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This question already has an answer here:

If we launch an object on one side of the moon (the velocity being tangent to the surface of the moon), how big must the velocity be to let the object arrive at a position on the moon diametrically opposite to the launching site (in some device to catch the mass)? The mass is very, very small compared to the moon's mass.

The acceleration due to the moon's gravitation is, of course, dependent on the distance to the moon (inverse square law).

Because of this dependence on the distance of the acceleration, it's more difficult to compute a trajectory, especially this particular one that also involves the curvature of the moon. I guess integration is involved, but I don't see how. For sure the velocity has to be higher than the velocity (which is easy to calculate) to let the mass make a circle around the moon. If we didn't catch the mass on the other side, the trajectory would resemble something that looks like an ellipse. This trajectory doesn't consist of two (opposite) parabola's because the trajectory in the question isn't a parabola, again because the acceleration depends on the distance of the mass to the moon.

Is there someone who knows how to do the calculation?

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marked as duplicate by sammy gerbil, Jon Custer, descheleschilder, stafusa, John Rennie homework-and-exercises Dec 12 '17 at 9:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The minimum velocity will place the particle in circular orbit close to the surface of the Moon. With any other velocity (provided less than escape velocity) the trajectory is an ellipse, except where obstructed by the Moon. With a smaller velocity the particle will hit the Moon before reaching the other side. ... The general case is dealt with in the linked question above. $\endgroup$ – sammy gerbil Dec 12 '17 at 0:52
  • $\begingroup$ See also How to calculate a ballistic trajectory for a suborbital flight? $\endgroup$ – sammy gerbil Dec 12 '17 at 1:00
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The moon is not a easy example body because it's gravitational field has significant higher multipoles, so I will answer for an idealized spherically symmetric body.


Your first issue is to recognize that—unless you intend to power the craft somewhere along the trajectory—you are constructing an orbit, and as such it has the usual features:

  • It is elliptical (of semi-major axis $a$ and semi-minor axis $b$).
  • It has the center of the body at one focus.

You want to go from one size of the spherically shaped body to the antipodes which means that you want to connect two points on the diameter of a circle around the same center.

We expect a whole family of such curves. We need a reason to prefer one over all the others. More on that later.

Now, the line perpendicular to the major axis and through a focus of a conic section is called the latus rectum. The length of the semi-latus-rectum of an ellipse is $\ell = \frac{b^2}{a}$. In this case $\ell = r$ where $r$ is the radius of the moon or planet, leading to $ar = b^2$ for orbits meeting our needs.

We might want to select the orbit with the lowest energy demand.

The energy needed $E$ is the difference between the energy of the orbit $$ E(a) = -G\frac{Mm}{2a} \;,$$ and the potential energy of craft at rest on the surface $$ U_0 = -G\frac{Mm}{r} \;.$$ We get \begin{align} E &= E(a) - U_0 \\ &= GMm\left[ \frac{1}{r} - \frac{1}{2a} \right]\\ &= GMm\left[ \frac{a}{b^2} - \frac{1}{2a} \right] \;. \end{align} Minimizing with respect to $a$ we get $$ 0 = GMm\left[ \frac{1}{b^2} + \frac{1}{2a^2} \right] \;, $$ which is only satisfied in the limit that $a$ and $b$ both are allowed to increase without bound. This represents a demonstration of what StephenG says: our lowest energy-demand orbit is, oddly enough, an escape trajectory meaning infinite time will be required, and making this option unfeasible for most purposes.

But this is a minimum meaning that any bound curve won't meet the requirements.


Discussion: Our calculation actually assumed we wanted to come down opposite the launching point relative the fixed stars. On a rotating body a we can aim at some other point relative the fixed stars by rigging our time-of-flight so that our target geography is where we land when we get there. For locations on the equator a straight-up and straight down trajectory will actually do the job, assuming you get the timing right.

Further discussion: This oddity might be seen as a problem for the designers of ballistic missile systems except that the missile don't accelerate instantaneously as (also) assumed here, but instead have a non-trivial boost phase. The height and down-range travel attained while boosting means that antipode-targeting remains an option.

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  • $\begingroup$ This is a great answer! I do have fewer problems with visualizing conic sections than "visualizing" the math. But I understand what you wrote. I know that the orbits of a mass (in the ideal case) making its way around a much heavier mass are conic sections but I thought it was difficult to prove that because of the varying force (with distance) acting on the small mass. $\endgroup$ – descheleschilder Dec 12 '17 at 7:37
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    $\begingroup$ The proof is moderately lengthy (it's half a chapter in Marion and Thornton, though they do establish a number of facts that are more general than $1/r$ potentials along the way), and involves some less than obvious tricks, but it is not terribly difficult to follow along the path blazed by others. Or we can just take Kepler's analysis of Copernicus' data as definite. $\endgroup$ – dmckee Dec 12 '17 at 7:54
  • $\begingroup$ Just one more question. Why has a mass that is sent to infinity the lowest energy? Doesn't the mass has the highest velocity in this case? $\endgroup$ – descheleschilder Dec 12 '17 at 8:44
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    $\begingroup$ @descheleschilder It has the lowest energy out of tracks with the return geometry you requested. It means that the return geometry exists only as a limiting case. As Stephen says you can hit spots near the antipodes cheaper (probably a lot cheaper). $\endgroup$ – dmckee Dec 12 '17 at 17:33
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As described this is impossible.

All paths will be the usual conic orbits if we can treat the Moon as an ideal spherically symmetric lump (or this gets really, really complicated).

Clearly elliptical orbits are the only ones that "return", but if you start with a velocity tangential to the ground then the only viable orbits will simply be ellipses that get further away and only fall back when they return to the starting point. (Ignoring rotation and other gravitational influences).

To "hit" the opposite side of the moon you need to point your projectile "up" so that it's elliptical orbit has a closest approach "inside" the Moon and it will hit the surface (on the opposite side) before it reaches (of course).

However the limiting minimum velocity for such an orbit is going to be the one for a circle of the same radius as the Moon itself. In other words the limiting orbit is the circle, which will (sort of) hit every point on the surface, not just the one you want. All other orbits that only hit the two points will require a greater starting velocity but you can (in this ideal theoretical model) get as close as you like to that minimum.

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  • $\begingroup$ An elliptical orbit has two focus points (the moon lies not at the center of the ellipse). On which of the two focus points lays (the center) the moon? Or do the two focus points lie almost at the same point because of the big difference of mass? $\endgroup$ – descheleschilder Dec 12 '17 at 1:45
  • $\begingroup$ Unless you've a very massive projectile then the focus point for the orbit will be the center of the Moon. Of course the real Moon is not ideally spherically symmetric and that would perturb the orbit from an ideal ellipse as would the effect of reduced mass. $\endgroup$ – StephenG Dec 12 '17 at 1:50
  • $\begingroup$ @descheleschilder If (like me) you are weak on visualizing relationships between conic section you can examine the math in my answer. $\endgroup$ – dmckee Dec 12 '17 at 1:50

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