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In the scene where Katherine Johnson/Taraji Henson has to calculate "Go / No-Go," which I take to mean the operational go-no-go orbital-insertion criteria.

She calculates a value for R, using this formula: $R = \dfrac{v^2}{g}\sin (2 \psi)$

Taking $V = 25731 \ \dfrac{f}{s}$ (note: the digits got transposed in the actual scene -- the value on the board is $25371 \ \dfrac{f}{s}$), $g = 32.2 \dfrac{f}{s^2},\ \sin (2 \psi) = 0.99848$,

I get the same value that Katherine comes up with: R = 20,530,372 feet.

Here's my question: is this the distance from the launch window to the target in the Bahamas? And why is it important to the orbital insertion?

Here's the movie scene: Calculating Go / No-go / R

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    $\begingroup$ That scene is talking about reentry, not insertion. The R figure you give works out to 3888 miles, slightly less than the radius of Earth at 3959 mi. $\endgroup$ – Russell Borogove Mar 6 '17 at 0:07
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    $\begingroup$ The equation is that for ground travel distance for a ballistic projectile launched at a given elevation angle upward, which could have some use in working out ballistic reentry distance. $\endgroup$ – Russell Borogove Mar 6 '17 at 0:14
  • $\begingroup$ Just out of curiosity, are those points calculated in imperial units? back then/today still? $\endgroup$ – AtmosphericPrisonEscape May 19 at 1:34
  • $\begingroup$ @AtmosphericPrisonEscape They are using English units which =/= imperial units. en.wikipedia.org/wiki/English_Engineering_units en.wikipedia.org/wiki/Imperial_units $\endgroup$ – Organic Marble May 19 at 2:38
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    $\begingroup$ You asked if imperial units were used. The answer is no, they used English units back then, not imperial. I gave you links that show the difference between English and Imperial units. Whats left? Nasa does not now and never did use imperial units and that's specifically what you asked. $\endgroup$ – Organic Marble May 19 at 5:00
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Looks like they've used some figures that are very close to the sample calculations shown in the original paper with Katherine Johnson, which is available from NASA archives.

enter image description here

The figure is for the distance of the vehicle from the center of the earth at the time of the retro rocket firing (or burnout) which initiates reentry. They've changed the original variable $r$ (or $r_1$ since it's the burnout position) to $R$ for the blackboard in the movie.

In the scene, $R$ is just part of the calculation necessary so that she can get to the desired information, the location where the retro rockets have to fire so that landing is at the planned location..

enter image description here

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  • $\begingroup$ Nice find! Well sourced. $\endgroup$ – Organic Marble May 19 at 11:34
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This I believe is in reference to John Glenn's flight. It seems that this is some real numbers, but simplified in a way that doesn't let everything be seen.

The speed is almost orbital. I believe what is actually being calculated is the location where a very late abort would put John Glenn landing, although it is somewhat difficult to tell from the limited details available.

The distance is actually much greater than the distance to the Bahamas, so this could be an alternative landing location being planned as well, the distance being the distance from the moment of retroburn until landing occurs, or something like that.

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