Gravitational potential energy versus free surface energy

Classical Mechanics Level 3

The radius of a water bubble is 1 1 m., which is doubled. What is the ratio of the increase in it's free surface energy and it's gravitational potential energy?

Given :

Universal Gravitational Constant = =

6.67408 × 1 0 − 11 6.67408\times 10^{-11} S. I. Units

Surface Tension of water = =

72 × 1 0 − 3 72\times 10^{-3} S. I. Units

Density of water = =

1000 1000 S. I. Units


The answer is 9273.44.

A Former Brilliant Member by A Former Brilliant Member

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

The gravitational potential energy of a spherical shell of mass m m and radius R R is G m 2 2 R \dfrac{Gm^2}{2R} .

The free surface energy of a spherical shell of radius R R whose material has a surface tension γ \gamma is 8 π γ R 2 8π\gamma R^2 .

Substituting values we get the required ratio as ≈ 9273.44 \approx 9273.44 .

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...