# Homework Help: If there was a zero net acceleration at the equator?

1. Nov 20, 2011

### PotentialE

1. The problem statement, all variables and given/known data
How long would a day have to be on Earth if there was zero net acceleration at the equator?

2. Relevant equations
Ac=v^2/r
g = 9.8m/ss
T = earth's circumference / rotational velocity
V= SQRT(Ac*R)
Earth's circumference = 40023890.41

3. The attempt at a solution
The real acceleration of earth due to it's rotation is .03m/ss, but it doesn't really matter to us because g=9.8m/ss so we can stay on Earth. so the net acceleration in 9.8 - .34

if there was a zero net acceleration, then the acceleration of earth due to it's rotation would have to be 9.8m/ss, right?

so doing V= SQRT(Ac*R), i get 7901.013m/s for the rotational velocity of the Earth.

T = earth's circumference / rotational velocity, i get 506.56s, or, .141hrs. Is that correct?

2. Nov 20, 2011

### PeterO

I take it the question relates to how quickly the earth would rotate if an object released is not observed to fall, as we currently see.

That is what happens in an orbitting satellite/space station.

They are not very far above the surface [compared to the Radius of the Earth [R = 6000+ km, but many satellites are only ~300 km above the surface] so they give a good indication of what is necessary.

Many of those satellites go around about once every 90 minutes, or a bit less, so we can imagine a day on that fast spinning world would be about 85-90 minutes long.

3. Nov 20, 2011

### PeterO

You might get there quicker if you used Ac = 4π2R/T2 rather than Ac = v2/R

4. Nov 20, 2011

### PotentialE

ok, i checked my math by doing my way again and you're way and i got 1.407 hours, my bad. but is 1.47hours now correct?

5. Nov 20, 2011

### PeterO

Why did you just change 1.407 to 1.47 ?

It is certainly an answer around the right size!

EDIT: If you use two methods and get the same answer, that answer is most likely correct.