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I was looking into Parker Solar Probes trajectory and was wondering what would be the most optimal way to utilize a retrograde thrust at perihelion to reduce the number of needed gravity assists. I think it would be best to apply the speed reducing thrust at perihelion since, by the Oberth effect, the faster you are moving the more you can benefit from a thrust.

If PSP were to apply a thrust at its first perihelion it would require an impossible amount of fuel for it to get it to its target distance of 0.05 AU from the Sun. If it applied this thrust at its second perihelion since it is closer it would require slightly less fuel per mass to get it to its target distance and also since it is moving faster, for the same thrust we should need less fuel (still an impossible amount for the given mass though).

So say we are able to add enough fuel, keeping the mass the same, such that if we were to burn it all at PSP's 3rd perihelion then we wouldn't need to go back to Venus for another assist (PSP would achieve its target perihelion distance). I was wondering if it would require less fuel to instead of burning all the fuel we have at the 3rd assist to instead burn 1/3 of the fuel on the 1st, 1/3 on the 2nd, and 1/3 on the 3rd such that the velocities are increasing but less fuel is being burned.

This seems like a simple problem to me but I am unsure how to solve it. For example, are there any formulas where I could estimate the effect of a certain retrograde thrust on the resulting aphelion velocity and then the perihelion velocity of the following orbit and then perihelion distance?

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Retrograde thrust at periapsis doesn't lower the periapsis, it lowers the apoapsis. If you're trying to lower the periapsis, you need to apply thrust at apoapsis.

If you're trying to lower your periapsis to a point absurdly close to the Sun, the most efficient option that avoids gravity assists is a bi-elliptic transfer: raise your apoapsis as high as possible, then apply a very small retrograde thrust at apoapsis to lower your periapsis. The downside to this is travel time: a bi-elliptic transfer with the apoapsis around Neptune will require over a decade of flight before you can start observing the Sun.

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  • $\begingroup$ But if the apoapsis were lowered wouldn't this then lower the next periapsis? If not, I guess the same question would apply for a burn at aphelion, would it be more efficient to burn all the fuel at once at the 3rd aphelion or burn 1/3 of the fuel on the first, 1/3 on the second and 1/3 on the third? Also, in that case since the burn would be happening at the same place as the assist, I was wondering if it is more efficient to enter the assist faster or slower? Since for gravity assists, a slower velocity in is generally better and for the Oberth effect, a higher velocity would be better. $\endgroup$ Commented Aug 19, 2020 at 7:03
  • $\begingroup$ @AlexanderIvanov Nope, lowering your apoapsis does nothing to your periapsis. An extreme example I've seen out of Kerbal Space Program, I was running low on fuel returning to Kerbin from the outermost planet. I entered Kerbin's atmosphere with my booster still attached, burned all my fuel close to periapsis, then the fire destroyed my booster and used up most of my heat shield--I was left with a periapsis in the upper atmosphere and an apoapsis not much below the Hill Sphere. continued... $\endgroup$ Commented Aug 19, 2020 at 14:58
  • $\begingroup$ I then turned around and went round and round, pointy end first so I didn't pick up too much heat and to save the last of my heat shield. Every orbit my apoapsis dropped, my periapsis didn't move by more than a few km over the whole thing. On the final loop I turned around again and re-entered normally. $\endgroup$ Commented Aug 19, 2020 at 15:01
  • $\begingroup$ @Mark I was also wondering, why is it most efficient to have an aphelion as high as possible? I thought that because of the Oberth effect it is better to thrust when moving faster rather than slower. $\endgroup$ Commented Aug 21, 2020 at 4:52
  • $\begingroup$ @AlexanderIvanov, in a highly elliptical orbit, the velocity change needed to lower your periapsis is very small. This means that a short burn with a weak Oberth effect can be more fuel-efficient than a long burn with a strong Oberth effect, even after you factor in the cost of getting into the orbit in the first place. $\endgroup$
    – Mark
    Commented Aug 21, 2020 at 20:00
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As Mark already pointed out:

Retrograde thrust at periapsis doesn't lower the periapsis, it lowers the apoapsis. If you're trying to lower the periapsis, you need to apply thrust at apoapsis.

This will be very important: every maneuver you plan effects basically the opposite site of your orbit.. want to lower Pericenter? -> retrograd thrust at Apocenter. So starting from an elliptical orbit and you want to lower your pericenter you need to give thrust where you are the slowest, at your Apocenter.

But back to your question(s):

But if the apoapsis were lowered wouldn't this then lower the next periapsis?

NO

For example, are there any formulas where I could estimate the effect of a certain retrograde thrust on the resulting aphelion velocity and then the perihelion velocity of the following orbit and then perihelion distance?

Simple: YES, there is a set of quite easy Formulas for that, you will find them in Wikipedia Hohmann Transfer Orbit

I would recommend you to use the vis-viva-equation (not the solutions for the hohman transfer):

v= sqrt(n*((2/r)-(1/a))),

with a= (r_apo + r_peri)/2

what you need to do:

play through both scenarios... the equation is giving you the velocity you have at a defined altitude of your orbit, you want to change to a transfer orbit? calculate the transfer orbits velocity at the same altitude. the difference between both velocities is your needed velocity-change. so you can calculate how much fuel you need. after that you need a second manouvre calculated the same way, etc...

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