PeriapsisPrograde
Wannabe addon dev
In my previous blog entry, I found the delta-v required for an Earth-Mars in a simplified scenario. In this entry, I will discuss how to get the timing right.
In the first entry, we determined the semi-major axis of our transfer orbit to be 1.885e11 m, and the standard gravitational parameter of the Sun is 1.327e20 m^3/s^2. The period of the orbit is then [math]2\pi \sqrt{\frac{a^3}{\mu}} =[/math] 4.464e7 s, or about 516.7 days. We will only travel half the orbit, so the time of flight is 258.3 days
We may want to account for the time it takes to leave Earth's sphere of influence before going into solar orbit. The hyperbolic ejection orbit doesn't really have a period, but we can find the time to get out to the edge of the Earth's sphere of influence. The sphere of influence is given by [math]r=a\bigg( \frac{m}{M} \bigg)^{2/5}[/math] = 9.245e8 m.
The true anomaly at periapsis is zero, while the true anomaly at a given time is found by [math]\nu = \arccos \bigg( \frac{a(1-e^2) - r}{er} \bigg)[/math]. For r=9.245e8 m, we get [math]\nu_{" \infty " }=[/math] 149.67 degrees. From this, the eccentric anomaly at "infinity" can be found: [math]F =\cosh ^{-1} \bigg(\frac{e+\cos \nu}{1+e\cos \nu}\bigg)[/math] =.
In the first entry, we determined the semi-major axis of our transfer orbit to be 1.885e11 m, and the standard gravitational parameter of the Sun is 1.327e20 m^3/s^2. The period of the orbit is then [math]2\pi \sqrt{\frac{a^3}{\mu}} =[/math] 4.464e7 s, or about 516.7 days. We will only travel half the orbit, so the time of flight is 258.3 days
We may want to account for the time it takes to leave Earth's sphere of influence before going into solar orbit. The hyperbolic ejection orbit doesn't really have a period, but we can find the time to get out to the edge of the Earth's sphere of influence. The sphere of influence is given by [math]r=a\bigg( \frac{m}{M} \bigg)^{2/5}[/math] = 9.245e8 m.
The true anomaly at periapsis is zero, while the true anomaly at a given time is found by [math]\nu = \arccos \bigg( \frac{a(1-e^2) - r}{er} \bigg)[/math]. For r=9.245e8 m, we get [math]\nu_{" \infty " }=[/math] 149.67 degrees. From this, the eccentric anomaly at "infinity" can be found: [math]F =\cosh ^{-1} \bigg(\frac{e+\cos \nu}{1+e\cos \nu}\bigg)[/math] =.
