# Homework Help: Loss of GPE of a linearly growing raindrop

1. Jul 26, 2016

### Happiness

• Moved from a technical forum, so homework template missing
Consider a spherical raindrop that falls at a constant velocity and whose radius $r$ is proportional to the distance $h$ fallen, i.e., $r=kh$. Find the loss of gravitational potential energy (GPE) after it has fallen a distance $h$.

The given answer is $mg\frac{h}{4}$ but my answer is $mg\frac{h}{5}$. (The mass $M$ of the raindrop is $\frac{4}{3}\pi r^3\rho$, where $\rho$ is the density of the raindrop.)

After the raindrop has fallen a distance $h$, the centre of mass is a distance $h$ below the starting point. But since the mass of the raindrop is different at different positions below the starting point, the loss of GPE cannot be $mgh$. Since the mass is greater at positions further below the starting point, the loss of GPE should be smaller than $mgh$. The "initial" height of the raindrop has to be weighted according to its mass at different positions. This weighted-average initial height

$\bar{h}=\frac{\int_0^hMh\,dh}{\int_0^hM\,dh}=\frac{\int\frac{4}{3}\pi k^3h^4\rho\,dh}{\int\frac{4}{3}\pi k^3h^3\rho\,dh}=\frac{4}{5}h$.

Thus the "initial" height of the raindrop is $\frac{4}{5}h$ below the starting point, giving us a loss of GPE of $mgh-mg\frac{4}{5}h=mg\frac{h}{5}$.

Is this correct?

The suggested answer is attached below. (The loss of GPE is the first term of (4.209).)

2. Jul 26, 2016

What you need is an integral of g(h-x)dM where the M is a function of x. This one is tricky because the mass M which you have as a function of h does not fall the distance h. (The distance back to the starting point is h.) That's why in this one, it should be helpful to introduce a position coordinate x along with the final destination which is located at h from the starting point. (I do agree with the book's 1/4 answer.) editing... You also seem to be considering the raindrop as being finite in size compared to the distance h that it falls. The radius of the raindrop r=kh where k is a very small number so that for purposes of the gravitational potential (i.e. the distance that it falls) the raindrop can be considered as a point at any given time.

Last edited: Jul 26, 2016
3. Jul 27, 2016

### Happiness

Even if we consider that the upper part of the raindrop has higher GPE than its lower part, the answer is still the same I suppose. Assuming that the gravitational field is uniform, the center of gravity is at the center of the raindrop. And so we can consider the change in GPE from the change in position of the center of the raindrop. Right?

4. Jul 27, 2016

Yes. That's correct. The raindrop is also considered to be almost infinitesimal (r=1/4" at most) and the distance h may be 100 yards or more.

5. Jul 27, 2016

### Happiness

If the radius of the raindrop is larger than the distance it falls, then the loss of GPE is not $mg\frac{h}{4}$? Why not?

6. Jul 27, 2016