In 1984 JPL engineer Paul Penzo described a ZRVTO (Zero Relative Velocity Transfer Orbit) between an elevator anchored on Deimos and an elevator anchored on Phobos.

Here is a Penzo’s illustration:

enter image description here

Release a payload from the top of Phobos’ 940 kilometer elevator and the payload will follow an elliptical path and arrive at the bottom of Deimos 2960 kilometer elevator. On arrival the relative velocity between payload and elevator terminus is zero.

It works the other direction as well. Drop a payload from the bottom of Deimos’ 2960 kilometer tether and it will arrive at the top of Phobos’ 940 km elevator. Again the relative velocity will be zero at the instant of rendezvous.

With less than 4000 kilometers of tether, the two moons can exchange payloads using nearly zero reaction mass.

Given two tidelocked moons in circular, coplanar orbits, is there a general method for finding elevator lengths that allow a ZRVTO between the moons' elevators?


1 Answer 1


Yes. Given the radii of the upper and lower moons and knowing their angular velocities, there are simple expressions that give the periapsis and apoapsis radius of the Zero Relative Velocity Transfer Orbit.

The specific angular momentum, h, of an orbit is constant. At any point on the orbit the specific angular momentum $\overrightarrow{h}$ is the cross product between the position and velocity vectors:

$\overrightarrow{h} = \overrightarrow{v}$ X $\overrightarrow{r}$

And so

$\overrightarrow{h} = \overrightarrow{v}_{periapsis}$ X $\overrightarrow{v}_{periapsis}=\overrightarrow{v}_{apoapsis}$ X $\overrightarrow{v}_{apoapsis}$

Position and velocity vectors are at right angles to one another at periapsis and apoapsis. So a cross product gives us the same magnitudes with ordinary multiplication.

$h = v_{periapsis} * r_{periapsis} = v_{apoapsis} * r_{apoapsis}$

What’s the length of $r_{periapsis}$ and $r_{apoapsis}$?

And what’s $v_{periapsis}$ and $v_{apoapsis}$?

Speed of a point on a vertical elevator is $\omega * r$ where r is the distance from center of rotation and $\omega$ is angular velocity of the elevator. In this case the Phobos elevator has the angular velocity of Phobos and the Deimos elevator has the angular velocity of Deimos.


Recall that
$v_{periapsis} * r_{periapsis} = v_{apoapsis} * r_{apoapsis}$

$v_{periapsis}=\omega_{phobos}*r_{periapsis}$ and $v_{periapsis}=\omega_{deimos}*r_{apoapsis}$,
$\omega_{phobos} * r_{periapsis} * r_{periapsis}= \omega_{deimos}*r_{apoapsis}*r_{apoapsis}$
$\omega_{phobos} * r_{periapsis}^2= \omega_{deimos}*r_{apoapsis}^2$

$r_{periapsis}=(1-e)a$ and $r_{apoapsis}=(1+e)a$,
$\omega_{phobos} *((1-e)a)^2= \omega_{deimos}*((1+e)a)^2$
$\omega_{phobos} *(1-e)^2= \omega_{deimos}*(1+e)^2$
$((1-e)/(1+e))^2= \omega_{deimos}/\omega_{phobos}$
$(1-e)/(1+e)= \sqrt{\omega_{deimos}/\omega_{phobos}}$

And now we can find eccentricity of the ZRVTO given angular velocities of each moon:


Now we need to find the periapsis and apoapsis of the ZRVTO.

At peri and apoapsis, magnitude of specific angular momemtnum is $v*r$.
Since $v = \omega*r$ we can say $h=\omega*r^2$

Specific angular momentum is also twice the area swept by the orbit over an orbital period or

$h=2\pi ab/T$
$h=2\pi ab/(2\pi\sqrt{a^3/\mu})$

The angular momentum of the ZRVTO from the top of the Phobos tether is


so $a=r/(1-e)$.
Substituting for a…

$\omega_{phobos} = 2\pi/T_{phobos}$
Where $T_{phobos}$ is Phobos' orbital period.
where $r_{phobos}$ is the distance from Phobos to the center of Mars. So
$\omega_{phobos} = 2\pi/((2\pi\sqrt{r_{phobos}^3/\mu})$
The $2\pi$ in the numerator and denominator cancel out
$\omega_{phobos} = 1/\sqrt{r_{phobos}^3/\mu})$

Substituting this Value for $\omega_{phobos}$ into
we get
$r_{periapsis}^3=(1+e)\mu r_{phobos}^3/\mu$
The $\mu$ s cancel out

Which brings to the the periapsis radius in terms of the e we've already found and the orbital radius of Phobos.


In a similar fashion apoapsis of our Zero Relative Velocity Transfer orbit is (1+e)a and


I put these expressions into a spreadsheet ZRVTO_finder.xslx. A screenshot:

ZRVTO spreadsheet

Here's a pic drawn from the spreadsheet's numbers:

enter image description here

It is reassuring my numbers are close to Penzo's. I'm just an amateur but Penzo was a very competent aerospace engineer working for JPL.

There are many tidelocked moons in nearly circular, coplanar orbits. The above expressions can be generalized:




The upper and lower moon could also be artificial objects, man made anchors for vertical orbital tethers. If we know the moons' angular velocities and orbital radii, it is easy to figure the tether lengths to accommodate Zero Relative Velocity Transfer Orbits between the two.

  • $\begingroup$ h=ωphobos∗r2periapsis=(a(1−e2)μ)1/2 - this doesn't look correct. Also have you tried calculating real tether lengths with final formulas? $\endgroup$
    – ZuOverture
    Commented Nov 12, 2017 at 3:52
  • $\begingroup$ @ZuOverture Is it the $h=(a(1-e^2)\mu)^{1/2}$ that bothers you? I added in some additional steps that I hope are helpful. As for calculating real tether lengths, yes. I will add to my answer as I have time. For me doing the markup for these equations is time consuming and difficult. $\endgroup$
    – HopDavid
    Commented Nov 12, 2017 at 6:20
  • $\begingroup$ ω*r^2 = (μ / r^3) ^ 1/2 * r^2 = (μ * r) ^ 1/2 = (μ * (1 - e) * a) ^ 1/2. Note that there's no square at e. Also calculations for Phobos-Deimos pair gave e=3, which means trajectory is hyperbolic. And object on hyperbolic trajectory can't have zero radial velocity anywhere in planet's SOI, which should mean ZRVTO is impossible. $\endgroup$
    – ZuOverture
    Commented Nov 12, 2017 at 7:14
  • $\begingroup$ Ok, found mistake in my calc - usage of incorrect formula for ω(r). Question about e=3 still remains. $\endgroup$
    – ZuOverture
    Commented Nov 12, 2017 at 8:19
  • 1
    $\begingroup$ seems like I've just divided the contents of brackets in wrong order. I confirm the value of 0.33, my bad. Maybe one last comment. Tethers like those probably won't be straight. Due to difference in ω(r) they should bend. The formulas above should be correct for rigid towers though. $\endgroup$
    – ZuOverture
    Commented Nov 13, 2017 at 3:51

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