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While reading my physics book. I came across a line that says that:

Rain drop falls with a constant velocity because the weight(which is the force of gravity acting on body) of the drop is balanced by the sum of the buoyant force and force due to friction(or viscosity )of air. Thus the net force on the drop is zero so it falls down with a constant velocity.

I was not satisfied by the explanation So I searched the internet which too had similar explanations:

The falling drop increases speed until the resistance of the air equals the pull of gravity, at which point the drop begins to fall at a constant speed, its terminal velocity.

My confusion regarding the matter is that if the net force acting on a body (here the rain drop) is zero then it should remain suspended in air rather than falling towards the earth. So how come the rain drop keeps falling when net force acting on it becomes zero? How the air resistance and other forces stops the rain drop from acquiring accelerated downward motion?

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    $\begingroup$ Except that rain drops don't fall at a particular constant velocity. Any particular rain drop will, once it reaches terminal velocity, but that velocity depends on the size of the drop, and other factors like elevation. But ANYTHING falling through a gas or liquid medium will eventually reach a terminal velocity that depends on factors including size, shape, and density. $\endgroup$
    – jamesqf
    Commented Oct 1, 2021 at 15:24
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    $\begingroup$ If you want to get REALLY technical, terminal velocity will be different depending on air pressure as well, of course this difference will be so slight as to not really matter... 6.99993 m/s and 7.00012 m/s are basically the same speed. $\endgroup$
    – aslum
    Commented Oct 1, 2021 at 16:09
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    $\begingroup$ @AdilMohammed: of course, including skydivers. They're capable of reconfiguring their bodies into poses with different terminal velocities, however. $\endgroup$
    – Nick Matteo
    Commented Oct 1, 2021 at 18:12
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    $\begingroup$ @AdilMohammed I was curious so I did some quick calculations, even with a perfectly streamlined dive, you will reach your top speed (about ~180 m/s or 310 mph) about 14 seconds into your jump, having traveled about 1000m (3240 feet). Typical jumping heights start at about 4200 m (14,000 ft) so you still have a ways to go after reaching terminal velocity. All these numbers are a bit off since I didn't properly account for wind resistance, but they are in the right ballpark. Of course in full drag configuration, you will reach TV much faster. $\endgroup$
    – eps
    Commented Oct 1, 2021 at 18:42
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    $\begingroup$ As counter-intuitive as it sounds, it is a key concept of physics that any object moving with constant velocity, regardless of what that velocity is, experiences zero NET force. In other words, it does not take a net force to keep an object moving at constant velocity. $\endgroup$
    – David White
    Commented Oct 1, 2021 at 19:31

14 Answers 14

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Here is a slightly different way to think of this. If the net force is zero, the acceleration of the droplet is zero- even though its velocity is not zero. With the acceleration zero, the velocity remains constant as it falls.

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If the forces on an object add to zero (in an inertial reference frame) it experiences zero acceleration and hence moves with a constant velocity (but not necessarily zero velocity). Thus, this is a particular case of the Newton's second law (not to be confused with the Newton's first law, which defines the intertial reference frames.)

Mathematically, the movement of a drop can be described as $$ \frac{d\mathbf{r}}{dt}=\mathbf{v},\\ m\frac{d\mathbf{v}}{dt}=m\mathbf{g} + \mathbf{F}_{drag}, $$ (for simplicity I neglect the bouyancy.) The drag force is directed against the drop velocity and can be modeled as $$\mathbf{F}_{drag}=-m\gamma\mathbf{v}.$$ Choosing our $y$-axis along the vertical direction and assuming that the initial drop velocity is zero, we have: $$ m\frac{d v_y}{dt}=-mg-m\gamma v_y\rightarrow \frac{d v_y}{dt}=-g-\gamma v_y $$ The stationary velocity can be obtained readily by setting the right-hand side of this equation to zero: $$ v_0=-\frac{g}{\gamma}. $$ The above differential equation can be also easily solved with usual methods as $$v_y(t)=\frac{g}{\gamma}(e^{-\gamma t}-1).$$ This equation describes a drop that starts with initial velocity zero, accelerates towards the earth, and eventually approach velocity $v_0$. enter image description here

(the image is taken from this link)

The same phenomenon is often studied as a laboratory excercise in physics programs, where small balls are dropped in a viscous liquid, such as glycerol: due to high viscosity of the liquid, the ball quickly reaches its stationary velocity and then moves with a constant speed. E.g., see this video

Update
While the calculation above assumed that the drag force is linear in speed (see Stoke's law), quadratic drag is more appropriate for the actual speeds of raindrops (see Terminal velocity). See also discussion about the speed and diameter of raindrops, as well as the list of useful references on this page.

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  • $\begingroup$ note that mass is not in the equation for drag $$\mathbf{F}_{drag}=-\frac{1}{2} \rho A c_d v \left|v\right|$$ $\endgroup$
    – Joost Döbken
    Commented Oct 5, 2022 at 15:07
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    $\begingroup$ Recently worked this out here for a constanc $c_d$ and $v\geq0$; you can write: $$\frac{\partial v}{\partial t} = g - k v^2$$ where $k=\frac{\rho A c_d}{2m}$. Rewrite to $$\partial t = \frac{\partial v}{g-kv^2}$$. which integrated gives $$t=\frac{1}{\sqrt{gk}}\tanh^{-1}\left(\sqrt{k/g} \cdot v \right)$$. Rewritten as $$v(t)=\sqrt{g/k} \tanh \left(\sqrt{gk} \cdot t\right)$$ $\endgroup$
    – Joost Döbken
    Commented Oct 5, 2022 at 15:17
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    $\begingroup$ So the maximum velocity is $$v_{max}=\sqrt{\frac{2gm}{\rho A c_d}}$$ since $\tanh$ converts to 1 $\endgroup$
    – Joost Döbken
    Commented Oct 5, 2022 at 15:21
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This is Newton's first law: if the force vanishes the velocity is constant. Constant but not necessarily zero.

The resistive force increase as long as speed increases. When it is equal to the gravitational force the speed no longer increases and the total force vanishes leading to constant speed.

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If the net force is zero, it means the body continues to be in a state of uniform motion (Newton's first law or the Law of inertia). When it attained terminal velocity, the net force on the droplet became zero and it continued its uniform motion (motion along a straight line with constant velocity).

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My confusion regarding the matter is that if the net force acting on a body (here the rain drop) is zero then it should remain suspended in air rather than falling towards the earth.

The friction force only exists when there is relative movement between a drop and the air.
If the drop is not moving relative to the air the only force on it is the gravitational attraction of the Earth which will accelerate the drop.
The frictional force increases as the relative speed between the drop and the air increases which means that the acceleration of the drop decreases but it is still speeding up.
Eventually the speed of the drop relative to the air is such that the frictional force is equal and opposite to the gravitational attraction of the Earth and so the net force on the drop is zero and it moves with constant velocity.

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The force of gravity is unchanging (mostly) as a water drop falls, if the object was in a vacuum it will continue to accelerate as it falls, moving faster and faster until it hits the ground.

However the object is not in a vacuum, as it starts to fall it feels a force from the air rushing past it. The force of gravity is not increasing or changing but the forces of the air on the droplet is increasing as the speed increases.

At some point the drag force will catch up to the force of gravity as only the drag force changes relative to velocity. Because the forces are equal and opposite the two forces will cancel out each other. As we know an object in motion stays in motion until a force is applied to it. Because the net force is zero the speed is unchanging at terminal velocity.

If the velocity happened to be magically pushed a tad faster as it was falling the drag force (up) would then be a tad higher then the forces of gravity (down) and it would start to slow, keeping perfect balance as no other forces exist to alter the speed.

Another perspective is if you shoot a bullet straight down at the ground from say a hot air balloon. The bullet will feel the force of gravity pulling it down but it will also have drag, the drag force starts higher then the force of gravity as the bullet was fired initial way faster then terminal velocity. Because the force of drag is opposite the direction of travel the bullet will slow. Until again the forces zero out and the velocity is unchanging as the bullet falls.

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Apart from the other answers which provide a physical explanation, perhaps a mathematical proof might convince you: $$\sum \vec F = 0 \rightarrow m \vec a = 0$$ $$m\frac{\text{d} \vec v}{\text{dt}} = 0 \rightarrow \text{d} \vec v = 0.\text{d}t $$ $$\int_{v_i}^{v_f}\text{d}\vec v = \int_{t_i}^{t_f}0.\text{d}t$$ $$\Delta \vec{v} = 0 \rightarrow \vec{v}_f = \vec{v}_i$$ Hence, if the net force is zero, the velocity will necessarily be constant, but it need not necessarily be zero.

Hope this helps

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    $\begingroup$ I think if he could handle these concepts, he'd not have needed to ask the question! $\endgroup$
    – Trunk
    Commented Oct 1, 2021 at 15:10
  • $\begingroup$ @Trunk, To understand how an object in a fluid moves with a constant velocity (a.k.a terminal velocity), one needs to solve a first-order differential equation, similar to the one in Roger Vadim's answer. The math involved here is almost nothing compared to that. $\endgroup$
    – Cross
    Commented Oct 1, 2021 at 16:05
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    $\begingroup$ Stokes Law was empirical. $F_v \propto v$ , $F_w = mg = $ const . $\endgroup$
    – Trunk
    Commented Oct 1, 2021 at 16:16
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My confusion regarding the matter is that if the net force acting on a body (here the rain drop)is zero then it should remain suspended in air rather than falling towards the earth.

This may be what your intuition based on everyday experiences suggests, but that's not how things actually work. Remember Newton's first law: things that already move want to keep moving the same way - and they don't need a net force to do it. As the droplet is already falling, it's not enough for the net force on it to become zero. Instead, the net force would have to reverse direction and decelerate the droplet until it stops. If after that the net force becomes zero, then the droplet will remain suspended.

E.g. Suppose you were an astronaut in a zero-gravity environment, say in a spaceship in outer space. You are free-floating next to a wall, and you push off of it with your arms. As soon as your arms leave the wall's surface, the net force on you is zero - but you are not going to stop moving. You are going to float away, and hit the wall on the other side. Unless you have a personal jet pack, or something to throw, to generate a reaction force on you, there's nothing you can do to slow down, speed up, or change direction.

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For intuition, think of curling (except the rain falls down, curling stones move horizontally). Once the stone is released, there are no horizontal forces on the stone (except a slight friction) yet it moves at a roughly constant speed (the slight friction slows it down over time). The player first accelerates the stone, just like gravity first accelerates the raindrops. Then, if the net force is zero, the velocity doesn't change. Whether this is due to a near-lack of forces (a released curling stone), or forces adding up to zero (raindrops), is irrelevant. A net force of zero preserves the prior state: either standing still or a constant speed.

enter image description here

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The whole journey of the raindrop can be explained as follows:

Firstly, the raindrop departs from the cloud towards the Earth with zero initial velocity. There exist two forces, buoyancy force and gravitational force conflicting with each other. The gravitational force of Earth accelerates the raindrop downward, and the buoyancy force is very small (as the density of water is much greater than air). So the speed of drop increases.

As speed increases, the raindrop will experience a new force acting upwards: air resistance. This is a frictional force. Unlike friction between rigid bodies, this fluid friction depends on the relative velocity between the surfaces. That means as the speed of raindrops increases, the fluid friction also increases. So the forces on the raindrop now are buoyancy and gravitational force which are constant and fluid friction which is increasing.

When the speed of raindrop increases so high that the fluid friction is equal to the gravitational force-buoyancy force, the net force on the raindrop finally becomes zero. Now the drop does not accelerate anymore, hence the speed of the drop is saturated.

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    $\begingroup$ Great answer, very beginner friendly, very educational, very well-stated. Searching "saturated velocity" shows topics on electron behaviour. Is "saturated" also a normal term to use for describing terminal velocity? $\endgroup$
    – minseong
    Commented Oct 1, 2021 at 18:46
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There is a really nice answer by @Farcher, I am submitting my answer because I believe in certain cases a picture is more then words, and to add some interesting facts.

enter image description here

In your case, the raindrop does the same thing, it skydives, first, it accelerates, until increasing air resistance cancels gravity (acceleration), after which point the raindrop falls at a steady speed because air resistance and gravity (acceleration) are in equilibrium.

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The net force doesn't become zero at the start ,it takes some time.If you consider theoretically the drag is like an exponential decay and it dies out at infinity .but if we consider practically ,the acceleration of say $0.00000002\ \mathrm{m/s^2}$ can be taken as zero.

Practically:

The equation for Drag Force is given by

$F_d = \frac 12\rho DAv^2$

$\rho$ is the density of the medium through which rain drop is moving.

$F_d$ is the drag force.

$D$- Drag coefficient

$A$- Surface Area of Raindrop

$v$- velocity with which it's falling .

As the body starts from rest ,it's velocity increases due to $F_g=mg$ (Gravitational Force)

As the velocity increases, Drag force increases as a square term. So at one point, the net force acting is zero, it won't accelerate, it's falling, it's velocity doesn't change .

Consider the moment when the net force becomes zero, at that moment, it is falling, if it was to stay put at that moment, then a huge force must act in the direction opposite to which it is falling ,but there is no such force, drag doesn't achieves this, because it has an important job of tackling the gravity. So there is no new force (ie) only two forces are acting Drag and Gravity.

Consider a space where there is no atmosphere and no gravity due to any planet, if we throw a ball does it stop or does it move continuously?

We can certainly say that if it were to stop, it is due to some kind of force, we aren't giving any force, so it must move.

This is what happens in the case of raindrop, it continues to fall because the two forces acting on it cancels out and some other force must stop it, since there isn't any other force it continues to fall. Newton's First Law beautifully says this.

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The answer to the question lies in Newton's first law of motion:

An object at rest remains at rest, and an object in motion remains in motion at constant speed and in a straight line unless acted on by an unbalanced force.

https://www1.grc.nasa.gov/beginners-guide-to-aeronautics/newtons-laws-of-motion/

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  • $\begingroup$ Hello and welcome to Physics SE. Rain drops are acted upon by a constant force, gravity. Even if we consider constant gravity regardless of height from earth's centre it should provide constant acceleration (Newton's second law). The reason rain drops fall with constant velocity (assume zero wind) is drag, which is (roughly) proportional to speed. Thus at some specific speed, the drag will become equal to gravity, thus balancing the forces acting on the rain drop. Not sure this is what you intended to describe but if this is the case you should try to be a bit more clear and maybe explanatory. $\endgroup$
    – ZaellixA
    Commented Oct 20, 2021 at 9:07
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    $\begingroup$ Well, the OP already figured out that the forces are balanced. What I wanted to point out is, that remembering the Newton's first law would already help to clarify the dilemma. $\endgroup$
    – Gasper
    Commented Oct 21, 2021 at 7:38
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There is a lot going on in the cloud as the raindrop forms. Firstly you have warm moist air rising, as this column of air rises it expands (pressure reduces as altitude gained) and cools eventually the moisture in the air will precipitate onto any aerosols in the rising air and form a cloud. The water droplets will coalesce and accumulate until the weight of the water droplet exceeds the updraft that is keeping it aloft - NB a kilometre wide fluffy cloud weighs in at around 6 tonnes!!

Then the fun begins with the other answers!!

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