Phys.org;s Dynamic orbital slingshot: A cool idea to catch up with an interstellar visitor references the MIT press release To catch an interstellar visitor, use a solar-powered space slingshot and both discuss NASA's Dynamic Orbital Slingshot for Rendezvous with Interstellar Objects found in the April 8, 2020 announcement NIAC 2020 Phase I, Phase II and Phase III Selections.

The Dynamic Orbital Slingshot link shows the image below, which from an orbital-mechanical perspective is much better than the one shown in the Phys.org link!

Question: If an object approached the Sun from far away on a $C_3=0$ (zero energy, parabolic eccentricity=1) trajectory for example, and a statite parked at radius $R$ wanted to pass by it at close range, how soon would it have to "release" and fall in a straight line to intercept it as a function of the object's perihelion. Would the distance that the object was detected $R_D$ have to be much farther than $R$ from the Sun for this to work?

If instead it used its solar sail throughout the trajectory rather than just "going ballistic", would the required detection distance be smaller? Could it be smaller than $R$ in this case?

A static satellite would be "parked" far from the Sun by balancing the attractive gravitational force with a large, very low mass solar sail, roughly 650 square meters per kilogram at any distance since both forces scale as $1/r^2$.

$$\frac{GM \ c}{2 \times 1 \text{AU}^2 1361 \text{W/m}^2} \approx 650 \ \text{m}^2/ \text{kg}$$

where the factor of 2 comes from perfect reflection. See also Statites - Are they possible in anything but theory?

Rendering of the Dynamic Orbital Slingshot concept. Credits: Richard Linares and NASA

Rendering of the Dynamic Orbital Slingshot concept. Credits: Richard Linares and NASA

  • $\begingroup$ A good source for an answer might be the NIAC proposal itself, but I don' t know if those are publicly available after the award is made. $\endgroup$
    – uhoh
    Commented May 12, 2020 at 2:54
  • 3
    $\begingroup$ Are we assuming the statite just happens to be in the visitor's orbital plane? $\endgroup$ Commented May 12, 2020 at 3:31
  • $\begingroup$ @RussellBorogove yes that's an excellent simplifying assumption, thank you! Of course in the real world we can't and would have to tile a sphere of statites around the Sun. $\endgroup$
    – uhoh
    Commented May 12, 2020 at 3:33
  • $\begingroup$ "Once an ISO is detected, the system can deliver a CubSat on either a flyby trajectory or on a rendezvous trajectory (with propulsion) ". So the statite itself will not have to be released ? $\endgroup$
    – Cornelis
    Commented May 12, 2020 at 8:40
  • $\begingroup$ @Cornelisinspace That's an interesting point! I think the NIAC grant is seed money and they will analyze several different situations and options. Dropping a cubesat with some kind of on-board propulsion certainly sounds like one option and perhaps cheaper than sacrificing the whole statite. $\endgroup$
    – uhoh
    Commented May 12, 2020 at 9:08

1 Answer 1



The formula above is for the free-fall time to a point source of gravity, with $\mu$ being the Standard gravitational parameter for the Sun.

So this could be applied to the trajectory of a "released" statite.

Filling in the numbers it appears that for $R$ being the radius of the orbit of Jupiter the minimum lead time would be more than 2 years !


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