Intersecting Orbits Rendezvous Method
- Space Station: Circular Orbit
- a = 7000 km
- v = 7.546 km/s
- P = 5828 s
- Shuttle: Circular Orbit near
but not directly under the Station. (exact position to be determined)
- radius 6990 km
- v = 7.552 km/s
- P = 5772 s
The Shuttle is moving faster and will pass the Station in its present orbit. We know
that adding velocity to the shuttle will raise it into a higher, slower orbit which will can
cross the orbit of the Station.
What velocity will
- raise the Shuttle past the Station orbit
- give the Shuttle a velocity matching that of the Station at the cross over.
Determine a neww elliptical orbit for the Shuttle with v=7.542 km/s at a distance of r=7000 km from the center of the Earth. The orbit's perihelion with be its present circular orbit distance rperi = 6990 km
Use the velocity equation: v2 = vc2(2a/r - 1)
and the fact the mean velocity of the ellipitcial orbit is given by vc = √[kE/a]
- So a = 1/[2/r - v2/kE] = 6999.1 k
- eccentricity = 1 - rperi/a = 0.001286
- raphelion = (1 + 0.001286) 6999.1 = 7008.1 km
The shuttles orbit rises to 8 km above that of the Station.
- the angle, φ, from perihelion at r = 7000 is given by cos(φ) = [1 - a (1-e2)/r]/e
So... φ = 96.5°
- The time of flight to this point from perihelion is 1559 s = 26 minutes.
This was determined by calculating the eccentric anomaly and then the mean anomaly.
- vc = 7.547 km/s
Note that this new orbit has a mean velocity very similar to that of the Station's.
- vperi = vc √[(1+e)/(1-e)] = 7.557 km/s
In the same time the Station will revolve (1559/5828) 360° = 96.3° about the Earth.
In order to rendezvous precisely, the Shuttle should
- increase its velocity by 7.557-7.552 = 0.005 km/s
- when the Station is 0.2° ahead of it. (24 km)
14-Jun-2005 -- L.Bogan