g. = g`, it is maximum. At the equator, g. is minimum.
Variation of g with depth: Consider a particle of mass m which is in a deep mine on the Earth. (Example: coal mines in Neyveli). Assume the depth of the mine as d. To calculate g. at a depth d, consider the following points. The part of the Earth which is above the radius `(R_(e) - d)` do not contribute to the acceleration. The result is proved earlier and is given as
`g. = (GM.)/(R_(e) - d)^(2)` ...(1)
Here M. is the mass of the Earth of radius `(R_(e) -d)` Assuming the density of Earth `rho` to be constant,
`rho M/V` ...(2)
where M is the mass of the Earth and V its volume, Thus,
`rho (M.)/(V.)`
`(M.)/(V.) = M/V` and `M. M/V V.`
`M. = ((M)/(3/4 pi R_(e)^(3))) (4/3 pi(R_(e) - d)^(3))`
`M. = (M)/(R_(e)^(3)) (R_(e) - d)^(3)` ...(3)
`g. = G (M)/(R_(e)^(3)) (R_(e) - d)^(3) (1)/(R_(e)-d)^(2)`
`g. = GM (R_(e)(1 (d)/(R_(e))))/(R_(e)^(3)) = GM ((1 (d)/(R_(e))))/(R_(e)^(2))`
Thus
`g. = g(1 - (d)/(R_(e)))` ...(4)
Here also `g.lt g` . As depth increases, g. decreases. It is very interesting to know that acceleration due to gravity is maximum on the surface of the Earth but decreases when we go either upward or downward.
