I want to transfer between two planets using Hohmann transfers and the patched conic approach. However, I am struggling to figure out how long I have to wait at the starting planet (say Uranus) until I can return home (to Earth). I am choosing Uranus because earth's orbit is many times shorter than it.

What I have so far is that I need the initial angle between the planets when I arrived at Uranus on the transit from Earth, and I need something about the relative angular rates of both Earth and Uranus. But, nothing that I am coming up with gives me a reasonable answer. Many give me negative time values.

From Eq. 22 on page 11 of lecture #6 notes from Virginia Tech's Astromechanics course AOE4134, I was able to find an equation:

$$ wait\ time =\frac{\phi_0+\omega_1*t_{transfer}-\pi}{\omega_2 - \omega_1}, $$


  • The initial angular separation between Earth and Uranus when the craft arrives at Uranus is $ \phi_0$

  • $\omega_1$ is the angular rate of Earth.

  • $\omega_2$ is the angular rate of Uranus.

  • $t_{transfer}$ is the time to transfer between the planets on the Hohmann orbit.

The problem with this is that it gives me a negative time and I don't know how I am misunderstanding it to get the wrong time.

TLDR: I need an equation that will give me how long I have to wait at a planet to transfer to another planet.

  • $\begingroup$ Interesting question! Can you add a mention, citation, or link to where you found the equation? Thanks! $\endgroup$
    – uhoh
    Commented Feb 23, 2018 at 0:37
  • $\begingroup$ I pulled the equation from Lecture 6 Page 11. The main website is a class on astromechanics that was taught at virginia tech. the link is: www.dept.aoe.vet.edu/~lutze/AOE4134 $\endgroup$ Commented Feb 23, 2018 at 16:32
  • $\begingroup$ Thanks, these are great! "add" means add to the question, since in SE comments should be considered temporary and could be removed in the future. This doesn't happen often, but if it's important information always put it back into the original post. $\endgroup$
    – uhoh
    Commented Feb 23, 2018 at 23:56

2 Answers 2


I'm merely a KSP player, not NASA but I think I understand the problem here:

Looking at the numerator of your equation you are working backwards from the planned arrival. Hence the -Pi term because the departure is Pi radians before the arrival. The other terms could be small--thus driving the numerator negative.

What does this mean? You missed the launch window. (Yeah, you might not have had a chance to catch it.)

Solution? Wait for the next launch window. I'm going on shaky ground here but I think the answer is to add 2 * Pi to the numerator.


I looked at those lecture #6 notes. You aren't quoting equation 22 as it is written in the notes. Your denominator will always be negative (earth's angular rate being greater than Uranus'), In equation 22 the pi term is (1+2k)pi, where k can be chosen positive or negative to make the signs work out. Does that help?


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