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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).)

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).)