# Levitation/orbital speed and rotating objects

1. Jun 6, 2015

### federiconitidi

It is known that 8000m/sec is the speed required for an object travelling parallel to the earth's surface to orbit (i.e. so that the centripetal force counteracts the gravity force).

I am now thinking to an object linked to an axis of rotation (e.g. with a 1m string) and rotating about that axis with an angular speed (e.g. 80kRPM) so that the linear velocity of the object reaches the above mentioned 8000m/sec. Would that object start to levitate?

2. Jun 6, 2015

### Staff: Mentor

Welcome to PF!

Can you explain why it is that an object moving at 8km/s is in orbit? That will highlight the difference between the two scenarios.

3. Jun 6, 2015

### Noctisdark

Rotating around what ? if about the axis, then the centripital force point towards some point on the axis, anyway the object will become a helicopter, if you consider a planet with no air in it and a string that can hold a great force without breaking, then no, the centripital acceleration will point to point at which the string is attached but its weight will point to the planet's center of mass, these won't cancel out, it will help anything fly only if that acceleration points at the CoM and larger than g !

4. Jun 6, 2015

5. Jun 6, 2015

### Noctisdark

Do you understand what is the centripital force ? It's the force that bind a rotation object to the axis of rotation, and in earth it's the force if weight, if you want to rotate then your v^2/r must be equal to g, but but it doesn't stop there, force is a vector so the direction of the centripital force must point to the earth's center of mass, one last thing it isn't the centripital force that let you fly, it's what bind you to the earth while rotating

6. Jun 6, 2015

### federiconitidi

yes I consider rotation about the axis (say the axis is vertical)

The point I'm trying to make here is different.
• we say that the 8000m/sec is the linear velocity of a mass required to balance out the gravity force applied to it (orbital speed)
• a mass rotating about a vertical axis at 1m and 80kRPM has a linear velocity of about 8000m/sec
• this mean that at a give point in time, the mass has a linear velocity equal to the orbital speed
Question: would this counteract the force of gravity? if not, why?

7. Jun 6, 2015

### Noctisdark

Centripital force isn't responsible for preventing any rotating object from falling into earth, it doesn't cancel gravity in any case, and your case is very different, the centripital force is the string tension there !

8. Jun 6, 2015

### federiconitidi

Apologies I haven't been clear enough - please find a sketch which should explain better my question

9. Jun 6, 2015

### Noctisdark

Oh, it's different now, but no the object won leviate, you have the centripital force is the string tension and the weight force, sum them up and you'l get a nice vector, but still you'll have the same weight force, It will leviate if you try this expirement, because air is forcing you upword, a helicopter

10. Jun 6, 2015

### jbriggs444

In the case of an object in a normal orbit, the force of gravity on the object changes continuously. For half the orbit it is pointing generally in one direction and for the other half of the orbit it is pointing generally in the other direction. The net is zero. That is one way of understanding how an object can stay in orbit -- the force of gravity on it averaged over the entire orbit sums to zero.

In the case of the ball-orbiting-a-stick, the force of gravity is always in (very nearly) the same direction -- down. The force of gravity averaged over the entire circular path sums to a net downward force. No levitation.

11. Jun 6, 2015

### A.T.

No, gravity is the centripetal force for orbits. But in your scenario the string provides the centripetal force, while gravity just pulls the whole thing down.

12. Jun 6, 2015

### DaveC426913

Are you conjecturing that a speed of 8,000m/s in any direction (such as around a pole) should result in levitation?

13. Jun 7, 2015

### federiconitidi

correct,
• this tought being based on the fact that in any moment in time, the mass has a linear velocity equal to the orbital speed, so enough to "orbit"
• the fact that the object also rotates around a vertical axis should not change anything
I understand this is weird, however I have troubles finding where this reasoning is wrong

14. Jun 7, 2015

### DaveC426913

When velocity (including direction) is unchanging, the mass (which is trying to follow a straight line tangent to the Earth's surface) will find its altitude increasing as the curve of the Earth drops away. The mass is not free of gravity, in fact it continues to fall under the influence of gravity.

Your mass going around a pole will also try to move tangential to the Earth's surface, but unlike the first case, will never find the curvature of the Earth dropping away. Like the above, it continues to fall under the influence of gravity.

15. Jun 7, 2015

### Noctisdark

8000 m/s is the speed needed to orbit the EARTH, not some axis around it and the centripital force has nothing to do with cancelling the weight force, in fact if and object is orbiting the earth, the centripital force acting on it is it's weight, Fcentripital = mv2/R = mg if the object is near the surface if the earth, solving this will tell that the speed needed to orbit is 8000 m/s !