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Terminal Velocity of rain drops

  1. Jun 12, 2013 #1
    Please, explain the concept of terminal velocity on the help of rain drops... Is there is any retardation when rain falling to the ground.?
  2. jcsd
  3. Jun 12, 2013 #2


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    Yes, raindrops have a terminal velocity.

    The weight of a raindrop scales with the volume of a raindrop -- the cube of its size. The air resistance of a raindrop scales roughly with the area of a raindrop -- the square of its size. If you assume that air resistance also scales with the square of velocity, this gives a relationship.

    air resistance = gravity
    v2r2 = k r3
    v2 = k r3 / r2
    v = √(kr)

    where r is the raindrop radius, v is the fall rate and k is a constant of proportionality that includes the density of water, and acceleration of gravity and the viscosity of air.

    So the fall rate will be smaller for smaller raindrops. If the raindrop size is very small, you get fog.
  4. Jun 12, 2013 #3
    Rain drop having free fall. Am I right..? That is it get accelarated. Means, velocity changing... But. Terminal velocity is a constat velocity. Then how it happens..?
  5. Jun 13, 2013 #4
    In short without math...

    The rain drop falls because of gravity,they get accelerated towards earth ground at 9.8 m/s^2 and this acceleration is constant,but their are also many other forces acting upon the rain drop which resist the accelerating motion of the rain drop ,one of them is air resistance ,air resistance apply an opposing force or a force opposite in direction of that of acceleration due to gravity i.e.. its force is upwards on the rain drop .When these two forces ,acceleration due to gravity and force due to air resistance becomes equal on the rain drop the rain drop stops accelerating any further and achieves a constant velocity known as terminal velocity.

    For example one of your friend is pulling you on the left side with a certain force,you will be start moving towards left as he is pulling but then another friend of your's comes and from the same side he pushes you towards right with the same amount of force that the other friend is trying to pull you,what will happen you will stop moving because two equal in magnitude but opposing forces cancels each other.

    Why they are pulling you and pushing you I have no idea...


    Good Luck
    Last edited: Jun 13, 2013
  6. Jun 13, 2013 #5
    The force caused by air resistance is proportional to its velocity. The higher the speed, the more force you need, why? because it is "moving" more air, and that is energy. How the surface is two dimensional and motion is exerted perpendicularly to the amount of air moved or impacted drop is measurable volume per unit second. The energy required to move the particles within this volume (subsequently dispersed by butterfly effect in contact with adjacent surfaces) comes from the kinetic energy, ie, its velocity, which in turn comes from the gravitational potential energy (as that advances in the gravitational force field is gaining speed at a rate of g, for the small scales can be considered constant, but you should realize that a field is ALWAYS unchanged "small" which is their difference, these are practical considerations, more than theoretical). The higher the speed, the volume of molecules per second and higher colliding energy transfer due to their impact and friction layers. The impact between the surface and the air is much more complicated than a simple impact because they involve different particles in motion, but you can get an idea intuitively. Since energy can neither be created nor destroyed, it just flows. On one side we gain speed thanks to the potential energy of gravity, but this gain is constant and on the other side have a loss because the consequences of interacting with other particles and dissipate energy, which is proportional to speed. As you can see the speed plays the role of intermediary in the process, and there is an equilibrium in which a special rate equals this "transfer" of energy, which is the terminal velocity. At this speed the energy dissipated is equal to the energy supplied by the gravitational field and hence remains constant.

    Sorry for my English, I'm Spanish and I'm not very good with languages.
    Last edited: Jun 13, 2013
  7. Jun 13, 2013 #6


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    I imagine retardation to mean the object is slowing down.

    The rain drops do not slow down, they merely cease speeding up, when they reach terminal velocity.

    When objects fall near the Earth Surface, they have an initial acceleration of 9.8 ms-2, since gravity supplies the only Force acting on the drop - the weight of the rain drop.

    As soon as the drop begins to fall, some air resistance develops, reducing the effect of the weight of the rain drop, and reducing its acceleration from 9.8 ms-2.

    Air resistance has a higher value for an object moving at higher speeds, so eventually the rain drop will reach a speed where the size of air resistance exactly matches the weight force of the rain drop. At that time there will be no unbalanced Force acting on the drop, so it ceases to accelerate. It doesn't slow down, it just doesn't speed up; it has reached terminal velocity.
  8. Jun 13, 2013 #7


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    Air resistance.

  9. Jun 13, 2013 #8
    Thank you all....:-)
  10. Jun 16, 2013 #9
    The force caused by air resistance is given by the equation D=kv where D is the drag force, v is the speed at which the drop is falling, and k is the proportionality constant between them.
    The force of gravity is given by W=mg where W is the force caused by gravity, m is the object's mass, and g is the acceleration due to gravity, approx. 9.8 m/s^2.
    Terminal velocity happens when D and W cancel each other out i.e. when they are equal. This happens when kv=mg or v=mg/k.
    The big kicker about terminal velocity, though, is that it is never actually reached: only approached because as you approach terminal velocity, you accelerate less quickly.
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