# 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.

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.

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• 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