# Boyant force on an immersed body, when system is in free fall?

1. Jul 2, 2007

### Mr Virtual

1. The problem statement, all variables and given/known data
This is not a numerical problem.
A body floats in a liquid contained in a beaker. The whole system falls freely under gravity. The upthrust on the body due to liquid is:
1. zero
2. equal to weight of the liquid displaced
3. equal to the weight of the body in air
4. equal to the weight of the immersed portion of the body.

2. Relevant equations

3. The attempt at a solution

Since system is in free fall, the body's apparent weight becomes zero. Since the body is not exerting any force on the liquid, therefore the liquid will not be displaced, and no boyant force will act on the body.

However, I may be wrong, as boyant force depends on density (mass/vol).

Mr V

2. Jul 2, 2007

### Dick

The liquid will certainly be displaced. But there is no buoyant force in free fall (g is zero which is also in the formula). There is no up or down. How would it know which direction to point? You are correct.

3. Jul 3, 2007

4. Jul 3, 2007

### nrqed

Just a comment. Your answer is correct, but it is incorrect to say that in free fall, the weight of an object is zero. This is not true.
For example, consider the astronauts floating around in a space shuttle in orbit around teh Earth. Is their weight zero? No!!! They are in free fall but the weight is not zero.
Saying that the weight is zero would be saying that when an object is in free fall, the force of gravity on it suddenly disappear. This is not the case at all.

(I am talking about the Newtonian view of gravity here. In GR, the interpretation is completely different but that's another story!)

5. Jul 3, 2007

### Mr Virtual

Yeah, you are right. I meant "apparent" weight of body becomes zero, meaning it is exerting zero force on the liquid. This is because both liquid and body are accelerating at the same rate.

Mr V

6. Jul 3, 2007

### nrqed

Yes, stated that way I agree 100%.

cheers!

7. Jul 3, 2007

### Mr Virtual

So, thanks again...

Mr V

8. Jul 4, 2007

### Staff: Mentor

I think one needs to realize that astronauts in space are constantly accelerating, i.e. they are in orbit, and have a tangential orbital velocity, which is much greater than the rotational speed of the earth's surface.

In contrast, one most often relates 'free fall' to an object falling vertically at an acceleration of g and a lateral velocity equal to that of the earth's surface.

9. Jul 4, 2007

### nrqed

I agree completely that they have a tangential orbital velocity.

I personally never thought that free fall was only used for when the object remains above the same point above the surface of the Earth. So you would say that if the space shuttle would be in geosynchronous orbit, the atsronauts would be in free fall but if the orbit is not geosynchronous, they are not in free fall? So they would have to look out the windows at the surface of the Earth to know if they are in free fall or not?

I always thought that "free fall" just meant that no other force than gravity was acting on the object.

Regards

10. Jul 4, 2007

### Staff: Mentor

In "free fall" near the earth, one is falling in the gravity field, but one still has the tangential velocity (speed) approximately equal to the velocity of the earth's surface. At the equator that speed is 1037.51 mph (1670 km/h). The rotational speed decreases as one travels toward either pole, i.e. to higher latitudes.

One can then compare the centripetal acceleration v2/r to g (9.81 m/s2).

v2/r = (1670000/3600 m/s)2/(6,356,750 m) = 0.034 m/sec2. So centripetal acceleration, 0.034 m/sec2 << 9.81 m/s2.

In contrast, ISS revolves about the earth at ~ 7700 m/s, at an altitude of ~342 km above earth's surface. So,

v2/r = (7700 m/s)2/(6,700,000 m) ~ 8.85 m/s2, which should be about acceleration of gravity at 342 km above earth's surface.

The astronauts are "free falling", but they have tangential velocity that keeps them 'falling' in orbit.

11. Jul 4, 2007

### Dick

I was also not aware of this limited definition of 'free fall'. Like nrqed, I thought it was the same as 'inertial frame'. So if I fall out of a plane, the plane has to be stationary before it is truly 'free fall'.?