Centripetal Acceleration

[lauriecherie]

1. The problem statement, all variables and given/known data

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


2. Relevant equations

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




3. 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.