How would you calculate a Hohmann transfer going towards the sun?

Question

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?

Answer 1

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.

Answer 2

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).

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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.

Answer 3

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)

Answer 4

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.