# Delta V to get to the Sun-Earth Lagrange Point 1?

For any of these starting positions, GEO, GTO, EM-L2 or EM-L4/5, which would require the least delta-V to get to Sun-Earth L1? How much delta-V would it require?

Would this chart help in identifying which delta-V would most likely fit the requirements stated?

$$\rm L_1$$ is a similarly circular orbit, and according to this source, the Earth-Sun $$\rm L_1$$ is $$\approx$$ 1.5million km from the Earth. Between circular orbits, the cheapest transfer is the Hohman transfer.

According to the Wiki page, the required $$\Delta v$$ for Hohman-transfer orbits is

$$\Delta v = \sqrt{\frac{\mu}{r_1}} \left( \sqrt{\frac{2 r_2}{r_1+r_2}} - 1 \right) + \sqrt{\frac{\mu}{r_2}} \left(1 - \sqrt{\frac{2 r_1}{r_1+r_2}}\right)$$

Where $$\mu$$ is the "gravitational parameter" ($$\rm GM$$, the mass of the central body multiplied with the gravitational constant), and $$r_1$$, $$r_2$$ are the radii.

Although this formula could be significantly simplified using that now $$r_1 \approx r_2$$, substituting the values ($$M=2\cdot 10^{30} \rm kg$$, $$r_1=1.5\cdot 10^{11} \rm m$$, $$r_2=1.515\cdot 10^{11} \rm m$$, $$G=6.67\cdot 10^{-11} \frac{\rm Nm^2}{\rm kg^2}$$), we can get the required $$\Delta v$$, which is $$\approx 148 \frac{\rm m}{\rm s}$$. This adds to the escape speed from the Earth ($$\approx 11.2 \frac{\rm km}{\rm s}$$) or from LEO ($$\approx 3.4 \frac{\rm km}{\rm s}$$).

• 1) In a 2 body system, going from a 1 A.U. to a 1.01 A.U. heliocentric orbit takes about .14 km/s. 2) You would not add the inaccurate 5.65 km/s to the speed needed for earth escape. Speed hyperbolic orbit = sqrt( Vescape^2 + Vinf^2 ) 3) The 2 body Hohmann equation isn't appropriate for a 3 body scenario. – HopDavid Apr 18 '19 at 16:14
• @peterh I checked your work. Your 2.515e11 denominator is wrong. 1.5e11+1.515e11=3.015e11 – HopDavid Apr 18 '19 at 19:53
• @HopDavid Thank you very much! I improved the answer. Sorry for the explosive mood, your fixes were very useful. – peterh Apr 19 '19 at 16:08
• A comment cleanup was requested, and I trimmed what I found reasonable. As far as I can see, @HopDavid's main concern was not addressed. Applying the Hohmann transfer equation here ignores that the Earth exists, and you can't just add the cost of maneuvering in Sun orbit together with the Earth escape cost. The first burn of the transfer is combined with the escape in a single burn, at a smaller cost. You can calculate this cost from the hyperbolic equation provided. – SE - stop firing the good guys Apr 19 '19 at 19:23
• @Hohmannfan Ok, thanks. However, the final result will be a roughly static probe around 1.5million km far away. The required potential energy for that is practically the same as in the infinity. The overwhelming majority of the required $\Delta v$ will be spent to reach this near-escape speed, independently if it is a surface or LEO start. Thus, although the hyperbolic equation could get a better result, the difference will be negligible, as I can see. – peterh Apr 19 '19 at 20:18