0
$\begingroup$

$g$ effective at any latitude changes due to rotation of earth. Does it also change due to rotation when the body is at height or at depth or does it change only at the surface?

Edit: I am referring to variation in acceleration due to axial rotation of earth which changes with the latitude. Does it depend on the height or depth of the body also?

Improve this question
$\endgroup$
3
  • $\begingroup$ Are you referring to centripetal acceleration? That’s going to depend on radius, so it’ll change as you move further away from and closer in to the body $\endgroup$
    – Tal
    Mar 29 at 13:31
  • 1
    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/8074/2451 and links therein. $\endgroup$
    – Qmechanic ♦
    Mar 29 at 16:02
  • $\begingroup$ Related: en.wikipedia.org/wiki/Geoid $\endgroup$
    – PM 2Ring
    Mar 30 at 9:00

1 Answer 1

Reset to default
1
$\begingroup$

Acceleration due to gravity at a certain latitude is given by : $$g_{new}=g-R\omega^{2}cos\theta$$ where R is radius of earth(that is valid if you are on surface) , $\omega$ is angular velocity and $\theta$ is latitude of earth

It is certain that "R" changes with height or depth because its the distance from center of earth and hence the $g_{new}$ must change

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.