Apr 3, 2018

The value of $g$ would change almost not at all. There is a small reduction in the downward force acting on a body due to the rotation of Earth, but its magnitude is far smaller than the gravitational force at the surface.

Explanation:

In response to this question, I calculated the answer to this question, if we were at the equator, where the change would be greatest: https://socratic.org/questions/5667107b7c01493f09503ec0?source=search

It turns out that the rotational motion's contribution is about $\frac{1}{3000}$ that of the gravitational field. I maintain that that means there is essentially no change.

Apr 3, 2018

$\text{(3) increases}$

Explanation:

Acceleration due to gravity at a latitude ($\phi$) is given as

"g"_(phi) = "g"_"e" - "R"ω^2cos^2(phi)

where

• $\text{g"_"e} =$Acceleration due to gravity at equator
• $\text{R =}$ Radius of Earth
• ω =Angular Velocity with which earth is rotating
• $\phi =$Angle between the equatorial plane and the line joining the point where acceleration due to gravity is ${\text{g}}_{\phi}$ to the centre of the earth. It ranges from 0° at equator to 90° at poles.

When Earth was rotating

"g"_1 = "g"_"e" - "R"ω^2cos(phi)

Note: ${\cos}^{2} \left(\phi\right)$ can’t be negative. As $\phi$ ranges from 0° to 90° only.

When Earth stops rotating (ω = 0)

$\text{g"_2 = "g"_"e} - 0$

$\text{g"_2 = "g"_"e}$

In second case we are not subtracting any term from $\text{g"_"e}$. Therefore, ${\text{g"_2 > "g}}_{1}$. Acceleration is increased.

If you’re interested to know the derivation of that equation, click here.