I am a keen but novice skydiver.  In free fall when working on a
formation with others it is important to equalise one's fall speed
with the other jumpers.  It is common to see lighter jumpers strapping
on weight belts in order to help them "keep up" with their heavier
(larger?) colleagues.  I have suggested that this is ineffective as it
is cross sectional area and drag which affects fall speed not weight. 
The response is "look I have been jumping for years and have x
thousand jumps - what do you know about it when you have only 200
jumps?"

It is clear that in a vacuum weight will have no affect (Galileo etc.)
 Every molecule will be equally accelerated at 32ft/sec/sec.  Bring
the air into the calculation and we are dealing with air resistance or
drag - mass has no effect on this.  However if we think of the
skydiver's body as a wing, does the concept of wing loading have any
relevance?

Have I missed something or are thousands of skydivers wrong??

Hi Christopher99,

   Sometimes too many answer can obscure the truth.
   
   As you have probably realized, this is one of those 
   frequently mis-answered questions (FMAQs?) a sub species
   of urban legends, like the infamous


   Is glass a liquid or a solid?
http://answers.google.com/answers/threadview?id=527647
http://answers.google.com/answers/threadview?id=419211

So, first, just BTW, the legend:
 The Galileo story is apocryphal. While some of his predecessors
actually performed this experiment, Galileo did not. 

http://www.endex.com/gf/buildings/ltpisa/ltpnews/physnews1.htm 



In your case, friction is the air resistance, governed by Stokes law:

http://www.cord.edu/faculty/ulnessd/legacy/fall1998/sonja/stokes.htm

 Force of friction is constant * speed (of body relative to medium)

So, when you step out of the plane, your (vertical) speed is zero and
for a moment you are right.At this point, friction is zero and mass plays no
role.

 1) Here the equation of motion is
 (* means multiply )

M *g = M *a 

which means (M cancels out)

 a = g   (acceleration is equal to  9.81 m / s*s)

However, when you get up some speed,  the situation changes.

  2)When you reach terminal velocity so that there is no further
acceleration, then the equation of motion is

(* means multiply, constant depends on size and shape etc)

               M*g = constant * speed


 You see here that your speed depends on mass M. It is proportional to mass.

http://hypertextbook.com/facts/JianHuang.shtml

3) In the intermediate regime, there are 3 effects: gravity, acceleration,
and friction.  Therefore, the equation. of motion has three terms

              M*g + M*a = constant * speed

and mass does not cancel. So speed will depend on mass. 
The more experienced divers are right.

Solution of this third or general case,
  the physics of 'free fall with friction'
is provided by this applet 
http://lectureonline.cl.msu.edu/~mmp/kap4/cd095a.htm 

 Use the control sliders to turn friction on and off.

If you need any further clarification, please feel free to ask.



Search terms:        Stokes law
                     Terminal velocity
                     Free fall with friction

A weight belt will make you fall faster.  Terminal velocity is reached
when drag equals weight.  A weight belt increases weight without
increasing drag (much).  Drag increases with speed, so a body with a
weight belt on must fall faster before drag matches weight.

Look, take two balloons and fill one with air and the other with
water.  Which falls faster?  The drag is no different, only the
weight.

The factors affecting a body falling through a fluid (and we can treat
air as afluid for this) are
Gravitational force- this will not change enough to be of importance
The viscosity of the fluid: again we do not need to consider this if
everyon eis jumping from the same height
The shape of the object: in skydiving this is important but for this
question I assume everyone stays in the same position
The difference in density between the object and the fluid. This is
the crux of the matter. As techtor points out, adding a weight belt
increases mass without significantly altering the volume, thus the
density increases

You can ignore the density of the air.  The buoyant force of the air
is insignificant whether or not the diver wears a weight belt.  A
large (100kg) diver would experience of a buoyant force of only 1 N or
less (100 grams).  At terminal velocity, the drag force is at least
1000 times bigger than the buoyant force.  On the other hand, buoyancy
may be significant in my balloon example.

Perhaps one of the following may be helpful to you.

* Effect of a large weight addition (tandem jumping). 
The site below doesn't say anyone actually tested this, but I'll
presume that they did. It also doesn't explain the reason behind the
effect of extra weight - but you may be familiar with this effect.

"Just after jumping out, the instructor throws out a large (approx.
4-foot/1.2-m diameter) drogue chute, and this drogue is out during the
entire free fall. Without this drogue, the combined weight of the
instructor and student would cause the pair to fall at 180 to 200 mph
(290 to 320 kph) -- much faster than the normal 120 mph. The drogue
slows the pair down to the normal falling speed."
http://entertainment.howstuffworks.com/skydiving1.htm

* Free fall; the skydiver accelerates until reaching a constant
velocity. At this point, there is no more net acceleration
(approximately) and the net force on the skydiver is zero, that is,
the force of gravity is balanced by the opposing force of drag. This
is different from a fall in a vacuum where the falling object is
always picking up speed.

"The skydiver accelerates to 120 mph (190 kph) in about 10 seconds and
is then in free fall."
http://entertainment.howstuffworks.com/skydiving1.htm

* Terminal velocity and equations for it.
There are two forces affecting terminal velocity: the force of gravity
and the opposing force of drag.
There is a NASA site that has a good overview of this; unfortunately,
the top part of the site is a gif and I cannot paste it here. You may
wish to take a look at the equations and statements in color on this
site:
http://exploration.grc.nasa.gov/education/rocket/termvr.html

My interpretation re the "F = D - W" equation:

In a vacuum, only the F=W (aka F=mg) part needs to be considered.
Although gravity exerts a stronger force on a more massive object, a
larger force is needed to accelerate ("budge") a more massive object.
However, these effects balance out, and all objects accelerate at the
same rate and have the same velocity at a given point when falling in
a vacuum.

In air, an object reaches terminal velocity when the air resistance
(drag) equals the force exerted by gravity. A more massive object
experiences a stronger force due to gravity (but not a greater
acceleration) and so requires a greater drag (counterforce) to balance
the force of gravity. (Gravity is still acting, but velocity becomes
constant at this point during free fall.) As shown in the NASA
equation, drag depends on a number of things, but I'm only going to
cover area and velocity. I will assume that the cross-section area is
not too different between different jumpers and is subject to
modification, e.g., amount of arching. So I'm left with a
consideration of velocity as a major variable determinant of drag; as
the equation shows, drag increases as the square of velocity.

In air, gravity affects all falling objects equally in terms of
acceleration and velocity; however, drag increases as the skydiver
falls until it equals the force of gravity, and the force of gravity
is larger for more massive objects. Thus, a more massive object will
require more drag to cancel out gravity; the falling speed that
produces enough force to cancel the force of gravity on a smaller
object is not great enough to cancel all the force of gravity on a
more massive object, so the more massive object will still accelerate
until it reaches a higher speed, at which point, the force exerted by
the drag at a higher speed balances the force exerted by gravity.

I think the key to the problem is that there is uniform acceleration
due to gravity, but in skydiving, you achieve (after about 10 seconds)
a zero acceleration state due to an opposing force, drag. Acceleration
due to gravity is constant between objects, but the force of gravity
is not. The opposing force of drag depends on the velocity.

Perhaps these equations will help:

Force (massive object) = mass (massive object) * acceleration due to
gravity (constant between objects)
Force (less massive object) = mass (less massive object) *
acceleration due to gravity (same as above)

Opposing drags:
Drag (massive object) = ..... velocity (massive object) squared
Drag (less massive object) = ...... velocity (less massive object) squared

Since the total force exerted by gravity is larger for the more
massive object, it will fall faster until its drag finally balances
the force of gravity.

I  would like to restate my last paragraph.

Since the total force exerted by gravity is greater for a more massive
object, it will take a greater drag to balance this force than for a
similar less massive object. Assuming equals areas for contact with
air (between two objects - skydivers in this case), the drag that
balances the force of gravity on the less massive object will not
fully balance the force of gravity on the more massive object and the
more massive object will continue to accelerate until it reaches a
greater speed where the drag then equals the force of gravity on the
more massive object. (I hope that this is a bit clearer.)

Thank you everyone.  This has all been very helpfull. I think that I
have got it now!

This has always bothered me too. I've never been skydiving but i guess
at some point during the fall (10 seconds from the observations listed
here) that this is not true:

"The skydiver accelerates to 120 mph (190 kph) in about 10 seconds and
is then in free fall."

They skydiver isn't in free fall nore does he feel free fall, friction
sees to that. I guess that means that once terminal velocity is
reached you feel almost like you are laying on your stomach.

Stoke's Law does not apply to a skydiver.  It is only approximately
valid for Reynold's numbers less than 10.  As the Reynold's number of
a skydiver is over 1,000,000, Stoke's Law would predict a drag that is
perhaps 4 orders of magnitude too low.  In this regime, the drag force
not equal to a constant times the velocity.  It comes much closer to
being proportional to the square of the velocity.

rracecarr-ga says in
Subject: Re: How does weight affect a skydivers fall speed? 
From: rracecarr-ga on 27 Feb 2006 15:19 PST 	  	

Stoke's Law does not apply to a skydiver.  It is only approximately
valid for Reynold's numbers less than 10.

That is true. The equation I consider in the answer may apply to something like
bale of cotton.  I did concentrate on the physical mechanism, effect
of mass, not on complex question on how drag depends on velocity.
Details of that are discussed in a reference I quoted:

http://hypertextbook.com/facts/JianHuang.shtml

Speed of a Skydiver (Terminal Velocity)

Hedgie