## Intersecting Orbits Rendezvous Method

Starting Conditions:

- 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 r_{peri} = 6990 km

Use the velocity equation: v^{2} = v_{c}^{2}(2a/r - 1)
and the fact the mean velocity of the ellipitcial orbit is given by v_{c} = √[k_{E}/a]

- So a = 1/[2/r - v
^{2}/k_{E}] = 6999.1 k
- eccentricity = 1 - r
_{peri}/a = 0.001286
- r
_{aphelion} = (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-e
^{2})/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.
- v
_{c} = 7.547 km/s

Note that this new orbit has a mean velocity very similar to that of the Station's.
- v
_{peri} = v_{c} √[(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)