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If you were going from a lower orbit to a higher orbit such as a Earth-Mars transfer, you would use the periapsis of the transfer orbit in the vis-viva equation to calculate the delta V. But if you were going from Mars back to Earth, would you use the apoapsis to calculate the Delta V because you are trying to slow the the spacecraft down when going around the sun?

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    $\begingroup$ Are you assuming that at the destination, the transferring spacecraft will aerobrake, flyby, or impact, instead of using its engines to brake into orbit or land? $\endgroup$
    – notovny
    Oct 3 at 22:51
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    $\begingroup$ I would just like to know the velocity to get into the transfer orbit. Don't need. to know the circularisation DeltaV. $\endgroup$
    – Jack Pryde
    Oct 4 at 8:58
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    $\begingroup$ @JackPryde Going from Mars to Earth, you start at the aphelion of the Hohmann ellipse & finish at its perihelion. The Wikipedia article explains the delta V calculations, and how they're derived from the vis viva equation. But if there's something specific you need help with, please edit that into your question. $\endgroup$
    – PM 2Ring
    Oct 4 at 21:33
  • $\begingroup$ Also the main point I want to know is if there are any differences in calculations between going away from and going towards the sun in a transfer. what replaces 2/r in the vis-viva equation because the burn is starting from apoapsis not periapsis? $\endgroup$
    – Jack Pryde
    Oct 4 at 22:33
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A Hohmann transfer requires two burns.

You already appear to know how to calculate the burns going out. For coming in do exactly the same calculations, just reverse the order and signs on the burns.

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  • $\begingroup$ How would I reverse the order of the burns? Is the equation the same or is it different? As I said in a previous comment, I only want to know the DeltaV required to get into the transfer orbit and don't need to know about the second circularisation burn. $\endgroup$
    – Jack Pryde
    Oct 4 at 20:30
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    $\begingroup$ @JackPryde Figure out the burns to move from the lower orbit to the higher orbit. To move from the higher orbit to the lower orbit you do the exact reverse. Point the engine the other way, do the second burn first. $\endgroup$ Oct 5 at 3:14
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    $\begingroup$ @JackPryde Newtonian mechanics in general works the same going forwards and backwards, this comes in handy from time to time. However, it does not work when friction or other irreversible processes are involved (e.g. atmospheric reentry) $\endgroup$
    – uhoh
    Oct 5 at 3:31
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A burn is required to make up a difference in velocity.

In the case you have already figured out: The difference in velocity is between the velocity of the lower circular orbit, and the periapsis velocity of the transfer orbit (which you get from vis-viva).

In this case: The difference in velocity is between the higher circular orbit and the apoapsis velocity of the transfer orbit (which you also get form vis-viva).


Quick warning: what you are currently calculating is a Hohmann transfer betweeen two orbits at Earth-distance and Mars-distance from the Sun. That is, escaping the gravity of Earth and Mars are not accounted for! For proper interplanetary transfers, the transfer orbit insertion burn and planetary escape burns are combined into a single manoeuvre.

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I'll adapt ai-solutions.com/_freeflyeruniversityguide/hohmann_transfer.htm:

I'll go from 20,000 km to 7,000 km

First find the velocity of the starting orbit: $$\begin{align} v_{starting} &= \sqrt{\mu \left({2\over r}-{1 \over a}\right)}\\ &= \sqrt{\mu \left({2\over 20000km}-{1 \over 20000km}\right)}\\ &=4.464 km/s\\ \end{align} $$ (Unsurprisingly, this is the velocity of the end orbit in the origninal) Next find the semi major axis of the transfer orbit: $$ \begin{align} a &= {a_{from} + a_{to}\over 2}\\ &={a_{starting} + a_{transfer}\over2}\\ &={7000km+20000km\over2}\\ &=13,500 km\\ \end{align} $$ (Unsurprisingly, the semi major axis of the transfer orbit is the same as the original example) Now, we can find the velocity at apoapsis (because we're going high to low) of this transfer orbit. In this case r = 20,000 km, and a = 13,500 km. We plug these into the Vis-Viva equation to get: $$\begin{align} v_{transfer\_apo}&=\sqrt{\mu \left({2\over 20000km}-{1 \over 13500km}\right)}\\ &=3.215 km/s \end{align}$$ (NB: unsurprisingly, this is the same value for the transfer orbit apoapsis as calculated in the original version)

Then, we can calculate the $\Delta v$ of the first maneuver: $$\begin{align}\\ \Delta v_1 &= v_{to} - v_{from}\\ &= v_{transfer\_apo} - v_{starting}\\ &= 3.2155 km/s - 4.464 km/s\\ &= -1.639 km/s\\ \end{align}$$ (As others have indicated, this is the same as $\Delta v_2$ from the original example, but in the opposite direction)

This first burn will put our Spacecraft into its transfer orbit. Next, we need to calculate the speed at the transfer orbit's periapsis (because we're going from high to low. For this calculation, r = 7,000 km, and a = 13,500 km. We plug these into the Vis-Viva equation to get: $$\begin{align} v_{transfer\_peri}&=\sqrt{\mu \left({2\over 7000km}-{1 \over 13500km}\right)}\\ &= 9.185 \text{km/s} \end{align}$$ (unsurprisingly this is the same $v_{transfer\_peri}$ as the original example)

Now, we must calculate the velocity of the target orbit. For the variables, r = 7,000 km, and a = 7,000 km. $$\begin{align} v_{ending} &= \sqrt{\mu \left({2\over r}-{1 \over a}\right)}\\ &= \sqrt{\mu \left({2\over 7000km}-{1 \over 7000km}\right)}\\ &= 7.546 km/s\\ \end{align} $$ (unsurprisingly, this is the same as the parking orbit in the original example)

Finally, we need to work out the second burn: $$\begin{align}\\ \Delta v_2 &= v_{to} - v_{from}\\ &= v_{ending} - v_{transfer\_peri} \\ &= 7.546 km/s - 9.185 km/s \\ &= -1.639 km/s \\ \end{align}$$ (As others have indicated, this is the same as $\Delta v_1$ from the original example, but in the opposite direction)

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I believe I know what I have done wrong. I have used a different equation given on this website, ai-solutions.com/_freeflyeruniversityguide/hohmann_transfer.htm, which is simpler than the one on wikipedia but only works when going up, e.g going to a geosynchronous orbit from a parking orbit. This equation I think doesn't work in reverse, going from a higher orbit to a lower orbit. The wikipedia version allows for the swapping of r1 and r2 to change between going to higher or lower orbits.

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    $\begingroup$ So long as you don't touch an atmosphere it's reversible. $\endgroup$ Oct 5 at 16:36
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    $\begingroup$ The equations on wikipedia will give a negative value for delta-V if the destination radius is smaller than the source radius. This is something to watch out for if you're chaining maneuvers together algebraically/in code without looking at the results from each maneuver. I'd just take the absolute value of the equation and be done with it. $\endgroup$
    – Ingolifs
    Oct 6 at 1:42

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