# Simple question about gravity under water

1. Apr 1, 2017

### tomwilliam2

1. The problem statement, all variables and given/known data
Consider the same person at three locations. A: scuba diving underwater, B: standing on the beach, and C: standing on a mountain. Using <, >, =, make statements connecting the magnitude of their respective weights, i.e.
$W_A < W_B$

2. Relevant equations
Gravitational acceleration diminishes with distance from the mass causing the gravity.

3. The attempt at a solution
This was my 12-year-old son's homework question. He answered:
$W_A < W_B$
$W_A < W_C$
$W_B > W_C$
Using the reasoning that the higher up you are, the less you will weigh due to the diminishing gravitational acceleration. Under water, however, you would be effectively weightless due to the buoyancy. The teacher marked him wrong on the two containing the weight of A. Is he right?

2. Apr 1, 2017

### kuruman

The weight of a body is defined as the force with which the Earth attracts the body. The closer you move to the center of the Earth, the less your weight. Buoyancy is irrelevant. Is one weightless when sitting on a chair? Does it matter if the chair is holding one up instead of water?

3. Apr 1, 2017

### tomwilliam2

Are you sure? We thought it was the opposite. Isn't it based on an inverse square law?

4. Apr 1, 2017

### kuruman

I am sure. The inverse square law works for objects above the surface of the Earth. If you drill a hole straight through the center of the Earth, the deeper you go, the less your weight. That's because the only mass that counts is whatever is between you and the center of the Earth. At the center your weight is zero because you have equal amounts of Earth pulling you in all directions.

5. Apr 1, 2017

### PeroK

This is a very poor question, as there are all sorts of factors that might lead to different answers.

First, given that you are under water when scuba diving, you are below sea-level and could be said to be slightly below the surface of the Earth. So, you could argue that your weight has reduced from that on the beach. Not because of buoyancy, of course.

If you took the Earth to be of uniform density, then it would be:

$W_B > W_A > W_C$

However, the Earth is less dense near the surface, so as you go below the Earth's surface, gravity increases slightly for a while, before decreasing with depth (until eventually 0 at the centre)..

So, technically, I'd say it is $W_A > W_B > W_C$.

See:

https://en.wikipedia.org/wiki/Gravity_of_Earth#Depth

6. Apr 1, 2017

### tomwilliam2

Thanks for the answers. I should state that my son is only 12, so I'm guessing that the premises are simple, such as the density of the Earth is considered to be constant. I simplified the question a little just to get the idea I needed, but now I think the teacher himself got it wrong. I'm uploading a photo of the question:
It's not in English, so m = mass and P = weight (peso). If what you say is correct, then $P_A < P_B$ I guess.

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7. Apr 1, 2017

### TomHart

The photo is such a poor quality that it is not possible to make out some of the letters. Can you post a better quality photo?

8. Apr 1, 2017

### Bandersnatch

At 12 years old, I'd expect the question only to concern the distance from the centre of the Earth, so only height matters. The question seems to be designed to test the ability to understand the difference between the physical concepts of mass and weight.
If that's the case, then it's always lower=greater weight. Being immersed in water is a red herring.

I seriously doubt a class at this stage would have worked through the implications of varying mass, uniform density or not, especially since that'd require introducing the other consequence of shell theorem - telling kids that they can treat all of Earth's mass as if it were concentrated in the centre can usually fly without any actual explaining, but disregarding mass in the shell is a taller order.

9. Apr 1, 2017

### tomwilliam2

#### Attached Files:

• ###### Weight.jpeg
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10. Apr 1, 2017

### tomwilliam2

Thanks. I think you're probably right. The part about being immersed in water threw me too, when going through the test with him. I knew that the water exerts a buoyant force which counteracts the diver's weight, but I wasn't sure whether his "weight" was considered to be the balance of vertical forces or independent of other forces acting on him. The point made earlier about someone not being weightless sitting in a chair makes it clear, though.

11. Apr 2, 2017

### haruspex

Yes, but your original statement did not make it clear that you were comparing sea level with deeper. Indeed, the question has nothing deeper; in the context of the question there is only sea level (beach or scuba) compared with up a mountain (which, while still on Earth's surface, is above sea level).

Can we agree on "closer to sea level implies greater weight"? Though of course it is a bit more complicated than that.

12. Apr 2, 2017

### kuruman

I would agree with that. It's still not clear to me, though, what "below sea level" might imply in the context of this question for 12-year olds.