There are a couple discovered asteroids that orbit around L4 or L5 Sun-Earth Lagrange points



From what I read online, reaching these asteroids require a delta-V higher than the delta-V required for a moon landing, due to their orbit.

Is it feasible to stabilize those asteroids in their respective L4 and L5 Lagrange point, as a way to reduce the delta-V required for future missions?




another post from here!

The potential deltaV required to get to these objects could be substantially lower relative to other nearby objects of interest. Currently, the only known Earth Trojan asteroid, 2010 TK7, has such a large inclination (20.9°) that the delta-V required (9.4 km/s) would make it challenging to visit [1]. Investigating Trojan Asteroids at the L4/L5 Sun-Earth Lagrange Points.

  • $\begingroup$ Welcome to Space! "From what I read online..." Can you add a link to this? There may be some information there that needs some explaining within the answer. Any way that you can distinguish this from the generic "is what I read somewhere on the internet really true?" would be great. Thanks! $\endgroup$
    – uhoh
    Mar 1, 2019 at 4:41
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    $\begingroup$ Hi! thanks for the information! i've added some data in the main question $\endgroup$ Mar 1, 2019 at 7:28
  • $\begingroup$ Ah! thanks for the additional text! I found an error in my calculation so the delta-v's for plane changes match nicely now. I should remember to always show my math. I've added a link to the original paper in Nature where the 9.4 km/s comes from. $\endgroup$
    – uhoh
    Mar 1, 2019 at 8:15

1 Answer 1


tl;dr: Inclination change. Yikes! You would have to reduce their inclinations. It's better to consider a longer transit and tilt your own inclination with an Earth fly-by each time you go, than to change the inclination of the asteroids themselves.

The OP"s link in turn links to the paper in Nature Earth’s Trojan asteroid and a non-paywalled copy can be found at one of the authors' site http://www.astro.uwo.ca/~wiegert/papers/2011Nature.pdf

Earth Trojan asteroids have been proposed as natural candidates for spacecraft rendezvous missions. However, the large inclination of 2010 TK7 results in a Δv of 9.4 km/s being required, whereas other near-Earth asteroids have values of Δv less than 4 km/s. The reported absolute magnitude, 20.7 mag, puts the diameter of 2010 TK7 at 300 m with an assumed albedo of 0.1 (ref. 22), which makes it relatively large among the near-Earth asteroid population. No spectral or colour information is as yet available to determine whether the asteroid is in any other way unusual.

The Asteroids

Per Wikipedia:

(419624) 2010 SO16:

2010 SO16 has a horseshoe orbit that allows it to stably share Earth's orbital neighborhood without colliding with it. It is one of a handful of known asteroids with an Earth-following orbit, a group that includes 3753 Cruithne, and the only known asteroid in an horseshoe orbit with Earth. It is, however, neither an Aten asteroid nor an Apollo asteroid because the semi-major axis of its orbit is neither less than nor greater than 1 AU, but oscillates between approximately 0.996 and 1.004 AU, with a period of about 350 years.[5] In its ~350 yr horseshoe cycle, it never approaches Earth more closely than about 0.15 AU, alternately trailing and leading.

2010 TK7:

2010 TK7 is a sub-kilometer near-Earth asteroid and the first Earth trojan discovered; it precedes Earth in its orbit around the Sun [...] 2010 TK7 has a diameter of about 300 meters (1,000 ft).[4] Its path oscillates about the Sun–Earth L4 Lagrangian point (60 degrees ahead of Earth), shuttling between its closest approach to Earth and its closest approach to the L3 point (180 degrees from Earth).

So both asteroids are in heliocentric orbits with semi-major axis very close to Earth's 1 AU. This allows Earths' relatively weak gravitational attraction (relative to the Sun's) at order 1 AU to "lock" them into one kind of resonant behavior or the other.

2010 SO16 is associated with Lagrange point L3 but wanders so far behind and ahead of it that the orbit is called "horseshoe", and TK7 is associated with L4, but also wanders so far behind and ahead that it alternately approaches L3 and Earth. Either way, when you look at their orbits from above in an non-rotating frame, both are in inclined somewhat elliptical orbits with $a\approx \text{1 AU}$.

To go from LEO to escape from Earth orbit and start "walking" along a ~1 AU circle to approach the asteroids' longitudes, you need about 2.5 + 0.7 = 3.2 km/s of delta-v, according to the plot below.

The Problem

Today's osculating orbital elements relative to the Sun, from JPL's Horizons

               e         a (km)        i(deg)    delta-v (Δi, 1AU circ)
Earth       0.0167     1,495,974        0.003     
2010 SO16   0.0754     1,500,671       14.518      6.75
2010 TK7    0.1905     1,495,031       20.896      9.67

I'm assuming you can "walk slowly" around 1 AU and creep up on the longitude of your target asteroid with little delta-v if you have a lot of time. There is some significant eccentricity, and that takes significantly ore delta-v to acheive, However these astroids have a hefty inclination, and that's a deal breaker!

$$\Delta v_i = 2 v_o \sin \frac{\Delta i}{2} $$ Source

$$v_o = \sqrt{\frac{GM_{Sun}}{a}} \approx \text{29.8 km/s}$$

At 1 AU and still in your circular orbit escaped from Earth, you are moving at about 29.7 km/s. Without a trick like a flyby over one of Earth's poles a few years later, there's no simple way to make an inclination change besides burning a lot of propellant to deflect that 51.5 km/s velocity vector by 15 to 20 degrees. The additional propulsive inclination maneuvers would require 6.8 to 9.7 km/s of delta-v.

The Moon

To go from LEO to lunar orbit and then to a soft lunar landing, you need about *4.1 + 0.7 + 1.6 km/s** of delta-v, according to the plot below.

enter image description here Source

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    $\begingroup$ Thanks for the answer! My question was actually that. Supposing i have to send a lot of deliveries from/to those asteroids, and since they require a lot of delta-v to reach. How much delta-v is required to stabilize the asteroid's orbits? (probably the answer will be "too much to even consider"). That way we could make a couple "stabilization missions" and benefit from the reduced inclination change for all the other subsequent missions (example: Mining mission or space colony missions) $\endgroup$ Mar 1, 2019 at 7:43
  • $\begingroup$ @LeonardoCiferri If you have to use rocket propellant for delta-v and you only plan on transporting a few percent of the asteroids' mass, then it's always easier to use that propellant to change the inclination of your spacecraft, rather than the inclination of the asteroid. But like I say above, if you are not in a rush, there may be a way to include an Earth flyby in transit and get some amount of plane-changing for a much lower cost in propulsive delta-v. $\endgroup$
    – uhoh
    Mar 1, 2019 at 7:51
  • $\begingroup$ "2010 SO16 is associated with Lagrange point L3". Not really. L3 is unstable. Horseshoe orbiters are in effect "alternating trojans" that switch between L4 and L5, with L3 as a transit point. $\endgroup$ Mar 17, 2019 at 9:34
  • $\begingroup$ @OscarLanzi I don't think you can call it a trojan either if it doesn't even make it around L4 or L5 once before wandering off. Anyway, I've opened up some space to get to the bottom of this: Do horseshoe orbits have anything to do with Lagrange points? Do words fail us here? $\endgroup$
    – uhoh
    Mar 17, 2019 at 11:22
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    $\begingroup$ Hence "alternating Trojan" in quotes. One could equally say a Trojan is what would be a horseshoe orbital except it has not enough energy to make it to L3. $\endgroup$ Mar 17, 2019 at 12:05

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