Planetary rendezvous orbital mechanics

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SUMMARY

The discussion focuses on calculating the time required for Venus to align with Earth for a Hohmann transfer orbit, starting from specific angular positions. The initial angular difference is 190 degrees, and Venus must lag Earth by 36.0298 degrees. Using the mean motions of Venus (0.0279658 rad/day) and Earth (0.0172024 rad/day), the calculated time for this alignment is 366.516 days. The conversation also touches on the adjustments needed for different planetary configurations and the importance of visualizing the orbital mechanics involved.

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Dustinsfl
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Earth to Venus Hohmann transfer alignment time.

Earth is at periapsis and Venus is at 190 degrees at ##t = 0##.

For a Hohmann transfer, Venus must be lagging Earth by ##36.0298## degrees.

I am trying to find the time it takes for the planets to align from their current positions.

Is this correct: initial angular difference (190) = desired angular difference (-36.0298) + (nv - ne)t where nv and ne are the mean motion of Venus and Earth, respectively.

##nv = 0.0279658## rad/day and ##ne = 0.0172024## rad/day

$$
3.31613 = -0.628838 + (0.0279568 - 0.0172024)t\Rightarrow t = 366.516\text{ days}
$$

So from their initial configuration, it would take 366.516 days for Venus to lag Earth by 36.0298 degrees.
 
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If Venus is initially at +190° when the Earth is at 0°, what's the angular separation in terms of the direction of closing? (Venus is currently "ahead" of Earth by 190°, but closes on Earth from "behind", so to speak).
 
If you are asking what is the complement of 190, it is 170.
 
Dustinsfl said:
If you are asking what is the complement of 190, it is 170.

Yup. Isn't that the angular distance that Venus has to "make up" in order to catch the Earth? And you want to end up 36.0298° shy of that.
 
gneill said:
Yup. Isn't that the angular distance that Venus has to "make up" in order to catch the Earth? And you want to end up 36.0298° shy of that.

Yes but looking in Orbital Mechanics for Engineering Students by Curtis, we read the angle from planet 1 to 2. This wasn't a laid out equation in the book so I adjusted one slightly. So is it supposed to be done by catch up angular distance in separation angular distance and do you know or certain that the catchup method is how it is done?
 
Dustinsfl said:
Yes but looking in Orbital Mechanics for Engineering Students by Curtis, we read the angle from planet 1 to 2. This wasn't a laid out equation in the book so I adjusted one slightly. So is it supposed to be done by catch up angular distance in separation angular distance and do you know or certain that the catchup method is how it is done?

I just pictured the problem, as described, in my mind and drew conclusions from the given facts:

1. Venus' angular velocity is greater than Earth's, so Venus needs to catch up.
2. Venus is currently 190° ahead of the Earth, so it's 170° behind and catching up.
3. We want it to catch up to a spot that is about 36 degrees behind the Earth.

So I figure it needs to gain about 170 - 36 = 134 degrees. The angular closing speed is ω = ne - nv as you've written. Then it's a matter of finding the time it takes to cover hat 134° at angular speed ω.

Things would be different if the destination was an exterior planet (like Mars), where it would be the Earth catching up.
 
gneill said:
I just pictured the problem, as described, in my mind and drew conclusions from the given facts:

1. Venus' angular velocity is greater than Earth's, so Venus needs to catch up.
2. Venus is currently 190° ahead of the Earth, so it's 170° behind and catching up.
3. We want it to catch up to a spot that is about 36 degrees behind the Earth.

So I figure it needs to gain about 170 - 36 = 134 degrees. The angular closing speed is ω = ne - nv as you've written. Then it's a matter of finding the time it takes to cover hat 134° at angular speed ω.

Things would be different if the destination was an exterior planet (like Mars), where it would be the Earth catching up.


How does the equation change for an exterior planet?
 
Dustinsfl said:
How does the equation change for an exterior planet?

Then it would be the Earth that's closing the distance with the planet's position (or increasing the separation if that's required by the initial position). Really, it's the same underlying geometry, so the relevant equations are of the same form.

It would be useful to draw a diagram of the initial configuration and final configuration to see how one evolves to the other, and then write the appropriate expression from there.
 
I think my wait time is all messed up then.

I made a mistak prior the lag distance needed to be a -54 degrees not 36. 36degree is how far behind Earth is when the spacecraft arrives at Venus from the Hohmann transfer.

How do I determine the wait time from this position: Venus at 91.6987 degrees and Earth at 55.669 degrees.

I adjusted the angles for the actually location at present day. So the Hohmann trasfer occurred 275 days in the future from today and the above would be their current positions.

For the return trip, Earth would have to be ahead of Venus by 36.0298 degrees.

How long would it take to obtain this configuration from the current?

The equation I used was
$$
t_{\text{wait}} = \frac{-2\varphi_f' + 2\pi N}{nv - ne}
$$
where ##\varphi_f'## is the final angular separation on return.
 
  • #10
gneill said:
Then it would be the Earth that's closing the distance with the planet's position (or increasing the separation if that's required by the initial position). Really, it's the same underlying geometry, so the relevant equations are of the same form.

It would be useful to draw a diagram of the initial configuration and final configuration to see how one evolves to the other, and then write the appropriate expression from there.

I have a diagram: I had to readjust the positions 5 times due to mistakes but this one is up to date with corrections.

http://img805.imageshack.us/img805/5355/texdiagram.png
 
Last edited by a moderator:
  • #11
Are you saying:

The initial positions are:
Earth: 55.669°
Venus: 91.6987°​

and that for the commencement of the transfer orbit, you need Venus to trail the Earth by 54°, so what's the wait time until launch?
 
  • #12
gneill said:
Are you saying:

The initial positions are:
Earth: 55.669°
Venus: 91.6987°​

and that for the commencement of the transfer orbit, you need Venus to trail the Earth by 54°, so what's the wait time until launch?

The initial position is Earth III and Venus III since that is location after the Hohmann from Earth to Venus. I just checked my wait time the hard way and it worked out so I got it thanks.
 
  • #13
I'm glad that you've solved your problem. I must confess that I remain confused about just what it is you were trying to determine and from what starting information (the initial position is the position after the maneuver?) But as long as you're happy...
 

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