What Rotation Period Causes 9.8 m/s² Equatorial Acceleration?

[lauriecherie]

Homework Statement

What would Earth's rotation period have to be for objects on the equator to have a centripetal acceleration equal to 9.8 m/s^2?

____ min

Homework Equations

Centripetal acceleration is equal to velocity * (2PI/T),
where T is the period in seconds.

The Attempt at a Solution

I set centripetal acceleration equal to 9.8m/s^2 and solved for T. Then I took T and divided it by 60 so I could get the answer in minutes. I came out with 4.97 min which Webassign says is incorrect. Any ideas?

 

[LowlyPion]

One way to look at it is that if the object is not to fall into the center then calculate the orbit about a mass of Earth at 1 Earth radius.

So what is the period of such an orbit?

V²/R = GM/R² = ω²R

ω² = GM/R³ = (2π /T)²

T = 2π*(R³/GM)^1/2

 

[lauriecherie]

What does GM stand for?

 

[LowlyPion]

What does GM stand for?

That's sometimes written as μ which is the standard gravitational parameter for earth.

It is the product of Earth's mass and the Universal Gravitational constant.