If the earth stops rotating, the apparant value of g on its surface will?

No.

Yes, there is the effect of the rotation of earth on the value of g ! If you wan't an answer in detail then read further -

Lets us Resolve the various forces on acting on a mass m placed at a point P on the surface of the earth . The forces are the the centripetal force F_c and the Weight of the body =mg ! Let the resultant value of g at a place be g' and the angular speed of the earth's rotation be w ! Rest of the things will be clear from the figure -

Now apply vector law of addition in the Parallelogram PABO -

(PB)² = (PO)² + (PA)² + 2(PA)(PB) cos(180- lambda) , Or we can write it as -

(mg')² = (mg)² + (mrw²) + 2(mg)(mrw²) (-cos (lambda) )

Where F_c= mrw² = vector PA ,

Look at the figure , - cos(lambda) = r/R , therefore r=Rcos(lambda) !

(mg')² = (mg)² +(mRw²cos(lambda) )² - 2 (mg) (mRcos(lambda)w² ) cos(lambda)

(mg')² = (mg)² +(mRw²cos(lambda) )² - 2 (m²gRw² ) cos²(lambda)

Solving the equation and neglecting the higher order terms(who have got very less relative numerical value) we get the following relation between g and g' !

g' = g - Rw²cos²(lambda) !

Now it is clear that the value of g decreases on account of rotation of the Earth ! Thus if the earth stops rotating , then w=0

Hnece , g' = g at all the places on the earth !

Right now the value of g is not constant allwhere on the earths surface ! It depends on the cosine of the latitude (lambda) !

At the equator lambda = 0^o hence g' = g - Rw²

At the Poles , lambda = 90^o hence g' = g !

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