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## Drag on Spaceship in Atmosphere

Valued Contributor III
529 Views

As usual, Jacob Williams has put up some interesting Twitter Posts on Fortran usage in space.

His latest one includes a module for computing the drag on a spaceship,

``````pure function drag(density, v, area, cd)

use iso_fortran_env, only: wp => real64

implicit none

real(wp) :: drag !! drag force magnitude (N)
real(wp), intent(in) :: density !! atmospheric density (kg/m^3)
real(wp), intent(in) :: v       !! relative velocity magnitude (m/s)
real(wp), intent(in) :: area    !! spacecraft cross-sectional area (m^2)
real(wp), intent(in) :: cd      !! spacecraft drag coefficient

drag = 0.5_wp * density * v**2 * area * cd

end function drag``````

Here is the function, JW developed.  he references one of the many standard Fluid mechanics text books.  Although I used Streeter and Wylie.

The interesting bit, is the need for the complex Fortran model to get the density of the atmosphere as the spaceship rises or plunges.  He gives a list of the available models since 1976.

interesting stuff, the real problem would be the Fortran to look at the density at each location and I wonder how many time steps one takes for this sort of model.

I was reading the other day, that there was some math errors in the calculation for the locations for the first man in space from Russia.

5 Replies
Honored Contributor II
505 Views

I would have guessed at the potential speed ranges  encountered ( and hence reynolds numbers) the simple drag model would be quite inaccurate so fixating too long on time steps for density variation would be maybe not time well spent......

Valued Contributor III
498 Views

This article shows why re-entry is tough and it discusses the variations in the density.

I have seen some stuff from the 70's where NASA used models in water to estimate the drag,  using dimensional similitude.

Valued Contributor III
464 Views

IN thinking further about @andrew_4619' s comment, the real problem is the issue of space debris coming back to earth, if it is damaged then the drag coefficient is going to depend on the actual shape, and is it tumbling, if the face is not axisymmetric is it likely to tumble. It is going to heat the air in front of it, so there will be a pressure front.

Even on a timber beam vibrating at a constant temp, the thermal range can be 40% of the mean value.

Super User
458 Views

Wouldn't the drag also be affected by the rate of disintegration of the various materials/metals, which would cause the actual shape to change?  Not all material would disintegrate at the same speed/time.

Doc (not an Intel employee or contractor)
[Windows 11 is the new Vista]

Valued Contributor III
433 Views

Yes the shape would change, I suspect the shape factor for a given shape, speed and height above the earth would be the same.  But the real problem is a lack of data to predict the changes.

Experimental physics is a limited art, rather than a real science.

A German physics student in 1929 spoke of all the limitations in the experiment for measuring one of the basic physics constants.  We still see the same problems in labs today.

Given that NASA must have a lot of data on this topic, one could probably have a lot of fun playing with the results and a model.