# No gravity spin on Earth?

1. Apr 1, 2004

### 2Pac

"No gravity" spin on Earth?

I was recently asked this question in physics class:

If the day were shorter, you would weight less. How long would a fall day at the equator be (in hours) for you to be able to moon walk for the rest of your life?

After the ambiguity in the question was cleared up the question becomes: What would the length of a day be in order for an object, at the equator, to experience a net force of zero N towards the center of the Earth.

Any thoughts would we appreciated. I have come up with two different answers and would like to know which one is right.

2. Apr 1, 2004

### Staff: Mentor

It's the apparent weight--which is the normal force that the earth's surface pushes up with--that is reduced by a spinning earth. When the earth spins, objects on the surface are being centripetally accelerated. Applying Newton's 2nd law: mg - N = m ω2 R, you can solve for the angular speed ω that will give a zero normal force. From that you can figure out the length of the day.

3. Apr 1, 2004

### 2Pac

Ok. I solved for the velocity using: a=(v^2)/r

so... 9.8=(v^2)/6.378 X 10^6
v=7905m/s

then v=(pi*r*2)/t
t=1.4 hours? can anyone confirm if this is right?

4. Apr 1, 2004

### Staff: Mentor

Looks good to me.

5. Apr 1, 2004

### Janus

Staff Emeritus
Another way of looking at it is that the velocity an object has to have at the equator to be "weightless" is the same as the velocity it has to have to orbit the Earth at an altitude of sea-level.

So if you take the formula for gravitational force:

$$F_{g}= \frac{GMm}{r^{2}}$$

and the formula for centripetal Force:

$$F_{c}= \frac{mv^2}{r}$$

then equate them:

$$\frac{GMm}{r^{2}}=\frac{mv^2}{r}$$

solving for v will give you the needed velocity.

Divide the circumference of the Earth by this velocity, and you have your answer.

6. Apr 1, 2004

### 2Pac

thank you for the help

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7. Apr 1, 2004

### Janitor

A minor cough from Janitor

A geophysicist would probably predict that you should include the plasticity of the Earth in your calculations. A high rate of spin would involve more of a bulge at the equator. Some non-spherical model of the earth, such as an oblate spheroid, might make for an interesting calculation. I don't really have a feel for how much difference it would make from the spherical model.

8. Apr 4, 2004

### Andre

Which of course leads to the question, if an elastic spinning body can ever attain enough spinning to have zero gravity at its equatorial bulge.

9. Apr 25, 2004

### Divergent13

Janus I was wondering what should you take as M and m for the calculation you described?

10. Apr 25, 2004

### Divergent13

Nm that was a stupid question--- little m doesnt matter since they cancel. I apologize.

11. Apr 26, 2004

### Nereid

Staff Emeritus
... and the answer is ... ?

12. Apr 27, 2004

### Andre

The answer must be no. But how to prove it? Hey, there is so much mathematical power here. How about some action.

I'll start.

Consider a fully elastic (fluid) mass in a non gravitational environment. The gravitational force at the surface of the body, generated by it's own mass must be perpendicular to the surface, otherwise a flow would redistribute the mass.

Now when it starts to spin, a centrifugal force is added and the resultant total force pushes the spinning equator outwards.

But there is a second play of forces. The total gravity force on the poles is reducing because there is less effective mass directly under the poles and more mass to the sides that reduces effective gravity. Alternately on the equator, the total gravity component increases because there is more effective mass directly under the equator and less to the sides.

But we know that the gravity on the equator is less than on the poles so the resultant forces do not even out totally that way. I guess it's its not the forces that are in balance but (intuitively ) the surface pressure per angular segment or arc (what's the English term?). So the resultant force times the effective area of the arc (pressure) should be in balance. And the arc to the poles produces a smaller surface than a similar arc to the equator. So as long as the body has any geoid shape there is an area defined by an arc and a force. So the resultant force on the equator can't be zero

Who takes over with the math?

13. Apr 27, 2004

### Nereid

Staff Emeritus
Rather than just thinking of 'a planet' as we work through this, why not any object? I think it would be interesting to look at all scales of mass - except possibly (sub-)atomic - and states of matter (not only solid, liquid, gas, but also nuclear and electron degenerate states, and BECs; maybe even superfluids?).

Some regimes are easy - for objects with 'small' mass, the gravitational force is so weak that factors such as surface tension will clearly dominate. At the opposite end - dense, massive objects - the answer is also easy: a black hole cannot rotate fast enough, period.

So, no spinning blobs of gas, of any mass? How about a plasma?

Liquids (other than superfluids?) can exist as spinning objects, with zero gravity at their equator, provided they are small enough. How small? (perhaps it's a combination of mass and density - how different would the behaviour of a spinning blob of molten osmium be from that of a spinning drop of liquid hydrogen?)

14. Apr 27, 2004

### Stingray

It is possible for a body to spin fast enough that its surface is almost orbiting. This is called the mass-shedding limit.

MacLaurin found analytic solutions for a rigidly rotating incompressible fluid more than 200 years ago, which includes the maximally rotating case. Instead of having a smooth bulge, the "planet" develops a cusp at the equator. See this for pictures and current developments (its surprisingly complicated):

http://xxx.lanl.gov/abs/astro-ph/0208267

15. May 3, 2004

### Phobos

Staff Emeritus
Well I should say so!