This answer to the question "How to calculate delta-v required for a planet-to-planet Hohmann transfer?" gives among other things the 2 following equations:

$v_{inject}=\sqrt{\left(\sqrt{\frac{2\mu_S r_2}{r_1(r_1+r_2)}}- \sqrt{\mu_S\over r_1}\right)^2+\frac{2\mu_1}{a_1}}-\sqrt{\mu_1\over a_1}$

$v_{insert}=\sqrt{\left(\sqrt{\frac{2\mu_S r_1}{r_2(r_1+r_2)}}- \sqrt{\mu_S\over r_2}\right)^2+\frac{2\mu_2}{a_2}}-\sqrt{\mu_2\over a_2}$

I filled in the following numbers:

$\mu_S$ = 1.3271244e20 (GM of the Sun)
$\mu_1$ = 3.986e14 (GM of the Earth)
$\mu_2$ = 2.2032e13 (GM of Mercury)
$r_1$ = 149.6e9 (Earth's radius of its orbit around the Sun)
$r_2$ = 46e9 (Mercury's perihelion) or $r_2$ = 69.8e9 (Mercury's aphelion)
$a_1$ = 6678e3 (orbit altitude 300 km above Earth)
$a_2$ = 2600e3 (orbit altitude 160 km above Mercury)

Mercury at perihelion: $v_{inject}$ = 6659.7 m/s, $v_{insert}$ = 10457.4 m/s,Total $\Delta V$ = 17.1 km/s.
Mercury at aphelion : $v_{inject}$ = 4751.8 m/s, $v_{insert}$ = 5483.8 m/s, Total $\Delta V$ = 10.2 km/s.

The answer from @Notovny correctly states that instead of the velocity for a circular orbit at perihelion and aphelion, you need to have the velocities for the elliptic orbit of Mercury at those points !
After applying the correct velocities (from this factsheet) the results become:
Mercury at perihelion: $v_{inject}$ = 6659.7 m/s, $v_{insert}$ = 11653.6 m/s, Total $\Delta V$ = 18.3 km/s.
Mercury at aphelion: $v_{inject}$ = 4751.8 m/s, $v_{insert}$ = 4800.0 m/s, Total $\Delta V$ = 9.5 km/s.

Edit 2:
A miscalculation was made after applying the correct velocities, so the correct results become:
Mercury at perihelion: $v_{inject}$ = 6659.7 m/s, $v_{insert}$ = 5602.0 m/s, Total $\Delta V$ = 12.3 km/s.
Mercury at aphelion: $v_{inject}$ = 4751.8 m/s, $v_{insert}$ = 9832.5 m/s, Total $\Delta V$ = 14.6 km/s.

That's exactly the same result as the one in @BrendanLuke15's answer !

However, this answer to this question gives very different values:
Delta-v for a perihelion transfer: 4.37 km/s
Delta-v for an aphelion tranfer: 8.58 km/s

  • 1
    $\begingroup$ I think the formulation of the equations is specific to circular orbits, you would have to re-derive a more general elliptical case $\endgroup$ Jan 30, 2022 at 18:53

2 Answers 2


I think I have it sorted: this answer giving the lower values is "correct". The key part you have missed is emphasized (by me) here:

That means a delta-v cost of 4.37km/s to get captured into a Mercury orbit from a perihelion transfer, and a 8.58km/s for an aphelion transfer.

This is code for a fictitious $C3=0$ orbit. This is what the Ames Trajectory Browser gives you for arrival $\Delta V$'s. It is useful to use this for analysis when the final orbit around the destination planet (Mercury) is not well defined (which is the case for that question). It is fictitious because parabolic orbits don't really exist in the real world, it is merely signalling the transition from a $C3>0$ escape, hyperbolic orbit and a $C3<0$ captured, elliptical orbit. It makes sense that the $\Delta V$ values are lower here because the fictitious $C3=0$ orbit is a higher energy orbit than the proposed low circular one.

It also should be noted that those "lower values" are only for the capture at Mercury, not including departure from Earth (total $\Delta V$).

As for your results, the $\Delta V_{injection}$ values are correct, but the $\Delta V_{insertion}$ values are incorrect. I caution reliance on the large combined equations. It is far easier to sanity check your values when doing it in a piecewise like fashion using vis-viva.

The heliocentric transfer orbit must be correct (how else would you get the correct departure/injection values?). The perihelion transfer values are given first for all:

$V(r_2)=\sqrt{\mu_{S}(\frac{2}{r_2}-\frac{1}{a_{trans}})}$, $a_{trans}=\frac{r_1+r_2}{2} \to V(r_2)=66.4$ or $50.9$ $km/s$

The $v_{\infty}$ values needed for calculating the Mercury insertion are found by subtracting the orbital velocity of Mercury (about the Sun) from the transfer orbit perihelion velocity ($V(r_2)$). In the Hohmann transfer approximation, these two velocities are parallel (hence the simple subtraction):

$v_{\infty}=V(r_2) - V_{M} \to v_{\infty}=(66.4 - 58.98)$ or $(50.9 - 38.86) \to v_{\infty}=7.46$ or $12.1$ $km/s$

Note these values are supported in the other question.

From the Characteristic Energy Wikipedia we can find the semi-major axis ($a$):

$C3=v_{\infty}^2=-\frac{\mu}{a} \to -\frac{1}{a}=-\frac{v_{\infty}^2}{\mu}$

We then use those semi-major axes in the vis-viva equation to find the speed at periapsis in the Mercury hyperbolic encounter (since it is most efficient to burn at periapsis when inserting into orbit at Mercury):

$V=\sqrt{\mu_{2}(\frac{2}{a_2}-\frac{v_{\infty}^2}{\mu})} \to V=8.52$ or $12.7$ $km/s$

That is the speed that the spacecraft is going at Mercury close approach. The desired speed at this point is the circular orbit velocity (at radius $a_2$: $2.91$ $km/s$). Thus the $\Delta V$ for insertion/arrival is the difference between these two values:

$\Delta V_{insert} = 5.6$ or $9.8$ $km/s \to \Delta V_{total} = 12.3$ or $14.6$ $km/s$

  • 1
    $\begingroup$ Thank you for the extensive answer. It appeared that I made a miscalculation in the first edit, so I made another one. The new results are almost exactly the same as yours ! That gives some confidence again on large combined equations, but indeed, a miscalculation is easy made. $\endgroup$
    – Cornelis
    Jan 31, 2022 at 16:47
  • $\begingroup$ Are you sure the entry mass for the Venera 9 is the same as its landing mass ? Or is Wikipedia wrong ? $\endgroup$
    – Cornelis
    Feb 13, 2022 at 16:02

Your $v_{inject}$ and $v_{insert}$ equations assume that both the departure and destination worlds are in coplanar, circular orbits.

This is manifestly not the case with Mercury.

Mercury is in an elliptical orbit with an orbital eccentricity of $0.21$, and as a result, is moving significantly faster than the circular orbit velocity for its distance from the Sun at perihelion, and significantly slower than the circular orbit velocity for its distance from the Sun at aphelion. And because it's closer to the Sun, even the circular orbit velocities involved are fairly high.

Any Hohmann transfer that takes the spacecraft inward results in the spacecraft moving at a higher speed relative to the body being orbited. In the case of an in-ecliptic-plane-orbiting Mercury, the difference in speeds on spacecraft arrival is smaller for the perihelion transfer than than for the aphelion transfer, which results in the higher capture delta-v for the latter.

As such, while the provided equations are reasonably good fits for transfers from Earth to most of the planets (which are generally in more circular, lower inclination, and relatively slower orbits around the Sun), they don't work for Mercury, even neglecting the 7° orbital inclination an actual transfer would also have to deal with.

  • $\begingroup$ Mercury's orbital velocity appears in the $v_{insert}$ equation as $\sqrt{\mu_S\over r_2}$, right ? So for the 2 different distances from the Sun at perihelion and aphelion you have 2 different velocities. $\endgroup$
    – Cornelis
    Jan 30, 2022 at 14:22
  • 3
    $\begingroup$ @Cornelis No, that's the circular orbit velocity for the distance $r_2$. It's not going to be the orbital velocity on an elliptical orbit, which you would need to use the vis-viva equation for. $\endgroup$
    – notovny
    Jan 30, 2022 at 14:26
  • $\begingroup$ You're right, I've made an edit with the correct velocities at perihelion and aphelion, and it appears the results are equal to the ones in @BrendanLuke15's answer. $\endgroup$
    – Cornelis
    Jan 31, 2022 at 16:36

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