How can we conclude that the Earth is not a perfect sphere based on the acceleration due to gravity?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(N/A) The acceleration due to gravity is given by $g = \frac{GM_e}{R_e^2}$,which implies $g \propto \frac{1}{R_e^2}$. The Earth is slightly bulging at the equator,meaning the radius $R_e$ is larger at the equator,resulting in a lower value of $g$. Conversely,the Earth is flattened at the poles,meaning the radius $R_e$ is smaller at the poles,resulting in a higher value of $g$. Since $g$ varies with latitude,we can conclude that the Earth is not a perfect sphere.

Explore More

Similar Questions

Since the Earth is not a perfect sphere,what is the effect on the acceleration due to gravity $(g)$?

Difficult
View Solution

At what height above the surface of the Earth does the value of $g$ decrease by $2 \%$? [Radius of the Earth is $6400 \, km$]

There is a planet which is $8$ times more massive and $27$ times denser than the Earth. If $g^{\prime}$ and $g$ are the accelerations due to gravity on the surfaces of the planet and the Earth respectively,then:

How does the acceleration due to gravity $(g)$ at a location on Earth change with latitude?

How does the rotation of the Earth affect the acceleration due to gravity on its surface?

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo