The acceleration due to gravity on the surface of the Earth varies with latitude due to the centrifugal force resulting from the Earth's rotation. The effective acceleration due to gravity at a latitude
φ, denoted as
gφ, takes into account the centrifugal force and is less than the acceleration due to gravity that would be experienced if the Earth were not rotating (denoted as
g0 ). At the equator (
φ=0∘ ), this effect is maximal because the velocity due to Earth's rotation is maximal. As we go to the poles, the effect of rotation becomes minimal.
The effective acceleration due to gravity at latitude
φ is given by:
gφ=g0−ω2Rcos2φ Where:
g0 is the acceleration due to gravity without the Earth's rotation, at the equator
ω is the angular velocity of the Earth's rotation
R is the radius of the Earth
φ is the latitude
At the equator, the effective acceleration due to gravity,
g, is:
g=g0−ω2Rbecause
cos‌0∘=1.
At latitude
30∘, the effective acceleration due to gravity,
gφ, using
cos‌30∘=‌, is:
‌gφ=g0−ω2R(‌)2‌gφ=g0−ω2R×‌To find
|g−gφ|, we subtract the above two equations:
‌|g−gφ|=|(g0−ω2R)−(g0−ω2R×‌)|‌|g−gφ|=ω2R−ω2R×‌‌|g−gφ|=ω2R(1−‌)‌|g−gφ|=ω2R×‌So the value of
|g−gφ| is given by the option
A, which is:
‌ω2R