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I wrote "Hohmann-like" of course because Mercury's orbit is rather eccentric and has an inclination of 7 ⁰.
From this factsheet I've learned that perihelion is at 46 million km and aphelion at about 70 million km from the Sun with velocities at those points of 59 km/sec. and 39 km/sec. respectively.
Because there's such a big difference in the delta-v's for transfer orbits to Venus and Mercury of 2.5 km/sec. and 7.5 km/sec. respectively, I suppose the difference for the perihelion and aphelion of Mercury's orbit could also be considerable !

An additional factor to determine the most economic transfer orbit could be the choice of the best possible launch window for the aphelion, if there are such windows in the first place.

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    $\begingroup$ Are you looking for the $\Delta V$ just to encounter (flyby) Mercury, or also capture at Mercury? $\endgroup$ Dec 15, 2021 at 17:53
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    $\begingroup$ Are you just requiring a flyby, or are you intending an orbit or landing? When I made a similar estimation for transferring from KSP's analogous Kerbin to Moho several years ago, Moho's slower speed at apoapsis ate up all the savings from the Hohmann transfer when braking into orbit. $\endgroup$
    – notovny
    Dec 15, 2021 at 18:18
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    $\begingroup$ @BrendanLuke15 No, not a flyby but rather getting at the same velocity as Mercury at close proximity. I think as to delta-v a descent to the surface would not make much difference being at perihelion or aphelion. $\endgroup$
    – Cornelis
    Dec 15, 2021 at 18:19
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    $\begingroup$ @notovny I would like to see your calculations in an answer why that slower speed would "ate up" the savings ! $\endgroup$
    – Cornelis
    Dec 15, 2021 at 18:30
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    $\begingroup$ @BrendanLuke15 But now that I've read Notovny's comment I think it would be nice to be able to compare the delta-v's of EDL and transfer trajectory.. $\endgroup$
    – Cornelis
    Dec 15, 2021 at 19:21

1 Answer 1

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All the relevant velocities can be obtained from the vis-viva equation. The transfer/capture costs can then be derived from the $v^2 = v_e^2 + v_{\infty}^2$. I'm going to ignore Mercury's 7 degree inclination.

So let's look at the numbers.

Aphelion velocity for Mercury perihelion transfer: 20.43km/s
Aphelion velocity for Mercury aphelion transfer: 23.76km/s

That's a significant difference, resulting in an Earth $v_{\infty}$ of 9.35km/s in the former case, and 6.02km/s in the latter. Translated to transfer delta-vs, that's a 3.43km/s cost on top of Earth escape for a perihelion transfer, and 1.54km/s for an aphelion transfer.

So a 1.89km/s delta-v saving for a flyby mission.

But to get into an orbit around Mercury, the situation is the other way around.

At Perihelion, the $v_{\infty}$ of the spacecraft approaching is 7.45km/s
At Aphelion, the $v_{\infty}$ of the spacecraft approaching is 12.1km/s

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. That's a 4.21km/s difference, much greater than the difference in the other direction at the Earth side. It gets worse as this manoeuvre must be performed by lower efficiency storable propellant, while the Earth escape can be done with a hydrogen oxygen upper stage.

So:

Aphelion transfers are better for flybys, perihelion transfers are better for captures

No mission has gone to Mercury directly though, opting for Venus flybys to lower the cost to a manageable level.

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  • $\begingroup$ Thank you for all those numbers, but could you add some equations for the calculations so that we can understand how you got the results ? $\endgroup$
    – Cornelis
    Dec 16, 2021 at 9:19
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    $\begingroup$ Vis-viva (linked), orbital Pythagoras, spelled out. There are no others. $\endgroup$ Dec 16, 2021 at 9:21
  • $\begingroup$ For instance, you "translated" the aphelion velocities into transfer delta-v's, how do you do that ? $\endgroup$
    – Cornelis
    Dec 16, 2021 at 9:28
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    $\begingroup$ That would be the equation in the answer. $\endgroup$ Dec 16, 2021 at 10:50
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    $\begingroup$ For more details about applying it, see here: space.stackexchange.com/questions/56066/… $\endgroup$ Dec 16, 2021 at 10:51

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