Inspired by this answer on How to intuitively explain that reaching geostationary orbits consume more Delta V than escape orbits?, which states

Move to the edge of the Hill sphere, sit at Earth-Sun L1 lagrangian point and your orbital speed around Earth drops to 0.

I started wondering how the delta-v requirements for leaving Earth orbit change depending on where apogee is positioned (well, as long as you have an apogee, that is, since we're breaking orbit). Intuitively, it makes sense that an apogee right at L1 would give you the lowest delta-v requirement while an apogee through L2 would require more.

How much does the delta-v budget need to increase in order to break orbit from places other than Sun-Earth L1?

  • $\begingroup$ I made a slight change to the wording. You'd like to leave Earth orbit. "Earth's orbit" could be interpreted as Earth's orbit around something else (Sun, Earth-Moon Barycenter) and since your very interesting question is about the subtleties of multi-body gravity fields, it's best to watch the details. $\endgroup$
    – uhoh
    May 25, 2018 at 6:39
  • 1
    $\begingroup$ @uhoh Thanks for the wording edit; that's a good point. $\endgroup$
    – Brian
    May 29, 2018 at 15:18

1 Answer 1


Very, very little.

The two extremes of the equipotential plane, where pull of the Sun and of Earth are equal - the actual Hill sphere ("official" Hill sphere is an approximation, a perfect sphere with Earth in the middle) are L1 and L2 respectively. Using Lagrange Point Finder with semi-major axis = 1AU, Primary = 1 M_Sun, Secondary = 1 M_Earth, I'm getting L1 at 1491551km and L2 at 1501531km from Earth. That's 0.66% difference. That 0.66% would be your upper cap on delta-V had delta-V scaled up with orbit's major axis length linearly. But it doesn't! As explained in the linked answer - gravity drops with square of distance, and kinetic energy grows with square of departure burn velocity. Result: 4th degree exponent for the highest values permissible with this set (the very brink of escape).

Let's take a 1kg cubesat in LEO and try to send it on escape trajectory. It moves at 8km/s, so it has 32 megajoules of kinetic energy. To escape, it needs that velocity multiplied by $\sqrt2$ - so the kinetic energy will double ($E_k = {1 \over 2} m v^2$).

The difference in energy needed to depart is a result of 0.237 millinewtons difference in gravity force exerted by the Sun at these two points at the cubesat. (obtained using WolframAlpha gravity equation calculator plugging distances from the Sun to L1 and L2 from the prior calculator, mass of Sun and 1kg, then subtracting the two resultant forces.)

Energy = Work = Force x distance.

Approximating this force as a constant over the distance, and taking the difference between the two Lagrangian point distances - 1501531km - 1491551km = 9980km, we obtain the extra work needed to move the cubesat to the "higher energetically" point. 9980000 meters * 0.000237 Newtons = 2.36 kilojoules extra.

So, our baseline escape energy is 64000 kJ (2x32 MJ). For 1kg, that's 11313.71 m/s. Let's add the 2.36 kJ: 64002.36 kJ. The new velocity is 11313.91 m/s

Yay, the needed delta-V is about 20cm/s !


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