How could a 90 m/s delta-v be enough to commit the space shuttle to landing?

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

I think that something visual may be of help

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.

Answer 4

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.

Answer 5

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:

$$R_0=\frac{GM}{{V_0}^2}$$

$$\frac{E_0}{m}=\frac12{V_0^2}-V_0^2=-\frac12{V_0^2}$$

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.

Answer 6

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.

Answer 7

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.

http://www.nasa.gov/mission_pages/shuttle/launch/landing101.html

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