&brsp;Valles Marineris Skydiving on Mars

Valles Marineris, named after the Mariner 9 spacecraft team, whose vehicle returned the first close-up views of the Martian surface, is the largest canyon in the solar system. This complex equatorial canyon system measures 4,500 km from east to west and 150 km to 700 km from north to south. Individual canyons measure up to 200 km wide and 7 km deep. In comparison, our Grand Canyon only spans a maximum width of 28 km with a greatest depth of 2 km.

In this exercise we will locate and download a Viking Orbiter image showing a canyon wall nearly 4 miles high, from which we must jump! The goal is, of course, to survive the landing, and in order to do this we must design an appropriate parachute. Although Mars' surface gravity is relatively weak, 39% that of earth, its surface atmospheric pressure measures only 0.007 atm. (0.7% of earth's at sea level). It is immediately apparent that the very thin atmosphere of Mars will require a rather large parachute, but exactly how big?

To answer this question we`ll need to investigate some basic principles and concepts including freefall, terminal velocity, and aerodynamic drag.

Galileo demonstrated that an object falling only under the influence of gravity will experience a constant acceleration, ie., it gains the same amount of velocity for every additional second that it falls.

• On earth this amounts to 32.14 ft/sec/sec, meaning that it increases its downward velocity by 32.14 ft/sec for each second that it falls.
• On Mars the accereration of gravity equals 0.39 that of earth or 12.54 ft/sec/sec.

where V=velocity (m/sec)

t=time (sec)

g=acceleration due to gravity=.39 g_earth=12.54 ft/sec/sec for Mars

One can also show that the distance d fallen after time t is:

where d=distance fallen (ft)

g=12.54 ft/sec/sec for Mars

t=time (sec)

Furthermore, Galileo discovered that this acceleration is independent of the body's mass, but dependent only on the strength of gravity on the planet. So, in the absence of an atmosphere, heavy things don't fall faster than light things, but at moderate to large velocities even a thin atmosphere can have a significant effect on a falling body's motion due to the effects of aerodynamic drag.

As everyone knows, the atmosphere becomes increasingly resistant the faster one attempts to push through it. Skydivers know that at a certain velocity called "terminal velocity" one cannot fall any faster, and the actual velocity at which this occurs depends upon one's weight and the "shape" one assumes while falling. Terminal velocity is attained when one's weight equals the drag imposed by the atmosphere.

Weight is the force with which a planet's gravity attracts another mass:

where W=weight

m=mass

g=acceleration of gravity and

Where:

D=drag

=atmospheric density

v=velocity

S=cross-sectional area

c_d =drag coefficient, an empirically determined, dimensionless number, which relates to the shape of the object.

Let us briefly examine the factors which determine drag. Drag is directly proportional to density (), crossectional area (S), and drag coefficient (Cd), while it is proportional to the square of the velocity. This means that a change in density, area, or drag coefficient will effect an equal change in drag, but a change in velocity will have a much greater effect, eg, doubling velocity will quadruple the drag. At 100 mph drag is 16 times as great as at 25 mph.

On earth when a skydiver orients his body to maximize drag, he will soon reach a speed at which drag equals his weight; this occurs at about 120 mph. When terminal velocity is attained, acceleration stops and the diver continues to fall at a constant speed.

Acceleration only occurs in response to a net external force, and in this case the downward force (weight) is opposed by an equal but oppositely directed (upward) force, drag.

A Virtual Spacecraft

· Remember that URL's are case sensitive. 
· Home pages might look different now.
· Hyperlinks might have broken or changed since this was written, 
in which case you will be given alternate locations.

You will finally arrive at your jumping-off point, displayed in a special 
PDS (Planetary Data System) viewer. It looks like this:

                                                     
· Notice the fluvial features, evidence of extensive water flows in the past.

  Clicking the  I  button on the tool bar brings up a data box 
to the right of the image.

  Select a point on the cliff from which to take a running jump, 
then, assuming that the slope of the cliff is 15 deg. from horizontal, use 
the Image size, Resolution, and Scale data to determine  the height of 
the cliff. You can measure either on the screen or on a printed image.

· Height of cliff =______________ mi.  

    If you were to assume a streamlined shape during the jump, aerodynamic
 drag could be reduced to nearly zero, thus approximating a freefall 
condition. Convert the cliff height to feet, and assuming g= (.39) (32 ft/sec/sec), 
use the freefall equations above to calculate the velocity at impact.

· t =____________________sec. 

· v =___________________mi/hr  
  
 ( use 44 ft/sec = 30 mi/hr ) 

      Next we shall determine terminal velocity on first earth, then Mars.

    In the equation , is the mass density , which has a standard 
value of  0.00236 slugs/ft3 using English units. 

  Assuming a 150 lb. person, 9 sq.ft. effective projected area, and a drag 
coefficient of 0.7, the terminal velocity v  is 142 ft/sec or 98.8 mph for
 planet earth at sea level. Correcting the density for a 6000 ft. average altitude yields 
115.8 mph, a realistic number.

Now perform the same calculation for Mars:

      Mars has a thin (0.007 atm) atmosphere consisting mainly of CO2. 
Assuming the same values of area and drag coefficient as for earth we 
must correct the density and weight to their corresponding martian 
values. A reasonable approximation would be to multiply the 0.00236 slugs/ft3

 for earth by  0.007 (p mars / p earth) x 1.54 (density CO2 / density air) 

x 283oK / 210oK (temp.earth / temp.Mars)
resulting in: 

          mars =0.00003428 slugs/ft3 

   Weight is mg where gmars =0.39 gearth .

Therefore, one can divide 115.8 mph by

 (mars/ earth)

 then multiply  by 0.39  =0.624

The result is Vterminal = 597 mph

 Now calculate Vterm using your weight.
 
Use 597 mph  (your weight / 150 lb) mph

· Vterm =_____________mph

By first assuming a streamlined position, one could approximate a freefall
 condition  to reach a velocity equal to terminal velocity as quickly as possible, then 
reorient the body  for maximum drag and terminal velocity. 
The applicable expressions are: S = 1/2 g t2  and V = gt, 
where S= the cliff height, and V= freefall velocity after falling for time t.
 
Would you reach a freefall velocity = terminal velocity before crashing?

· Yes__________         No___________

      Finally, we are ready to design a special parachute, 
which will allow a nice soft landing, say, the equivalent 
of jumping from a 6 foot height on earth.
      Again using the freefall equations given above, 
one can calculate that a 6 foot jump takes 0.61 sec.
 with a final velocity of 19.6 ft/sec or 13.3 mi/hr,
which we will then use as a terminal velocity on Mars.

     Now using 
     Where D = your weight
                 V= 19.6 ft/sec
                  =0.00003428 slugs/ft 
                 S = area of the parachute ( ft2 )
                 Cd = drag coefficient

         The drag coefficient, a function of the shape of the object, 
is approximately 1.3 for a flat plate having an aspect ratio of 1 : 2 
                                           

                   X x 2X  = S            S =2X2            X =?  

Calculate the projected area S then determine the dimensions X and 2X.



· Indicate the dimensions of your parachute:        X=  _________ ft
                                                                                                   
                                                                                          2X=  _________ ft