Tuesday

October 13, 2015
Posted by **Nikita** on Monday, October 26, 2009 at 11:58am.

- Physics -
**Nikita**, Monday, October 26, 2009 at 12:01pmHow long would it take for the Earth to complete a full turn if a person at 49.2° northern geographical latitude floats apparently weightlessly across the room? Use REarth = 6,385 km for the radius of Earth.

- Physics -
**MathMate**, Monday, October 26, 2009 at 12:37pmNot sure if I understood your question.

I interpret it as saying "if the Earth is rotating at a yet unknown angular velocity ω such that a person would float weightlessly at latitude 49.2°N, find &omega."

It is not as simple as it sounds, because the acceleration due to gravity acts towards the centre of the Earth. On the other hand, the rotation of the Earth is around a N-S axis, causing the centripetal force to be at an angle θ with the vertical, where θ is the latitude.

Assuming that the vertical (towards the centre of the earth) components balance, and the person floating is restrained from flying south by a horizontal rope, then we can do the following calculations:

Acceleration due to gravity, g = 9.8 m/s²

Radius of the Earth, R = 6385 km = 6385000 m

Latitude = 49.2°

We will find r, the distance of the surface of the earth to the axis of rotation, i.e. measured along the equatorial plane.

r = Rcos(θ)

Centripetal acceleration, a

= rω² (perpendicular to axis of rotation)

Vertical component of centripetal acceleration, av

= a cos(θ)

= rω² cos(&theta)

= Rω&up2; cos^{2}(&theta)

Equate av and g, solve for &omega.

I get 0.0019 radians/sec. which translats to a full rotation in 55 minutes and 14 seconds.

- Physics-corr -
**MathMate**, Monday, October 26, 2009 at 12:40pmEditorial correction:

Vertical component of centripetal acceleration, av

= a cos(θ)

= rω² cos(θ)

= R ω² cos²(θ)

- Physics-supp. reading -
**MathMate**, Monday, October 26, 2009 at 12:51pmHere's an article complete with figures for supplementary reading:

http://galitzin.mines.edu/INTROGP/notes_template.jsp?url=GRAV%2FNOTES%2Flatitude.html&page=Gravity%3A%20Notes%3A%20Latitude%20Variations

- Physics -
**Pat**, Thursday, October 28, 2010 at 11:15pmhow did you translate 0.0019 radians/sec into minutes?