# Calculating the buoyant force of a rising bubble

I am solving the problem:

A gas bubble rising from the ocean floor is 1 inch in diameter at a depth of 50 feet. Given that specific gravity of seawater is 1.03, the buoyant force in lbs being exerted on the bubble at this instant is nearest to:

The given answer is 0.020 lbs.

I start from the equation for buoyancy:

$$F_b=(P_{fluid}-P_{bubble})Vg$$

$g=32.2\:\mathrm{ft/s^2}$

$V=(4/3)\pi(1\:\mathrm{in}/2)^3 * (1 \:\mathrm{ft}/12 \:\mathrm{in})^3=3.03*10^{-4}\:\mathrm{ft^3}$

$P_{fluid}=sp.gr(saltwater)*densityH_2O=1.03*62.4=64.272\:\mathrm{lb_m/ft^3}$

P_gas: Pressure, $P=62.4*50\:\mathrm{ft}=3120\:\mathrm{lb/ft^2}$

$PV=mRT$

$P_{bubble}=m/V=P/RT=3120/53.34*492R (STP)=0.11889\:\mathrm{lb_m/ft^3}$

When I substitute all of the values I only get 0.607 lbs. What am I doing wrong?

The buoyant force is equivalent to the weight of the fluid that the bubble displaces, not the difference of the fluid's and body's weight.

$$F_b = \rho V g$$

$$V = \frac{4}{3} \pi (1/2)^3 = \frac{1}{6} \pi\:\mathrm{in^3}$$

$$V = 3,03 \cdot 10^{-4}\:\mathrm{ft^3}$$

$$\rho = 1.03 \cdot \rho_{water} = 64.272\:\mathrm{\frac{lbs}{ft^3}}$$

And now you simply multiply $$\rho$$ with $$V$$ and get $$\sim 0.02$$.

As a matter of fact the information about the depth is irrelevant as the force is independent of the depth. As per comments: Generally speaking the depth is not relevant for the force as long as the object immersed is incompressible. In the case of a bubble, which is compressible, the volume will decrease with increasing depth. Hence the buoyant force will decrease. Also the object needs to be fully immersed for the first statement to be true.

• this answer is demonstrably incorrect. To see why that is, assume the density of the bubble material is 999.9 kg/m^3. Clearly the buoyant force on this bubble is much less than it would be if the bubble was air. The density difference is important, as stated by engineering.stackexchange.com/users/8246/patrick Jun 27 at 22:14
• sciencedirect.com/science/article/pii/S1738573316303667 - see equation 1. This does not answer OP's question by his answer is different, but there is no doubt that density difference is relevant. You shouldn't be posting answers unless you know they're right Jun 28 at 2:14
• @user192127 If you look at equation (1): they calculate the rising velocity. There is no doubt that the velocity is dependent on the bubble material's density. However the question was about the buoyancy force. And that force is not dependet on the density of the bubble. The dependency on the density of the bubble is introduced by looking at the force balance, where the gravitational force introduces said density. You are confusing the net force with the buoyant force. If this is still unclear to you feel free to ask a question on this SE :) Jun 29 at 8:29
• webhome.phy.duke.edu/~schol/phy151/faqs/faq11/node4.html - to quote: Buoyancy'' is a more specific term. It refers to the net force experienced by an object immersed in a fluid the "net" here is important. As to "confusion" and "unclear" - I was engineering design lead for the largest refinery in the world. Nobody has ever used such terms to refer to me. Precise definitions of terminology matters. See how the Duke article specifically states what they mean by including the word "net". Jun 30 at 16:32
• this is incorrect: "As a matter of fact the information about the depth is irrelevant as the force is independent of the depth." Bubble volume varies according to the depth since gases are compressible. The buoyant force (whether the absolute buoyant force or the net force) will be much different when the same bubble is 100m below liquid surface or 1 m below. Undergrad college problems usually neglect this effect, but deep-sea construction organizations consider this important. More realism: higher pressures will cause more gas to be absorbed in the liquid thus further affecting bubble size Jun 30 at 17:55

the difference in densities is essential. A bubble of air will have much more of a buoyant force than a bubble of oil in the water.

• This doesn't answer the question of why the OP is not getting the answer that was expected.
– Fred
Oct 1 '16 at 23:16