Numerical Example
Assume the following values for an Earth centred Kepler orbit
- r1 = 10000 km
- r2 = 16000 km
- α = 100°
These are the numerical values that correspond to figures 1, 2, and 3.
Selecting the parameter y as 30000 km one gets a transfer time of 3072 seconds assuming the gravitational constant to be = 398603 km3/s2. Corresponding orbital elements are
- semi-major axis = 23001 km
- eccentricity = 0.566613
- true anomaly at time t1 = −7.577°
- true anomaly at time t2 = 92.423°
This y-value corresponds to Figure 3.
With
- r1 = 10000 km
- r2 = 16000 km
- α = 260°
one gets the same ellipse with the opposite direction of motion, i.e.
- true anomaly at time t1 = 7.577°
- true anomaly at time t2 = 267.577° = 360° − 92.423°
and a transfer time of 31645 seconds.
The radial and tangential velocity components can then be computed with the formulas (see the Kepler orbit article)
The transfer times from P1 to P2 for other values of y are displayed in Figure 4.
Read more about this topic: Lambert's Problem
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