# Rate of reduction of the volume of a raindrop

• benzel20
In summary, Virga is a weather phenomenon where rain evaporates before reaching the ground in very dry regions. The volume of a raindrop is proportional to the 3/2 power of its surface area, which is reasonable because raindrops are typically spherical. The rate of reduction of a raindrop's volume is also proportional to its surface area, and the formula for the amount of time it takes for a raindrop to evaporate completely can be expressed in terms of these constants and the initial surface area of the raindrop. It is not possible for the rate of volume reduction to be proportional to the square of the surface area, as this would not align with the formula for the volume of a sphere.
benzel20
I have trouble solving the following question, would someone offer some help? thanks.

In very dry regions, the phenomenon called Virga is very important because it can endanger aeroplanes. [See Wikipedia]

Virga is rain in air that is so dry that the raindrops evaporate before they can reach the ground. Suppose that the volume of a raindrop is proportional to the 3/2 power of its surface area. [Why is this reasonable? Note: raindrops are not spherical, but let's assume that they always have the same shape, no mater what their size may be.]

Suppose that the rate of reduction of the volume of a raindrop is proportional to its surface area. [Why is this reasonable?]

Find a formula for the amoung of time it takes for a virga raindrop to evaporate completely, expressed in terms of the constants you introduced and the initial surface area of a raindrop. Check that the units of your formula are correct. Suppose somebody suggests that the rate of reduction of the volume of a raindrop is prpoportional to the square of the surface area. Argue that this cannot be correct.

The volume of a sphere (and a falling rain drop is very well modeled by a sphere, not a "raindrop" shape) is given by $V= (4/3)\pi r^3$. Its surface are is given by $A= 4\pi r^2$. If you solve the second equation for r, you get $r= \sqrt{A/4\pi}$ (r is positive, of course). Putting that into the first equation,
$$V= \frac{4}{3}\pi \left(\sqrt{\frac{A}{4\pi}}\right)^3$$
$$= \frac{4}{3}\pi \frac{A^{3/2}}{4^{3/2}\pi^{3/2}}= \frac{1}{(3)4^{1/2}\pi^{3/2}}A^{3/2}$$
and you can diferentiate that with respect to A.

Another way to do this, perhaps simpler, is to use the chain rule:
$$\frac{dV}{dA}= \frac{dV}{dr}\frac{dr}{dA}= \frac{\frac{dV}{dr}}{\frac{dA}{dr}}$$

## 1. What factors affect the rate of reduction of the volume of a raindrop?

The rate of reduction of a raindrop's volume is affected by several factors, including air temperature, humidity, wind speed, and the size and shape of the raindrop.

## 2. How does air temperature impact the rate of reduction of a raindrop's volume?

Air temperature plays a significant role in the rate of reduction of a raindrop's volume. Higher temperatures cause the raindrop to evaporate more quickly, leading to a faster reduction in volume.

## 3. What is the relationship between humidity and the rate of reduction of a raindrop's volume?

Humidity is another important factor in the rate of reduction of a raindrop's volume. Higher humidity slows down the rate of evaporation, resulting in a slower decrease in the raindrop's volume.

## 4. How does wind speed affect the rate of reduction of a raindrop's volume?

Wind speed can have a significant impact on the rate of reduction of a raindrop's volume. Higher wind speeds can cause the raindrop to break apart into smaller droplets, increasing the surface area and promoting faster evaporation.

## 5. Is the rate of reduction of a raindrop's volume consistent for all raindrops?

No, the rate of reduction of a raindrop's volume can vary based on the size and shape of the raindrop, as well as the environmental conditions. Larger raindrops tend to have a slower rate of reduction compared to smaller raindrops.

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