I'm trying to find out how much a double in height (making it more ovular or oblong in shape) of the parachute affects air resistance compared to a double in circumference or diameter. Can someone tell me if parachutes with the same circumference (therefore looking the same when looked from the top) but are taller or shorter have different air resistance?

# [Physics] How does height of a parachute affect air resistance compared to circumference or diameter

classical-mechanicsdragmathematical physics

#### Related Solutions

Suppose the balloon with air (A) weighs $m_a$ and the balloon with concrete (B) weights $m_b$. The force accelerating the balloons downwards is $m_a g$ for A and $m_b g$ for B, where $g$ is the acceleration due to gravity. In the absence of air the acceleration is simply this force divided by the mass, so both balloons accelerate at the same rate of $g$. So far so good.

Now suppose the air resistance is F. We don't need to worry exactly what F is. The force on balloon A is $m_a g - F$, so its acceleration is

$$ a_a = \frac{m_a g - F}{m_a} = g - \frac{F}{m_a} $$

and likewise the acceleration of balloon B is

$$ a_b = \frac{m_b g - F}{m_b} = g - \frac{F}{m_b} $$

So the balloons don't accelerate at the same rate. In fact the difference in the accelerations is simply

$$ \Delta a_{ba} = a_b - a_a = \frac{F}{m_a} - \frac{F}{m_b} $$

Since $m_b \gg m_a$ the difference is positive, i.e. balloon B accelerates at a much greater rate than balloon A.

**Response to comment**

I think there are a couple of possible sources of confusion. Let me attempt to clarify these, hopefully without making things even more confused!

Firstly the air resistance affects both the acceleration and the terminal velocity. I've only discussed acceleration because terminal velocity can get complicated.

Secondly, and I think this is the main source of confusion, the gravitational force on an object depends on its mass while the air resistance doesn't. However the air resistance depends on the velocity of the object while the gravitational force doesn't. That means the two forces change in different ways when you change the mass and speed of the object.

Incidentally, Briguy37 is quite correct that buoyancy has some effect, but to keep life simple let's assume the object is dense enough for buoyancy to be safely ignored.

Anyhow, for a mass $m$ the gravitational force is $F = mg$ so it's proportional to mass and it doesn't change with speed. Since $a = F/m$ the acceleration due to gravity is the same for all masses.

By contrast the air resistance is (to a good approximation) $F = Av^n$, where $A$ is a constant that depends on the object's size and shape and $n$ varies from 1 at low speeds to 2 at high speeds. In your example the size and shape of the ballons is the same so $F$ just depends on the speed and for any given speed will be the same for the two ballons. The deceleration due to air resistance will depend on the object mass: $a = Av^n/m$, so air resistance slows a heavy object less than it slows a light object.

When you first release the balloons their speed is zero so the air resistance is zero and they would start accelerating at the same rate. On the moon there is no air resistance, so the acceleration is independant of speed and the two objects hit the ground at the same speed.

On the Earth the acceleration is initially the same for both balloons, but as soon as they start moving the air resistance builds up. At a given velocity the **force** due to air resistance is the same for both ballons because it only depends on the shape and speed. However because acceleration is force divided by mass the **deceleration** of the two balloons is different. The deceleration of the heavy balloon is much less than the deceleration of the light balloon, and that's why it hits the floor first.

From the way you phrased your question I'm guessing that and answer based on calculus won't be that helpful, but for the record the way we calculate the trajectory of the falling balloons would be to write:

$$ \frac{dv}{dt} = g - \frac{Av^n}{m} $$

This equation turns out to be hard to solve, and we'd normlly solve it numerically. However you can see immediately that the mass of the object appears in the equation, so the change of velocity with time is affected by the mass.

I am sorry to say, but your colleague is right.

Of course, air friction acts in the same way. However, the friction is, in good approximation, proportional to the square of the velocity, $F=kv^2$. At terminal velocity, this force balances gravity,

$$ m g = k v^2 $$

And thus

$$ v=\sqrt{\frac{mg}{k}}$$

So, the terminal velocity of a ball 10 times as heavy, will be approximately three times higher. In vacuum $k=0$ and there is no terminal velocity (and no friction), thus $ma=mg$ instead of $ma=mg-F$.

## Best Answer

Air resistance or air drag is basically proportional to square of the velocity of the moving object ,that is,

F=bv^2since air has turbulent motion where b is the air drag constant.Drag constant depends on the dimensions of the object. So air drag is, rather itindependent of length or volume of object.depends on the surface area of the objectThe action of the parachute is based on the principle of air resistance. It dramatically lowers your terminal velocity by increasing your air resistance as you fall. It does that by opening out behind you and creating a large surface area of material with a huge amount of drag. Parachutes are designed to reduce your terminal velocity by about 90 percent so you hit the ground at a relatively low speed of maybe 5–6 meters per second (roughly 20 km/h or 12 mph)—ideally, so you can land on your feet and walk away unharmed.

Increasing the height of the parachute hence

air resistance. But increasing the circumference of the parachute in turndoes not affect.increases the surface area of the parachute and it increases the air drag exerted on the parachuteSimilarly for parachutes with same circumference, air resistance is

.same