# How to calculate delta-v required for a planet-to-planet Hohmann transfer?

How do you calculate the delta-v required to do a Hohmann transfer from a circular orbit around one body to a circular orbit around another?

I'm assuming you'd need to know the masses of the two bodies, and the mass of whatever they're orbiting around, also the radii of the different orbits. Let's assume those are all known.

Here is an approximate but quite accurate solution. This assumes that the orbit about the departure body is aligned with the outgoing asymptote of the Hohmann orbit, which is the standard practice when putting a spacecraft in a parking orbit by a launch vehicle before its escape maneuver. This also assumes that the bodies' orbits about the Sun are circular and that they are coplanar. This is pretty close for the planets (as currently defined) about our Sun.

We will also assume instantaneous maneuvers, which is a good approximation for chemical rockets, and what is implied by the use of the Hohmann solution. The transfer would be quite different for low-thrust systems.

The $\Delta V$ can be determined by repeated application of this equation that simply says that the total energy is the sum of the kinetic energy and the potential energy:

$\mathcal{E}=\frac{v^2}{2}-\frac{\mu}{r}$

where $\mathcal{E}$ is the total energy per unit mass of the object or the "specific energy", $v$ is the velocity of the object at the current position, $\mu$ is the GM of the central body, i.e. Newton's gravitational constant times its mass, and $r$ is the current distance from the center of the central body.

The key is that the total energy of the object is a constant of motion over the orbit.

We will also use the fact that orbits are ellipses, and this equation, which determines that constant of motion from the apses of the orbit, i.e. the radii of the closest and farthest points in the orbit, $r_1$ and $r_2$:

$\mathcal{E}=-{\mu\over r_1+r_2}$

For this problem we define:

$\mu_S$ = GM of the Sun.
$\mu_1$ = GM of the departure body.
$\mu_2$ = GM of the arrival body.
$r_1$ = radius of departure body orbit about the Sun, assumed to be circular.
$r_2$ = radius of arrival body orbit about the Sun, assumed to be circular.
$a_1$ = circular orbit radius of the spacecraft about the departure body.
$a_2$ = circular orbit radius of the spacecraft about the arrival body.

The Hohmann transfer orbit has apses $r_1$ and $r_2$. So its specific energy is:

$\mathcal{E}_H=-{\mu_S\over r_1+r_2}$

We then compute the velocity of that orbit at $r_1$:

$\mathcal{E}_H=-{\mu_S\over r_1+r_2}=\frac{v_{H1}^2}{2}-\frac{\mu_S}{r_1}$

which gives:

$v_{H1}=\sqrt{\frac{2\mu_S r_2}{r_1(r_1+r_2)}}$

For the circular orbit of the departure body about the Sun, we have for its velocity about the Sun:

$-{\mu_S\over 2r_1}={v_1^2\over 2}-{\mu_S\over r_1}$

or:

$v_1=\sqrt{\mu_S\over r_1}$

The geometry dictates that the Hohmann transfer orbit velocity at periapsis is in the same direction as the departure body velocity, and they are at the same radius from the Sun. So the velocity change from the departure circular orbit to the Hohmann transfer orbit is just the difference (this will be squared later, so the sign doesn't matter):

$v_{T1}=v_{H1}-v_1=\sqrt{\frac{2\mu_S r_2}{r_1(r_1+r_2)}}-\sqrt{\mu_S\over r_1}$

That is the velocity change to get from the departure body orbit to the Hohmann transfer orbit, but without the departure body being there. To do the maneuver from a circular orbit about the body, we will use the patched conic approximation that we need an escape velocity from the body equal to that transfer velocity. Here is where the plane of the orbit about the body is important, since it needs to line up with the direction of the transfer velocity to minimize the $\Delta V$. In this case, that is the plane shared by the orbits of the two bodies.

The specific energy of an escaped object sufficiently far from the body (where the sufficiency is at the heart of the patched conic approximation), is the energy equation with the second term going to zero as the distance goes to infinity:

$\mathcal{E_{escape}}=\frac{v_{\infty}^2}{2}$

Now to get the change in velocity for that escape and injection to the Hohmann transfer, we take the difference between the velocity of that escape energy at orbital radius of the spacecraft about the departure body and the orbital velocity of the spacecraft:

$\frac{v_{T1}^2}{2}=\frac{v_{escape}^2}{2}-\frac{\mu_1}{a_1}$

$v_{escape}=\sqrt{v_{T1}^2+\frac{2\mu_1}{a_1}}$

$v_{orbit}=\sqrt{\mu_1\over a_1}$

$v_{inject}=v_{escape}-v_{orbit}=\sqrt{v_{T1}^2+\frac{2\mu_1}{a_1}}-\sqrt{\mu_1\over a_1}$

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

That is the actual $\Delta V$ required by the spacecraft to go directly from the circular orbit about the departure body to to the Hohmann transfer orbit with a single instantaneous maneuver.

We can repeat all that for insertion into the arrival orbit to get:

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

The total $\Delta V$ is then the sum of the two maneuvers:

$\Delta V=\sqrt{\left(\sqrt{\frac{2\mu_S r_2}{r_1(r_1+r_2)}}-\sqrt{\mu_S\over r_1}\right)^2+\frac{2\mu_1}{a_1}}-\sqrt{\mu_1\over a_1}+\sqrt{\left(\sqrt{\frac{2\mu_S r_1}{r_2(r_1+r_2)}}-\sqrt{\mu_S\over r_2}\right)^2+\frac{2\mu_2}{a_2}}-\sqrt{\mu_2\over a_2}$

or simplified a little:

$\Delta V=\sqrt{{\mu_S\over r_1}\left(\sqrt{\frac{2r_2}{r_1+r_2}}-1\right)^2+\frac{2\mu_1}{a_1}}-\sqrt{\mu_1\over a_1}+\sqrt{{\mu_S\over r_2}\left(\sqrt{\frac{2r_1}{r_1+r_2}}-1\right)^2+\frac{2\mu_2}{a_2}}-\sqrt{\mu_2\over a_2}$

• This is exactly what I was looking for, thanks!
– Joe
Aug 23, 2013 at 6:18
• It's typical to give delta V from one low circular orbit about a planet to another low circular orbit about the destination planet. This gives an inflated delta V budget. If you have periapsis passing through a planet's upper atmosphere, a capture orbit suffices. Repeated periapsis passes can lower the apoapsis as much as you like. If you have a propellent source high on planet's gravity well (for example EML2) you can depart from near capture orbit. I talk about this at inflated delta Vs. Mar 15, 2014 at 21:27
• @Cornelis No, it is a subtraction of velocities. You can see it in the equations. Jan 26, 2022 at 17:59
• You don't seem to be looking at the equations. It is additions and subtractions of a bunch of square roots of specific energies, which are speeds. Jan 26, 2022 at 20:49
• And no, $v_\infty$ is not directly useful. You need the velocity at the radius of transition. You need to spend more time reading and understanding the answer. Work through the derivation yourself. Jan 26, 2022 at 20:51

There is no closed form solution to the n-body problem, so you will probably start by using a patched-conic approach. This will treat the trajectory as a simple Hohmann transfer between to spheres of influence (assuming your bodies are in coplanar orbits). The delta-v for a Hohmann transfer is well known. To this you will add (subtract) the delta-v for the "patched" trajectories inside the bodies spheres of influence which are solvable for two (n=2) bodies.

At this point you will probably need a refined answer -- especially if you are going to be onboard this mission. This means you will need to do some numerical integration with your patched-conic approximation as your starting point.

Good luck, and be sure to take enough food!

Atomic Rockets is another great source. Plain-language explanations with good graphics and enough math to be useful. Rocket and Space Technology is even better for the technically inclined - basically an online textbook with practice problems.