# How much runway does the space shuttle need to land?

A textbook question proposes the following scenario:

NASA needs to plan emergency landing sites for the Space Shuttle (RIP). There are a number of things to consider. The shuttle touches down no sooner than $300$ meters into runway track, and upon touching down deploys a parachute, which takes 5 seconds to be fully deployed. Once the speed decreases to $60\ \mathrm{km/hr}$, the brakes can be applied and require $200$ meters further to arrive at a complete stop. Here are the rest of the specs (not necessarily accurate in real life):

• Total Landing Mass: $104,328\ \mathrm{kg}$

• Velocity: $350\ \mathrm{km/hr}$

• Area of parachute: $68.0625 \pi$

• $C_D$, the drag coefficient of the parachute: $1.4$

• $\rho_\text{air}$, the density of air: $1.1774\ \mathrm{kg/m^3}$

The question is, asses NASA's decision to land the space decision at an airport with a runway of $11,000\ \mathrm{ft}$.

MY ATTEMPT

I begun my assesment by assuming the Shuttle touches down at the $300$ meter mark, at which it is travelling at $350\ \mathrm{km/hr}$. Immediately the parachute deploys and takes five seconds to deploy.

Once the parachute deploys, a negative Force is applied to the Shuttle in the form of $$m \frac{dV}{dt} = - \frac{1}{2} \rho_\text{air}C_DAV^2 = - \frac{1}{2}(1.1774)(1.4)(68.0625 \pi)(V^2) \approx-56.0958 \pi V^2$$

Out of this, I formed a function for velocity based on time by solving this as a differential equation and got: $$\frac{1}{V^2}dV = .005377 \pi dt \longrightarrow - \frac{1}{V} = .0005377 \pi t +C$$ and as a function of $t$: $$v(t) = - 185.977 \frac{1}{ \pi t} + C$$

Since my initial condition is $350$ when $t=0$, my function finally is: $$v(t) = - 185.977 \frac{1}{ \pi 0} + 350$$

But I'm dividing by zero! My goal was to see how far the shuttle travels in the time it takes to reach $60\ \mathrm{km/hr}$ and determine if it has enough track to make it. I'm not sure if what I'm doing so far, or my strategy is correct. Obviously what I'm doing is somewhat flawed.

I would really appreciate any help. Thanks!

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Is your definition of $C$ the same throughout the equations? – Bernhard Nov 9 '12 at 7:19

You're correctly saying:

$$\frac{dv}{dt} = const \times v^2$$

and therefore:

$$\frac{dt}{dv} = const \times v^{-2}$$

so:

$$t = -\frac{B}{v} + C$$

for some constants $B$ and $C$. however this gives you:

$$v(t) = -\frac{B}{t - C}$$

So $t$ may be zero at the moment the parachute deploys, but $t - C$ isn't. The constant $B$ is $2m/\rho C_d A$ and you need to find $C$ from the initial conditions. A qick bit of algebra will tell you that $C = B/v(0)$ so:

$$v(t) = -\frac{B}{t - B/v_0}$$

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How does $\frac{dv}{dt} = const \times v^2$ become $\frac{dt}{dv} = const \times v^{-2}$? – Imray Nov 9 '12 at 13:17
@Imray note that const just refers to a constant. They are not same in the two equation. If we use A = const in first equation then in second equation const = 1/A, but since is it still a constant, we just call it const. – mythealias Nov 10 '12 at 3:30
@Imray: as mythealias says, I just meant const to be some random constant and it isn't meant to be the same in both equations. Anyhow, did my answer help? – John Rennie Nov 10 '12 at 7:17
Yes! I'm nearly finished! I got $\frac{9}{875}$ as the constant. THen I solved for $v(0) = \frac{875}{9}$ since I converted to meters/second, followed by $v(t) = \frac{50}{3}$ and found the time it took to slow down: $29.42$ seconds. Thank you so much John! (and mythealias) – Imray Nov 10 '12 at 23:48