# Homework Help: Compression of a sphere underwater

1. Nov 9, 2008

### ultrapowerpie

1. The problem statement, all variables and given/known data
A solid metal sphere of volume 1.05 m^3 is lowered to a depth in the ocean where the water pressure is equal to 1.38 x 10^7 N/m^2. The bulk modulus of the metal from which the sphere is made is 9.9 X 10^9 N/m^2

Given: The atmosphereic pressure is 1.013 10^5 Pa.

What is the change in the volume of the sphere? Give in m^3

2. Relevant equations
The equation is:

$$\Delta V = (-1/B)*Pressure*V$$

B= Bulk modulus
V= volume

3. The attempt at a solution

I really just have one question: what does the atmospheric pressure have to do with anything? I know that F/A (which is the original equation) is pressure, but can I just use the water pressure, or do I somehow have to incooperate the atmospheric pressure?

2. Nov 9, 2008

### asleight

You probably have to use the change in pressure, $$\Delta\rho$$.

3. Nov 9, 2008

### ultrapowerpie

Hmmm, that makes a little bit of sense, let me try that real fast...

Nope, problem says it's wrong. I also tried adding the two together, that's wrong. And multiplying them makes the Delta V larger then the total volume of the sphere. >.>

4. Nov 9, 2008

### mgb_phys

Wether you should subtract the air presusre depends wether they mean that the total pressure is 1.38 x 10^7 ps or the pressure only due to seawater is 1.38 x 10^7pa. They almost certainly mean the total pressure in which case the atmospheric pressure is unnecessary.
But since the difference is much smaller than the accuracy you are given any figures for you can just ignore it.

5. Nov 9, 2008

### ultrapowerpie

Ok, I don't know what is going on here, but the stupid online homework keeps saying it's wrong. I don't know what the heck I'm doing wrong, and I really wish that the system had more then just a 1% margin of error >.>

1/B= 1.01e-10

(1/B)*P= .00139

.00139*1.05= .00146 m^3

Yet, for some reason, it says it's wrong. >.>

6. Nov 9, 2008

### mgb_phys

The simplest way is to just look at the units.
Bulk modulus has units of pressure.
The volume obviously has units of volume and the answer you want has units of volume.
You don't expect the volume of a metal ball to change very much so you are looking for a small number.

So it must be something like (bulk modulus/pressure)*volume.
(9.9 X 10^9 N/m^2/1.38 x 10^7 N/m^2) * 1.05 m^3 = 0.075 m^3 (check units balance)

Working it out this way is a lot more reliable than trying to remember the formula .

7. Nov 9, 2008

### ultrapowerpie

Ok, when I tried doing it your way, I'm getting 753.26 m^3, not .075 m^3, did I do something wrong?

8. Nov 9, 2008

### mgb_phys

After what I said about wanting a small number I typed them the wrong way up!
(1.38 x 10^7 N/m^2/9.9 X 10^9 N/m^2) * 1.05 m^3 = 0.0015 m^3

You always have to be careful with calculators and big/small numbers, so do an estimate
1.38 x 10^7 /9.9 X 10^9 = 1.38/9.9 x 10^(7-9) = (1.4/10) * 10^-2 = 0.14/100 = 0.0014