Problem 1 teacher own.

Nasser Abbasi

Problem: a probe has the following position and velosity rbo=1.1 J and vbo= sqrt(2) I. Finds its position and velosity vectors after 3 hours and sketch the orbit.

Assumptions:

Method:

Find e,a,p. since time is given, solve for Universal variable X from

___

\/ u (t-t0) =

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