The Space Shuttles glided to a landing in the atmosphere after reentry. They were not good gliders, aerodynamically speaking, so their glide slope angle was rather steep.
What was the glide ratio of the Space Shuttles in this phase of flight?
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1$\begingroup$ You should clarify what you mean by "in this phase of flight". The shuttle was constantly slowing down throughout reentry and landing, and its L/D changed as its speed changed. $\endgroup$– Bret CopelandCommented Feb 7, 2017 at 16:57
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$\begingroup$ After the so-called "alignment cone", the glide slope is minimal, about 20-22°. This can be guessed from the landing systems on the SLF runway: MBLS, and PAPI plus Ball-Bar, at 20°. Here the glideslope in the HUD (source Youtube, STS-129 landing). $\endgroup$– minsCommented Feb 7, 2017 at 23:06
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$\begingroup$ @mins it's always 20° on final (or 18° for orbiters returning with heavy payloads). I would agree that "cone" should be in quotes since it's not technically a cone, though it was called that. $\endgroup$– Bret CopelandCommented Feb 8, 2017 at 3:21
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3 Answers
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As far as I know the space shuttle is no longer in use. However, the glide ratios (more than one in different configurations) can be found on the Wikipedia page: Space Shuttle Wikipedia page. There it states that “The orbiter's maximum glide ratio/lift-to-drag ratio varies considerably with speed, ranging from 1:1 at hypersonic speeds, 2:1 at supersonic speeds and reaching 4.5:1 at subsonic speeds during approach and landing.”
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2$\begingroup$ "No longer in use" is an understatement. The last shuttle launch was in 2011, and the orbiters themselves were decommissioned and rendered unfit for flight. $\endgroup$ Commented Sep 29, 2020 at 15:34
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The NASA website quotes an approximate ratio of 1
The shuttle was designed with a low L/D ratio (~ 1) because during the descent the spacecraft must be slowed from about 17,300 mph to about 250 mph at landing. (Source)
Note that glide ratio is normally equal to the lift to drag ratio - Wikipedia article - Aviation.SE Question
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3$\begingroup$ This; they wanted drag, because drag was what allowed the spacecraft to slow down. Orbital velocity is blazing fast; you don't want to hit a runway that way, or even the thicker parts of the atmosphere (which in this case amounts to a few tens of km altitude). Compare how the Space Shuttle flew S-patterns during reentry to bleed off speed. $\endgroup$– userCommented Feb 7, 2017 at 16:27
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7$\begingroup$ The L/D increased substantially as it slowed down, especially in the subsonic region. You would need at least a 45° glide slope on approach with a L/D ratio of 1:1. The shuttle flew at either 18° or 20° on approach (depending on weight), and it did so on the front side of the L/D curve. $\endgroup$ Commented Feb 7, 2017 at 17:02
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7$\begingroup$ @MichaelKjörling sorry, it's just a nitpick I have anytime I hear the shuttle flew s-turns "to bleed off speed". That makes it sound like the point of the s-turns were to bleed off speed, and they weren't. The shuttle needed to use bank angle to control drag (by way of controlling descent rate). The s-turns simply existed to compensate for the fact that this method of drag control caused them to turn off course. I wrote a whole answer on it here. $\endgroup$ Commented Feb 7, 2017 at 17:05
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This paper apparently answers your question.
Eyeballiny the figure (I haven't read the paper itself):
Mach 4 ≈ 2
Mach 3 ≈ 2.5
Mach 2 ≈ 3
Mach 1 ≈ 4
Mach 0.6 ≈ 5
answered Jun 22 at 8:32
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3$\begingroup$ So far the most correct answer. Funnily enough, the selected answer is the least correct... $\endgroup$– sophitCommented Jun 22 at 13:56