Delta-v for Hohmann transfer from hyperbolic trajectory to circular orbit

[lujz]

I get different result than stated in the book.
What am I doing wrong?

Homework Statement

A spacecraft returning from a lunar mission approaches Earth on a hyperbolic trajectory.
At its closest approach A it is at an altitude of 5000 km, traveling at 10 km/s. At
A retrorockets are fired to lower the spacecraft into a 500 km altitude circular orbit,
where it is to rendezvous with a space station.
Verify that the total delta-v required to lower the spacecraft from the hyperbola into the parking orbit is 6.415 km/s.

r_Earth = 6378
Gravitational parameter μ = 398600

Homework Equations

r - radius
e - eccentricity
A - apogee
P - perigee

h - angular momentum
v - velocity

r = altitude + r_Earth
e = (r_A - r_P) / (r_A + r_P)
r_P = (h²/μ)*(1/(1+e))
v_A = h/r_A
v_P = h/r_P
v_circular = sqrt(μ/r)

The Attempt at a Solution

I get h = 58458,

Speed at apogee of the transfer orbit:
v_A = 5.1378 km/s,

Delta-v at apogee:
Δv_A = 10-5.1378 = 4.86219 km/s

Speed at perigee of the transfer orbit:
v_P = 58458/6878 = 8.499 km/s

Speed of the final orbit:
v_circular = 7.6127 km/h

Delta-v at perigee:
Δv_P = 8.499 - 7.6127 = 0.8866 km/s

Total delta-v:
Δv_T = 4.86219 + 0.8866 = 5.749 km/s

 

[gneill]

Hi Lujz, welcome to Physics Forums.

I don't see anything wrong in your calculations. Is there perhaps an accompanying diagram that might introduce some "quirk" of the setup that is not included in the problem statement? An orbital plane change perhaps?

 

[lujz]

Hi gneill,

The accompanying diagram is this:
[PLAIN]http://www.shrani.si/f/1l/r7/acGDhup/2/example62.png

The original question is: "Find the location of the space station
at retrofire so that rendezvous will occur at B."
It then proceeds with calculations for periods and the angle in question.
Nothing I can notice that would affect total Δv.

 

[gneill]

Okay, so I don't see anything there that would affect your solution method. I suppose that the text's proposed answer is in error.

 

[D H]

I used the vis viva equation to arrive at the answer of 5.75 km/s:

http://www.wolframalpha.com/input/?...)-1/(2750+6378)))-sqrt(398600*(1/(500+6378)))

In other words, I arrived at same answer as did the OP, but using a different method.