Loss of GPE of a linearly growing raindrop

[Happiness]

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).)

 

[Charles Link]

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.

 

[Happiness]

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.

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?

 

[Charles Link]

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?

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.

 

[Happiness]

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.

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?

 

[Charles Link]

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?

Then your calculation involves building a mountain of water. All kinds of things like the shape of the drop=shape of the huge blob of water would need to be considered. You would then need to reexamine your calculations very carefully. Perhaps for a perfectly spherical drop your calculations would hold precisely. For the calculation at hand you are basically assuming a drop that is very small compared to any distance traveled. There's really no reason to add any extra precision. You are then able to use the mathematics to treat the case of the falling drop that is shown to be very much non-spherical (the part that was at the bottom of your post #1) without any additional complications.