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Astronautics, Spring quarter 2003, HW 4, UCI

by Nasser M. Abbasi

1 problem 3.5 from BMW book

Analysis:

Since Vc2   is given as    DU-
0.5TU  , and Vc1 = 1DU ∕TU  , then service orbit must be the larger orbit (outside orbit), since the smaller the distance from the attracting body the larger the velosity of the orbiting body).

First find r1   and r2

Since         --
V   = ∘ μ-= ⇒ r  = -μ-=⇒  r  = 1-= 1DU
  c1     r1     1   V2c1     1   12

Similarly       μ           1
r2 = V2c2-=⇒  r2 = 0.52 = 4DU

From geometry, for the transfer orbit, 2at = r1 + r2 =⇒ at = 5DU

But ξ = - -μ =⇒ ξ  = - 0.2(DU-)2
 t    at     t         TU

Now, velocity in the transfer orbit is given by     ∘ --(-----)
Vt =  2  μr + ξt hence at point 1, r = r1 = 1  DU, so we get       ∘ -(1-----)-
Vt1 =   2 1 - 0.2  =⇒  Vt1 = 1.264911DU ∕TU

Similarly,      ∘ -(------)-
V  =   2  μ-+ ξ
 t2        r2    t hence at point 2, r = 4
 2  DU, so we get      ∘ ----------
V  =   2 (1- 0.2) =⇒ V  =  0.316227 DU-
 t2       4           t2           TU

So, ΔV1  = |Vc1 - Vt1| = |1 - 1.264911 | = 0.264911  DU/TU

ΔV2  = |Vc2 - Vt2| = |0.5- 0.316227| = 0.183773  DU/TU

hence, minumum ΔV  = ΔV1 + ΔV2  =  ____________________________0.448684  DU/TU

2 problem 3.8 from BMW book

Compute the minimum ΔV  required to transfer between 2 coplaner elliptical orbits which have their major axes aligned. The parameters are:

rp1 = 1.1     DU.   rp2 = 5  DU
e1 = 0.290   DU.   e2 = 0.412  DU

Assume both preigrees lie on the same side of the earth.

Assumptions:

Method: First find V1   and V2  , the velositites for ellips 1 at its perigree and for ellips 2 at its apegee.

Next find ξt  , the energy for the transfer orbit. From this, find V1t  and V2t  , the velosities in the transfer orbit at point 1 and point 2 respectively.

Finally, final ΔV  follows as from the sum of the ΔV  at point 1 and point 2.

This is the minumum, since the transfer orbit is a Homann orbit.

Analysis:


PIC


since a = rp+ ae  from ellips geometry.

rp = a(1 - e)

a = r1-pe

For ellips 1:

a =  rp1-⇒ a  = --1.1- = 1.5492957
 1   1-e1     1  1-0.29   DU.

      -μ-    ----1----                 2    2
ξ1 = - 2a1 = - (2)1.5492957 = - 0.322727 DU ∕TU

But ξ = V22-- μr  , hence, since ξ  is constant over the orbit, we can use this relationship to solve for  V  for different r  .

At point 1, for first ellips, r = rp1   , hence

             V2
- 0.322727 = -12-- 11.1 ⇒ _______________________ V1 = 1.082925  DU/TU.

For ellips 2:

Here we want to find the velosity V
 2   , the velosity at the apegee for ellips 2. So, need to find r
 A2   for ellips 2.

Since rp2 = 5  DU, and e2 = 0.412  , we get

a = r1-pe ⇒ a2 = 1-50.412-= 8.5034  DU.

Hence, since rA = a + ae = a(1+ e) ⇒ rA2 = 8.50034 (1 + 0.412 ) = 12.0068  DU.

Now find ξ
 2   the energy for ellips 2

      μ-    ---1---               2    2
ξ2 = - 2a = -(2)8.5034 = - 0.0588 DU ∕TU

but       2
ξ2 = V2---μ
     2   rA   then

           V2     1
- 0.0588 = 22-- 12.0068-⇒ ________________________ V2 = 0.2212968  DU/TU

For the transfer orbit:

From geometry, 2a = rp1 + rA2 = 1.1+ 12.0068 =⇒ at = 6.5534  DU

so ξ  = - --μ----= - 0.0762963
 t     (2)6.5534  DU  2∕TU 2

So,       ∘ --(-------)   ∘-----------------
V1t =   2  -μ-+ ξt =   2 (1- - 0.0762963) = 1.2905
           rp1            1.1  DU/TU

So,       ∘ --(-------)   ∘ --------------------
V   =   2  -μ-+ ξ   =   2(--1---- 0.0762963) = 0.11823567
  2t        rA2   t        12.0068  DU/TU

so ΔV  at point 1 = |1.082925 - 1.2905| = 0.207575  DU/TU

so ΔV  at point 2 = |0.2212968 - 0.11823| = 0.1030668  DU/TU

hence minumum ΔV  =  0.207575 + 0.1030668 = 0.3106418DU ∕T U