Wikipedia claims (although with no citation) that in order to make the space shuttle land, an initial powered delta-v of 322 km/h was applied in orbit, retrograde to the shuttle's orbit. 322 km/h is equal to 89.4 m/s. This caused the orbit to be lowered into the atmosphere, ultimately causing the shuttle to come to a full stop intact on the ground (or that was the general idea; as we know, it didn't work out perfectly every time).

Organic Marble points out the Shuttle Crew Operations Manual, which states that

The deorbit burn usually decreases the vehicle's orbital velocity anywhere from 200 to 550 fps, depending on orbital altitude.

where 200 fps is about 61 m/s, and 550 fps is about 168 m/s. Given this data and the space shuttle's operational range, 90 m/s seems a reasonable figure to use as a median-mission deorbit-burn delta-v.

What I fail to understand is how this relatively small (about 1% velocity change: in the case of Wikipedia's figure, 90 m/s out of the on the order of 7 km/s orbital velocity in a low Earth orbit) could be enough to sufficiently lower the orbit to commit the orbiter to landing, rather than being just a small change in the shuttle's orbit.

Why was such a small delta-v applied under power in orbit sufficient to commit the orbiter to landing?

I expect that good answers will draw on orbital mechanics and atmospheric gas density (aerobraking) to show why the small change was sufficient.

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    $\begingroup$ If you get a good answer with references, you should use the reference to update Wikipedia. $\endgroup$ Commented Sep 16, 2015 at 13:41
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    $\begingroup$ It helps to realize that the Shuttle is in a very low orbit. Specifically, the orbital radius is only a few percent larger than the radius of the Earth itself. And that's using the surface of the earth, not the top of the atmosphere. It's not like the Shuttle was in a geostationary orbit, with an orbital radius 500% larger than Earth's. $\endgroup$
    – MSalters
    Commented Sep 16, 2015 at 14:57
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    $\begingroup$ At 90 km altitude, the air is thick enough that aerobraking can de-orbit the shuttle. A 400 km altitude circular orbit is about 7.67 km/s. An orbit 90 km altitude perigee and 400 apogee is moving 7.58 km km/s at apogee. These speeds can be found with the vis viva equation. $\endgroup$
    – HopDavid
    Commented Sep 16, 2015 at 17:27
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    $\begingroup$ I too was surprised by this, specifically by playing with the real solar system mod in kerbal space program; LEO velocity is around 7.7 km/s but your periapsis doesn't actually rise out of the thick atmosphere until the last few dozen m/s. Very interesting. And frightening. $\endgroup$
    – Thomas
    Commented Sep 17, 2015 at 3:37
  • $\begingroup$ Imagine throwing a bowling ball at someone's head and just missing - like flying just over their head with a few hairs of clearance. Now throw the same ball on the same trajectory again, this time a little bit more slowly... same thing. Things in LEO are flying straight at the Earth and are missing by just a little bit. Slow them down only a tiny amount and they crash. $\endgroup$
    – J...
    Commented May 3, 2021 at 12:46

7 Answers 7


If you're just looking for an intuitive handle on it, try this:

In circular LEO, your orbital period is about 90 minutes.

If you apply a velocity change of 90 m/s, then wait half an orbit -- 45 minutes -- you should expect to be out of position by 90 m/s * 45 min * 60 s/min = 243,000m, or 243km.

The distorting effect of Earth's gravity means that the positional offset isn't in the direction you'd expect, of course, but it does explain the magnitude.

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    $\begingroup$ Another intuitive approach is to understand that the 400 km altitude actually means an orbital radius of 6778 km - now the difference between 90 km (6468 km) and 400 km (6778 km) is obviously much smaller, only around 5%. And we only have to lower the perigee, not keep a circular orbit. $\endgroup$
    – Luaan
    Commented Sep 17, 2015 at 14:59
  • $\begingroup$ Yeah, that's an even better approach to the question. $\endgroup$ Commented Sep 17, 2015 at 15:12

Page 331 in the Shuttle Crew Operations Manual, an official NASA astronaut training document, confirms that

The deorbit burn usually decreases the vehicle's orbital velocity anywhere from 200 to 550 fps, depending on orbital altitude.

The deorbit burn was not intended to reduce the Orbiter's velocity to a small value, but rather to change its orbital parameters, so that its orbit intersected the sensible atmosphere. Specifically, it significantly lowered the orbital perigee. This example from the old NASA Quest site states that on STS-82, the deorbit burn changed the orbit from 333x312 nautical miles, to 333x28.

Aerodynamic drag then performed the majority of the velocity reduction. This drag, by converting the kinetic energy of the Orbiter to heat, led to the high temperatures experienced on entry.

1 Page 33 in the pdf, not the internal document page numbering.

Edit: since your question may really boil down to "how can a small burn change the perigee so much?", here's a handy-dandy guide to orbital adjustment burns and their effect.

enter image description here

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    $\begingroup$ I understand that the deorbit burn was only a small part of the total velocity reduction from orbital velocity to full stop, and I'm not questioning whether the deorbit burn was on the order of 90 m/s. What I don't understand is how such a small change in velocity could be enough to commit the orbiter to landing. Let me see if I can clarify this further in the question. $\endgroup$
    – user
    Commented Sep 16, 2015 at 14:04
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    $\begingroup$ I recently learned a rule of thumb: 2 fps delta-v in LEO changes your altitude by 1 mile, $\endgroup$ Commented Sep 16, 2015 at 14:14
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    $\begingroup$ At 75 miles up, the Shuttle encounters enough drag that the deorbit becomes inevitable (from spaceflight.nasa.gov/shuttle/reference/shutref/events/deorbit). $\endgroup$
    – Hobbes
    Commented Sep 16, 2015 at 14:14
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    $\begingroup$ "This frictional drag," - are you sure about frictional? I was sure form drag plays more important part here. Can't find solid source for either. $\endgroup$
    – Mołot
    Commented Sep 16, 2015 at 19:58
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    $\begingroup$ This is a good answer, and it's a shame that I can't accept two because yours and Russell's complement each other really well, but I found Russell's to be slightly easier to grasp the concept out of. Still, at least have an upvote. $\endgroup$
    – user
    Commented Sep 17, 2015 at 9:53

I think that something visual may be of help

enter image description here

This is a bit more to scale than most peoples pictures, but the shuttle only orbits at 200 miles, while the earth itself is almost 8000 miles across, so the orbit its more like a thick skin on an orange.. we fly very close to it.

In the picture the red dot is the ship, the thick line is earth, thin lines show the expansion of the orbital trajectory, and the arrow is the direction of its burn, if it were not to continue to burn it would be in a suborbital trajectory and hit the ground, in fact even if it made trajectory very large (going across 7,500 of the earths miles, it would still hit the ground on the other side of the earth), its only those final 200 miles that actually will bring it above the earths surface on the other side.

So once it is in orbit, all it has to do is decrease its orbit circle until its just far enough into the atmosphere on the other side to be able to land (as the atmosphere will slow it down further). In the deorbit burn, it burns the opposite direction, which lowers its orbit by just enough to hit the atmosphere at the right angle (the red orbit in the second picture), this would never be more than the height of the orbit (200 miles in this case) which is much less than the 8000 miles of earth it had to lift itself over first.

I'm sure there is a lot of math to explain this, but i think the practical answer is simply one of being able to conceive the scale.

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    $\begingroup$ It may be worth noting that the "orbit" circles drawn in the first diagram (and especially the parts drawn inside the Earth) are so far from reality as to be absurd. The parts outside the Earth aren't too bad -- you could pass them off as approximations of proper Newtonian orbit ellipses, and nobody would likely notice -- but their continuations inside the Earth are just crazy. $\endgroup$ Commented Sep 17, 2015 at 8:36
  • $\begingroup$ @IlmariKaronen, can you clarify what's absurd in the first figure? Obviously an orbiting object cannot literally go through the earth. Perhaps the problem is that "orbits" don't have a focus at the earth's center? $\endgroup$
    – Deolater
    Commented Sep 17, 2015 at 15:27
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    $\begingroup$ The "inner earth orbits" are there to illustrate that such trajectories are an expansion of the original ballistic suborbital trajectory that would hit the earths surface rather than create an actual orbit in an effort to show that the majority of delta - v is spent pushing the periapsis through the earth out to the other side (and to contrast how little of it is actually needed to raise/lower the periapsis through those 200 km of atmosphere). $\endgroup$
    – Czyrek
    Commented Sep 17, 2015 at 15:59
  • $\begingroup$ @Deolater: Exactly. The center of any circular two-body orbit must be center of mass of the system. A minimal suborbital trajectory (like, say, that of a dropped rock) would have its nominal periapsis approximately at the center of the Earth. Certainly the periapsis can never be closer to the apoapsis than the center of the Earth is. $\endgroup$ Commented Sep 17, 2015 at 17:06
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    $\begingroup$ @ Ilmari Karonen: Updated first figure. is this more in line with the reality of it? (none of the orbits are closer than the center) $\endgroup$
    – Czyrek
    Commented Sep 17, 2015 at 20:38

After the burn, the orbiter goes into an elliptical orbit. To do the math for such an orbit, we can use the vis viva equation which relates the semimajor axis to the velocity of the orbiter:

$$v^2 = GM(\frac{2}{r}-\frac{1}{a})$$

Where G = gravitational constant, $M$ = mass of earth, $r$ is the instantaneous distance and $a$ is the semi-major axis.

We can use this to compute the change in $a$ when we change the velocity: since $r$ is essentially constant during an instantaneous burn, and $a=r$ at the moment the burn is done, so $v^2=\frac{GM}{r}$, we get

$$2v\;dv = \frac{GM}{a^2}da\\ \Delta a=\frac{2v a^2}{GM}\Delta v = \frac{2vr}{v^2}\Delta v = \frac{2r\Delta v}{v}$$

In other words, for every % change in velocity, you get a 2% change in the semi-major axis. And since your apogee is unchanged, that change must be entirely applied to the perigee. This in turn means that the distance to the center of the earth will change by 4% for every 1% change in velocity.

Plugging in numbers that you used in your question (7 km/s for orbit, 90 m/s deceleration, 7000 km semimajor axis) we obtain a change in height

$$\Delta h = 4 \frac{\Delta v}{v} r = \frac{4\cdot 90 \cdot 7000}{7000} \rm{km} = 360\;\rm{km}$$

Since the shuttle orbit varies from 300 to 500 km depending on the mission, that is indeed a good fraction of the height. According to this NASA link the shuttle experiences the force of atmospheric drag at an altitude of about 129 km (80 miles) - so for most of the range of orbits, dropping 360 km is indeed sufficient.

  • $\begingroup$ Yes; as pointed out in the NASA manual quoted in the answer by Organic Marble, the actual delta-v applied is "usually" anywhere between 200 and 550 feet/s (and thus can fall outside of that range), which translates to the range 61 to 168 m/s. $\endgroup$
    – user
    Commented Sep 18, 2015 at 13:35
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    $\begingroup$ Hmm I was gonna say you just beat to the punch, but it looks like you posted right after I started working on my answer. Did you mean force of the atmosphere rather than gravity for the 129km altitude? $\endgroup$
    – Eph
    Commented Sep 18, 2015 at 14:55

The retrograde burn removes energy from the orbit. The energy of orbits remain constant when not burning and can be described by: $$\frac{E}{m}=\frac12{V^2}-\frac{GM}{r}$$

The angular momentum also remains constant: $$\frac{L}{m}\approx rV$$ (This is only approximate because it ignores velocity towards or away from the focus of the orbit, but at the aphelion and perihelion it is exact and those are the only places we're going to use it.)

Starting from a circular radius $R_0$ we have:



So now let's apply a $90 / 7\,000= 1.3\% =\epsilon$ reduction in velocity:

$$V_1=(1-\epsilon) V_0$$ $$\frac{L_1}{m}=R_0 V_1=(1-\epsilon)R_0 V_0$$ $$\frac{E_1}{m}=\frac12 V_1^2-\frac{GM}{R_0}=\frac12 V_1^2-V_0^2=\left(\frac{(1-\epsilon)^2}2-1 \right)V_0^2$$ Now the angular momentum and energy will be constant through from the aphelion to perihelion.

$$r\,V=(1-\epsilon)R_0 V_0$$

$$\frac12{V^2}-\frac{GM}{r}=\left(\frac{(1-\epsilon)^2}2-1 \right)V_0^2$$

$$\frac12{V^2}-V\frac{V_0}{(1-\epsilon)}=\left(\frac{(1-\epsilon)^2}2-1 \right)V_0^2$$

Now this is a quadratic equation with two solutions for velocity. These correspond to the aphelion and perihelion:

$$V=\begin{matrix} V_1 \\ \left(\frac{2}{(1-\epsilon)^2}-1\right) V_1 \end{matrix}$$

Which means the radius at the perihelion would be:

$$R=\left(\frac{2}{2\epsilon-\epsilon^2+1}-1\right) R_0$$

Doing a Taylor expansion on $\epsilon=0$ yields:

$$R=(1-4\epsilon+ ...) R_0$$

This is equivalent to Flouris's Answer.

Thanks to TildalWave for doing the research for this last section: For $\epsilon=1.3\%$ this corresponds to a $5\%$ reduction in orbital radius. So for an initial orbit altitude of $400\ km$ this corresponds to an orbital radius of $6\,760\ km$ which corresponds to a drop of $338\ km$. This will put the perihelion at $62\ km$ which is well below where atmospheric drag will deorbit anything.

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    $\begingroup$ For small $\epsilon$ your result can be simplified significantly by simple expansion. This would get you to $R\approx\frac{1-2\epsilon}{1+2\epsilon}R_0\approx(1-4\epsilon)R_0$ which was the result I got. May be worth doing that to show the equivalence with the simpler relationship I derived. $\endgroup$
    – Floris
    Commented Sep 18, 2015 at 15:09
  • $\begingroup$ @TildalWave wow, I took the earths radius and rounded up to approximate the orbital diameter whoops. Thanks. $\endgroup$
    – Eph
    Commented Sep 18, 2015 at 15:26
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    $\begingroup$ @Rick That's OK, that's why we peer review posts. It took me a couple of edits to get the right number, too. :) Anyway, your previous perigee numbers were pretty close to skip reentry altitudes for LEO orbital speeds. But while that was one of the reentry modes for Shuttle, they never really used it. ;) $\endgroup$
    – TildalWave
    Commented Sep 18, 2015 at 15:30

The formula relating speed to orbital radius is:

$$ v_c = \sqrt{ \frac{G M}{r} } $$

Or, rearranging this, we get: $$ {v_c}^2r = C $$

for some constant C. If your shuttle's speed decreases by 90/7000 m/s = 1.3% approx, the required orbital radius would have to increase by approx 2.6% ( $= 1.013^2 - 1$ ). If the current orbital radius is 4000 miles, this means the shuttle is now 2.6% x 4000 = roughly 100 miles lower than where it needs to be to sustain circular orbit.

Now, I realise I've not accurately explained what will happen to the shuttle next, but you can see that's roughly the right order of magnitude speed change to bring it down into the atmosphere.


What you are describing is the deorbit burn. The short answer is that change in velocity allows the Shuttle to slow enough to act as a glider (way oversimplified). During descent the Shuttle brakes by adjusting its angle to further decelerate. The Shuttle also employed a chute.


NASA has plenty of information on the subject. Feel free to Google it.

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    $\begingroup$ The Shuttle's drag chute was only used for final braking assistance once landing was essentially done and only on-ground braking remained. See for example this photo of the STS-132 landing, NASA photo KSC-2010-3517 , which shows the drag chute being deployed around the time when the rear wheels hit the runway. $\endgroup$
    – user
    Commented Sep 16, 2015 at 14:02
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    $\begingroup$ An additional major benefit cited for its addition was its positive effects on directional stability during rollout. $\endgroup$ Commented Sep 16, 2015 at 14:17

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