Applications of Differential Equations (1): The Rocket Problem

Differential equations are among the most powerful and useful analytic tools known, and are applicable to vast areas of interest ranging from engineering to astrophysics and even cosmology. Therefore, it's useful to look at some of these applications.

For some readers it may be useful to first examine previous blogs to do with this topic, e.g.

http://brane-space.blogspot.com/2010/06/looking-at-basic-differential-equations.html

and http://brane-space.blogspot.com/2010/06/back-to-diferential-equations.html

One of the most powerful applications is to rocket flight, for which a diagram of the type shown is often used, say to define the launch angle q_o, as well as the final x, y- coordinates (e.g. at impact). Consider then the following problem:

Find the  x- and y-coordinates of the points on the trajectory of a rocket launched at an angle of 80 degrees with an initial velocity of 100,000 f/s if the air resistance is 0.01mv. Find the value of x and y after 10 seconds.

Data:    q = 80^o,   k = 0.01mv   v = 10⁵ fs^-1 and t = 10s

The differential equation can be written:

m(d²x/dt²) =  - k (dx/dt)/ v = - (0.01mv)x’/ v  = - 0.01mx’

Similarly:

my” = -mg – (0.01mv)y’/v  =   -mg – 0.01my’

Yielding the simultaneous pair:

x” + 0.01x’ = 0

y” + 0.01 y’ = -g

We know:

x’_o = v_o cos (q) = v_o cos (80) = (10⁵) cos 80

y’_o = v_o sin (q) = v_o sin (80) = (10⁵) sin 80

General Solutions:

x = c1 + c2 (e ^-0.01t)

y = c3 + c4((e ^-0.01t)  - 100 gt

x’ = -0.01c2 (e ^-0.01t)

y’ = -0.01c4((e ^-0.01t)  - 100 g

At t = 0, x = 0 so:

0 =  c1 + c2 (e ^-0.01t)

But:

x’_o  = v_o cos (80) = (10⁵) cos 80 =  -0. 01c2 (e ^-0.01t)

Then: c2 = - (10⁷) cos 80

So: 0 = c1 + ( - (10⁷) cos 80) or c1 =  (10⁷) cos 80

At t = 0, y = 0:

Then: 0 = c3 + c4 (e ^-0.01t)

And: y’_o  = v_o sin (80) = (10⁵) sin 80 = -0.01c4(e ^-0.01t)  - 100 g

So: c4 = -(10⁷) sin 80 – 10000g

\ c3 = - c4 (e ^-0.01t) = 10⁷ sin 80 + 10000g

To obtain the x and y-coordinates:

First, the x-coordinate:

x = c1 + c2 (e ^-0.01t) = 10⁷ cos 80 – (10⁷ cos 80)( e ^-0.01t)

x = 10⁷ cos 80 [1 - ( e ^-0.01t)] =  10⁷ (0.17365) [1 - ( e ^-0.01t)]

or:   x = 1736500 [1 - ( e ^-0.01t)]

The y-coordinate:

y = c3 + c4 ( e ^-0.01t)

Or:  y = 10⁷ sin 80 + 10000g – (10⁷ sin 80 + 10000g(e ^-0.01t))

y = 10⁷ sin 80 + 10000g(1 - e ^-0.01t) – 100 gt

y = [10⁷ (0.9848) + 320,000](1 - e ^-0.01t) – 100 gt

y = 1016800(1 - e ^-0.01t) – 100 gt

After 10 seconds:

x = 1736500 [1 - ( e ^-0.01t)] = x = 1736500 [1 - ( e ^-0.01(10))]

x = 165, 200 ft. (» 31. 3 miles)

y = 1016800(1 - e ^-0.01(10)) – 100(32)(10)

y =  930,580 ft.  ( » 177 miles)

Problem for the Math Maven:

Re-work the problem but obtain the coordinates in kilometers, instead of miles. (Hint: you will use an initial velocity of  30,300 m/s instead of 100,000 f/s. Also you will use g = 10 ms^-2   instead of g = 32 fs^-2) What differences did you find in your final values - from those in the blog problem - and how might you account for them? (Convert  either miles to km, or vice versa, to compare).