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Notice:

Although this question is similar to this one, it is different because asteroid 2012 XE$_1$$_3$$_3$ has a transitional path between Venus's Lagrangian points L5 point and L3 point, whereas 2013 ND$_1$$_5$ follows a tadpole orbit around the L4 point.

2013 ND$_1$$_5$ is an asteroid of the Aten group that is a co-orbital and a temporary trojan of Venus.

It is following a tadpole orbit around Venus's Lagrangian point L4 and is also a Mercury crosser and an Earth crosser. It comes to within 0.05 AU of Earth periodically and has a diameter in the range of 40 to 100 meters.

Edit: From the example of a tadpole orbit on Wikipedia I got the impression that its orbital plane is perpendicular to that of Venus's orbit around the Sun.
If not, couldn't this be done with 2013 ND$_1$$ _5$, giving it gradually an increasingly larger orbit, and wouldn't this allow to approach Venus more easily ?

Langrangian points

Credit:NASA

On the image above we can see that as the asteroid would approach Venus from L4 with an increasingly growing orbit in a plane perpendicular to the circular path of Venus around the Sun, in the end that orbit could go through both the L1 and L2 points.

Question: Can it be shown with equations and reasoning that a polar orbit around Venus would be the easiest attainable option ?

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    $\begingroup$ It may be possible for an asteroid with a diameter of less than 4 meters, but 40 meters is too heavy by a factor of 1000 or even more. But moving even the small 4 m asteroid will be enormously expensive. $\endgroup$
    – Uwe
    Jul 10, 2018 at 14:51
  • $\begingroup$ @Uwe Yes, it's good to see mankind's limitations. $\endgroup$
    – Cornelis
    Jul 10, 2018 at 16:22
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    $\begingroup$ This would be a much more interesting SE question if you asked "How could I estimate the delta-v necessary to move this asteroid into a high orbit around Venus?" since the answer would be in the form of a procedure, an equation or two, and some helpful insight. I think "Would it be possible to..." questions are low quality, as the answer always ends up being of the form "how much time or money do you have?" I could write a "yes" answer and a "no" answer and they could be equally true and equally false - because of course I would write them that way ;-) $\endgroup$
    – uhoh
    Jul 11, 2018 at 0:53
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    $\begingroup$ I think you are missing a critical thing: to become a moon, you need to brake around venus in order not to escape again. Earth sometime capture some objects, but that's mainly thanks to our moon gravity assist; and the objects endup ejected after a few orbits. Venus has no moon. $\endgroup$
    – Antzi
    Jul 11, 2018 at 5:56
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    $\begingroup$ Gawd, I thought it's actually co-orbital with Venus, following it while librating around L4, in a nearly the same orbit. In that case the work would be doable, push it out or the shallow L4 force well at such a point it doesn't crash into Venus, then capture into Venus gravity well. A couple m/s one way, a couple another. But it's Earth and Mercury crosser again, eccentricity 0.61 (Venus: 0.006) and doesn't pass anywhere near Venus. I don't even get how that orbit could be classified as tadpole. $\endgroup$
    – SF.
    Jul 11, 2018 at 8:32

1 Answer 1

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$$v_{in} = v_{out}$$

That's the rule when entering a single body system. You are going to leave with exactly as much velocity as you entered with. There are two notable exceptions:

  • More than one body in the system. The flybys of both bodies can very well change how much velocity you leave with. Earth-Moon is such a system. Venus is not.

  • Very small relative velocities. In that case, extended time is spent near the borders of a system, so $v_{in} = v_{out}$ is not strictly conserved. 2013 ND15 has a high velocity relative to Venus.

This conserved quantity means the asteroid must be slowed down after encountering the Venus system, in order to be captured as a moon.

Regardless of how you go about slowing it down however, the hyperbolic orbit it enters the system with can be picked in advance at a $\approx 0 \Delta v$ cost.

A tiny nudge months or years before the encounter has butterfly-effect-like impact on the periapsis and inclination of the flyby, while quantities like the semi-major axis and angle between apsis line and angle of entry is are preserved.

The important thing here is the "inclination" part. An arbitrary inclination can be picked at close to zero cost. Thus, a polar orbit is as easy (or difficult) as any other inclination.

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  • $\begingroup$ But when the inclination of the asteroid's orbit with that of Venus around the Sun would be small, when the asteroid would gradually approach Venus, wouldn't that make the inclination smaller and smaller, with a good chance of hitting each other ? $\endgroup$
    – Cornelis
    Oct 21, 2020 at 17:52
  • $\begingroup$ Inclination is not a function of distance. $\endgroup$
    – SE - stop firing the good guys
    Oct 21, 2020 at 17:54
  • $\begingroup$ I mean, the asteroid would be attracted to Venus, from above or from below, thus lowering the inclination ? $\endgroup$
    – Cornelis
    Oct 21, 2020 at 17:57
  • $\begingroup$ Yes, the heliocentric inclination can be reduced through Venus flybys. The Venus-centric inclination can be arbitrarily selected. $\endgroup$
    – SE - stop firing the good guys
    Oct 21, 2020 at 17:58
  • $\begingroup$ But when the tadpole approachhes Venus, at a certain moment it stops and reverses away. Whatever the forces are that causes this, wouldn't they be hard to overcome ? $\endgroup$
    – Cornelis
    Oct 21, 2020 at 18:11

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