# Falling to Earth

1. Oct 9, 2013

### pitchwest

I'd like to graph the function of and object falling to earth from a point high enough that air density has an effect on the rate the object falls due to gravity. I'm not really sure where to start though.

I'm thinking about that guy that jumped from the highest point ever free-falled from and broke the sound barrier.

2. Oct 9, 2013

### SteamKing

Staff Emeritus
First, figure out what function you are talking about. Velocity versus time? distance versus time?

3. Oct 9, 2013

### pitchwest

What do you think would best show the change in gravity due to the change in air resistance against an object?

I was thinking that I'd have to first find an equation for the amount of air at different altitudes. Then calculate the level of air resistance and apply it to the amount of air equation. Then figure out how that applies to gravity at different altitudes.

I hope that helps clear up what I'm getting at.

4. Oct 10, 2013

### SteamKing

Staff Emeritus
It's not clear what you mean by 'the change in gravity due to the change in air resistance against an object'.

You should review Newton's Law of Gravitational attraction: F = Gm1m2/r^2

5. Oct 10, 2013

### mephestopheles

I believe he is talking about rate of descent and the effect varying atmospheric densities would have on a falling object of something like the size of a person. So that rate of descent would become variable due to the area resistance would be placed upon.

6. Oct 10, 2013

### SteamKing

Staff Emeritus
Eventually, a terminal velocity should be reached when the weight of the body balances the drag force.

7. Oct 10, 2013

### Staff: Mentor

The terminal velocity depends on the density of the medium (air). An object falling from a very great height would first reach a rather high terminal velocity corresponding to the low density of air at high altitude. As it falls into denser air, the terminal velocity decreases. To put it another way, the object's drag coefficient increases with air density.

If the drag coefficient is constant, it's possible to solve the differential equation of motion exactly for certain cases. With a variable drag coefficient, you'd have to solve the differential equation numerically.

8. Oct 10, 2013

### pitchwest

At extreme heights there's less gravity. A falling person would fall slower, but there's less air resistance at those heights, so a person would fall faster. The guy that did a free fall from 128,100 ft broke the speed of sound. I'd like to calculate the change in speeds he fell and view it on a graph. I'd imagine he broke the speed of sound but slowed down as he got closer to earth.

I'd also like to know if he slowed down to the standard terminal velocity eventually.

9. Oct 10, 2013

### cjl

This is not true. Drag coefficient is independent of air density. In fact, that's a large part of the reason why you want a drag coefficient in the first place - it's dimensionless, and largely independent of both air density and airspeed (across a fairly broad range of speeds, though if you start approaching the speed of sound or higher, this can change), and is effectively only dependent on the geometry of the object in the flow.

Drag force increases with air density (all else equal), but not drag coefficient.

10. Oct 10, 2013

### Staff: Mentor

You can come fairly close assuming a constant coefficient of drag and gravity that decreases linearly with altitude:
$$F(h) = \frac 1 2 \rho(h) v^2 C_dA - mg(h)$$
Here, h is the person's altitude. I'll use meters rather than feet and force is a signed quantity (not a vector; I'm only looking at the "falling straight down" problem here). A negative value means the person is accelerating downward, positive means upward.

The first part is drag to air resistance. The term ρ(h) is the density as a function of altitude. You'll need some model of air density. I'd suggest using the 76 US Standard Atmosphere model. It's fairly simple, something you can easily implement in a spread sheet or simple computer program. Unfortunately, you might have to wait a few days to get your hands on that model. You might be able to find it with some good internet searching. The US government websites where you would normally find it are down right now The final part of that term, CdA, is the (unit less) coefficient of drag times the cross section to drag. It might be easier to make this one item rather than two in your model. If you do this, try using 1 square meter as a first cut.

The final term is just gravitational force. A simple linear model of gravitational acceleration g(h) will work quite nicely for this problem. The linear term is called the "free air correction". Subtract 3.084×10-6 m/s2 for every meter of altitude from the ground level gravitational acceleration. For New Mexico, I'd suggest using (9.796 -3.084×10-6h) m/s2, where height h is in meters.

11. Oct 10, 2013

### voko

At 128100 feet, the force of gravity is about 1% percent less than at the sea level. This can be ignored for any practical purpose, because the other errors and uncertainties in the model will most likely be far greater than that.

12. Oct 10, 2013

### 256bits

That is not true either.
Drag coefficient may be dimensionless but it does vary with speed, density, and viscosity of the medium. Drag coefficient can be plotted against another dimensionless number called the Reynold`s number which incorporates velocity, dynamic viscosity, density and a characteristic length. One can use kinematic viscoscity in favour of density and dynamic viscoscity.
In addition, temperature of the medium will also affect the viscosity, and for gases the kinematic viscosity will increase with temperature.

As the speed of the object increases through the medium particular flow seperation points are reached which will change the coefficient of drag. The range of CD will vary from attached laminar flow at slow speeds, where the coefficient can be quite high ( > 1.0 ), to that at detached turbulent flow where it can be much lower.

13. Oct 13, 2013

### cjl

This is true, but I didn't want to overcomplicate matters for the time being. For fairly high reynolds numbers and low mach numbers (which is the case for the majority of objects traveling in air between, say, 20 and 200 miles per hour), the drag coefficient is approximately constant (for a given shape). As you said though, it isn't a simple subject.