Problem 4.9 BMW

Nasser Abbasi

Problem: at burn out, given vbo= sqrt(2) I, and rbo= 1.1 J, how it long it will take for the probe to cross the x-axis?

Assumptions:

Method:

Find e,a,p. Note that change in ‘mu’ is 90 degrees. Relate this angle to the hyporabolic Eccentric angle F, then use the classical Hyporabolic solution for the kepler equation to find time duration.

Analysis:

At burn out, satellite is at perigee (this can be seen also from the fact that vbo is perpendicular to rbo)

rbo = 1.1 J, hence rbo = 1.1

vbo = sqrt(2) I hence vbo = sqrt(2)

v^2 u

Energy = ------ - ----

2 r

Hence Energy (calculated at perigee)= 2/2 – 1/1.1 = 0.0909091 DU^2/TU^2

Since E>0, this is a hyperbola.

u

Energy = - ----- ---(1)

2a

To find ‘a’, Plug Energy in (1), and noting that u=1, we get

1

0.0909091 = - -----

2a

hence a = -5.5 DU

| I J K |

h = r0 x v0 = | 0 1.1 0 | = I (0) – J (0) + K (1.1*sqrt(2))

| sqrt(2) 0 2 |

h = -1.555634 k

hence h = 1.555634

h^2

Now, p = ---- = h^2

u

Hence p = 2.42

To find e, from p=a(1-e^2)

2.42 = -5.5 (1-e^2)

e=1.2

Now we are asked to find TOF when it crosses the x-axis. This means the angle mu = 90 degrees.

Using classical method, angle F is found from

e + cos(90)

F = arccosh ------------- eq 4.2-21 BMW

1+ e cos(90)

hence F = 0.6223625 radian

now, using equation 4.2-19 in BMW, we find TOF

sqrt((-a)^3)

TOF = -------------- ( e sinh(F) – F )

u

TOF = sqrt( -5.5^3 ) ( 1.2 sinh(0.6223625) - 0.6223625 )

TOF = 2.2395 TU