Question

Please solve q 82?

Answer 1

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 `1/3000` that of the gravitational field. I maintain that that means there is essentially no change.

Answer 2

`"(3) increases"`

Explanation:

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

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

where

• `"g"_"e" =`Acceleration due to gravity at equator
• `"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 `"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(phi)` can’t be negative. As `phi` ranges from `0°` to `90°` only.

When Earth stops rotating `(ω = 0)`

`"g"_2 = "g"_"e" - 0`

`"g"_2 = "g"_"e"`

In second case we are not subtracting any term from `"g"_"e"`. Therefore, `"g"_2 > "g"_1`. Acceleration is increased.

_If you’re interested to know the derivation of that equation, [click here.](https://hemantmore.org/physics-1/p11202004/3882)_