## Orbital Rendezvous - A Challenge

• Situation: Space Station in circular orbit about the Earth, Shuttle is 10 km lower in another circular orbit.
• Challenge: Manuever the Shuttle with a single change in speed to rendezvous with the Space Station.
• Catch up with the Station
• Give the shuttle a velocity directly toward the Station
• Use a moderate velocity - say 0.02 km/s (72 kph)- time of travel (10 km/0.02 km/s) = 500 s
• Reality of Process:
• Direction of orbit velocity is changed by 0.15 degrees and changes the circular orbit into an elliptical one of eccentricity 0.00265
• The new orbit has apogee of 7000 km (1 + 0.00265) = 7000 + 18.55 km
• Semi-major axis (7000 km), period (5828 seconds) and orbit energy is unchanged
• The initial position of the shuttle is 90 degrees past perigee of the new orbit.
• The shuttle slows down as it moves toward apogee of it new orbit.
• Does the shuttle rendezvous with the Station?

• Determine the Details of Motion using Kepler's laws.
• Find the angle where the shuttle crosses the Station orbit
7010 = 7000 ( 1 - 0.00265 cos θ)
cos θ = -10/18.55
θ = 122.75°
This is 122.7° - 90° = 32.7° past the initial position
• Find the time it will take to reach that position
The eccentric anomalies are: cos E(90°)= 0.00265 and cos E(122.6°) = -0.5372 radians
giving E1 = 89.8° and E2 = 122.5°
t1 = 0.2496 P
t2 = 0.3402 P
time = 531 seconds
• Find where the Station is when the shuttle crosses its orbit
The period of the Station orbit is 5841 seconds
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Its motion in 531 seconds in its circular orbit is 32.72°
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What does 0.03° represent for a 7000 km radius orbit? 2.7 km
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The Station is still moving 7.541 km/s but catching up with the shuttle which is now moving at 7.536 km/s. Since the Shuttle is going toward apogee, it is slowing down. It will eventually slow to 7.527 km/s at apogee.
• What would have been the proper Shuttle position to rendezvous with the Station
The Shuttle should start fire its rocket to change its velocity when it is a little over 2.7 km ahead of the Station.

### Alternate Analysis of Rendezvous Dynamics - The Rotation Reference Frame.

Another method is to move within the reference frame of the Space Station. This is a rotating reference frame and one must include Coriolis acceleration in the dynamics. The Coriolis acceleration is in a dirction perpendicular to the velocity in the plane of the orbit. This causes the relative path to be nearly a circle. Centrifugal acceleration (some time called tidal acceleration) is also present and always is in the radial direction.

Coriolis Acceleration = aCoriolis = 2 vrelative ω

Centrifugal Acceleration = aCent = ω2rrelative

As you can see in the diagram to the right, the shuttle will veer behind the Space Station and miss it unless it ahead of the Space Station in its orbit. (Despite the fact that the Shuttle is travelling faster than the Space Station).

In the example above ω = 0.00103 radian/s
aCoriolis = 2 (0.02 km/s)(0.00103 /s) = 0.0000432 km/s2
This produces a relative circular motion with a radius of 9.3 km at the Shuttle orbit.
aCent = (0.00103)2 10 km = 0.0000106 km/s2 at the Space Station Orbit.
The Centrifugal will 'stretch' the 'height' that the Shuttle with reach.

### Orbital Equation for small eccentricity (e << 1)

 Period P = 2π√(a3/k) Radial Distance r = a [1 - e cos(θ)] Velocity in Orbit v = vc[1 + e cos(θ)] Velocity angle to circular tan(φ) = e sin(θ) Eccentric Anomaly cos E = cos(θ) + e sin2(θ) Time past Perigee t - T = P/(2π)*[E - e sin E]

Created on ... May 29, 2005 - L.Bogan