
#1
Oct2112, 03:45 PM

P: 17

1. The problem statement, all variables and given/known dataA lunar probe is given just escape speed at an altitude of .25 DU and flight path angle 30°. How long will it take the probe to get to the vicinity of the moon (r = 60 DU), disregarding the moon's gravity?
Circular orbit, radius = 1.25 DU θ = 30° gravitational parameter = μ = 1 DU^3/TU^2 true anomaly = v Eccentric Anomaly = E eccentricity = e tto = Time of flight a = semimajor axis 2. Relevant equationsEscape velocity = √(2μ/r) = 1.26 DU/TU tan(θ) = (esin v)/(1 + ecos v) tan (v/2) = √((1+e)/(1e))tan(E/2) tto = √(a^3/μ)(2k(pi) + (EesinE)  (Eo  esinEo)) θ(parabolic) = v/2 3. The attempt at a solution The answer is 223.2TU I just can't see how flight path relates to time of flight. The second equation in the relevant equations section seems like it would fit for equating flight path and true anomaly but I can't separate the true anomaly variable. I'm not sure if that last equation for a parabolic flight path applies (the orbit is parabolic but I don't know). Any help would be greatly appreciated. 



#2
Oct2112, 08:11 PM

P: 17

So I'll post a solution I've been working on:
I'm not sure how to think of flight path, but I assumed the transfer maneuver was like a Hohmann Transfer with 60 DU forming the apoapsis radius: e = (601.25)/(60 + 1.25) = .9592 Now I'm trying to find the semimajor axis from the periapsis radius Rp = a(1e) a = Rp/(1e) = 1.25/(1.9592) = 30.625 DU Assuming the probe reaches the Moon's vicinity at apoapsis radius, E = pi and: tto = √(30.625^3)(pi  sin pi) = 532 TU 



#3
Oct2112, 08:43 PM

Mentor
P: 11,416

If the probe is given escape speed at some point, it will not have a closed orbit; it'll be parabolic.




#4
Oct2112, 09:15 PM

Mentor
P: 11,416

Time of Flight using flight path and radius (Kepler's Equation)
Okay, some hints.
If you know the radial distance, speed, and flight path angle at some instant then you can determine the specific angular momentum h of the probe. That, in turn, gives you the length of the latus rectum p (since you also know the gravitational parameter μ by assumption). The eccentricity is a given since you know that the probe has escape speed. That means you can write the general equation for the conic orbit radius: ## r = \frac{p}{1 + e cos(\nu)}## and can solve for the two values of ##\nu## involved (at 1.25DU and 60DU). You should be able to work out the time of flight between those two values of true anomaly. Extra hint: The parabolic eccentric anomaly will be helpful. 



#5
Oct2112, 09:52 PM

P: 17

Awesome, I got the right answer! Thanks so much, I had completely forgot eccentric anomaly was solved differently depending on the orbit/trajectory.



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