Orbits – Hohmann Transfer

Suppose you and your rocket are in a circular orbit of radius r_0 about a stationary planet. You want to change its altitude from r_0 to nr_0 for n > 0. How could you do this? The diagram below shows how you might use a Hohmann transfer orbit.

hohmann transfer diagram

Suppose you want to increase your altitude. The procedure would be as follows: apply an impulse to your rocket at 1 in a direction tangential to your path; wait until you reach apoapsis at 2, your point of furthest approach from the planet; apply a second impulse to circularise your orbit. In cases such as this where n > 1, both impulses are delivered parallel to the rocket’s velocity vector. If you wanted to decrease your altitude instead (0 < n < 1), the impulses would be antiparallel, or in the ‘backwards’ direction. Throughout this post, we’ll assume we want to increase the radius of orbit.

We’ll now take a look at why this works. We will quantify the impulse required of the rocket’s engines by finding the fractional change in velocity \alpha in terms of n. That is, the impulse acts so as to increase the rocket’s instantaneous speed from v_0 to (1+\alpha)v_0.

First we need an expression for v_0. For an object of mass m moving in a circle of radius r, we know that the centripetal force F is related to its tangential velocity v by the equation

\displaystyle F=\frac{mv^2}{r}

For an orbit, this force is the mutual gravitation experienced by the rocket and planet, hence

\displaystyle \frac{mv_0^2}{r_0}=\frac{GMm}{r_0^2}


Here, GM is the gravitational parameter of the planet, the product of its mass and the gravitational constant. The total energy E of the rocket-planet system is the sum of its kinetic and potential energy, given by

\displaystyle E=\frac{1}{2}m\bold{v}^2-\frac{GMm}{r}

Hence the system’s energy after the first impulse has been delivered is given by

\displaystyle E=\frac{1}{2}m(1+\alpha_1)^2v_0^2-\frac{GMm}{r_0}

where the rocket’s velocity has been increased by \alpha_1v_0. Using the expression for GM above, we can eliminate r_0 entirely:

\displaystyle E=\frac{1}{2}m(1+\alpha_1)^2v_0^2-mv_0^2

We can equate this with the general expression for the system’s energy quoted in the previous post:

\displaystyle \frac{1}{2}m(1+\alpha_1)^2v_0^2-mv_0^2=\frac{1}{2}m\dot{r}^2+\frac{L^2}{2mr^2}-\frac{GMm}{r}

assuming energy is conserved between positions 1 and 2. Recall that L, the system’s angular momentum about the planet, is a constant. The angular momentum of the rocket has a particularly simple form immediately after the first impulse is delivered, since the velocity and position vectors are orthogonal:

\displaystyle L=mvr\sin\Big(\frac{\pi}{2}\Big)=m(1+\alpha_1)v_0r_0

\displaystyle \frac{1}{2}m(1+\alpha_1)^2v_0^2-mv_0^2=\frac{1}{2}m\dot{r}^2+\frac{m(1+\alpha_1)^2v_0^2r_0^2}{2r^2}-\frac{mr_0v_0^2}{r}

When the rocket is at apoapsis or periapsis, the points of furthest and closest approach to the planet respectively, \dot{r}=0 by definition. If the separation of the rocket and planet is R at these points, then R satisfies the equation

\displaystyle \frac{1}{2}m(1+\alpha_1)^2v_0^2-mv_0^2=\frac{m(1+\alpha_1)^2v_0^2r_0^2}{2R^2}-\frac{mr_0v_0^2}{R}

\displaystyle (1+\alpha_1)^2-2=\frac{(1+\alpha_1)^2r_0^2}{R^2}-\frac{2r_0}{R}

Multiplying through by R^2 and rearranging, we can solve a quadratic equation in R to get

\displaystyle R=\bigg(\frac{\pm\alpha_1(2+\alpha_1)-1}{(1+\alpha_1)^2-2}\bigg)r_0

R has two solutions, hence the apoapsis and periapsis are distinct; this means the orbit becomes elliptical after the impulse is delivered. If we take the positive root first, we find


This is the new periapsis. It coincides with the point at which the first impulse is delivered (point 1). This just means that the rocket will still pass through the point 1 at which the impulse was delivered, were it allowed to complete its new elliptical orbit. The negative root yields the value of the new apoapsis at 2 as a function of \alpha_1:

\displaystyle R=-\bigg(\frac{\alpha_1(2+\alpha_1)+1}{(1+\alpha_1)^2-2}\bigg)r_0

\displaystyle R=\bigg(\frac{(1+\alpha_1)^2}{2-(1+\alpha_1)^2}\bigg)r_0

This is the point of furthest approach the rocket eventually reaches the opposite end of the orbit. For the Hohmann transfer, we demand that R=nr_0:

\displaystyle nr_0=\bigg(\frac{(1+\alpha_1)^2}{2-(1+\alpha_1)^2}\bigg)r_0

Rearranging, the required fractional change in velocity \alpha_1 from the first impulse is

\displaystyle \alpha_1=\sqrt{\frac{2n}{n+1}}-1

So we can get the rocket to reach the height nr_0 for an instant, but we need a second impulse to keep it there and circularise the orbit. Say this requires an additional change in velocity \alpha_2v_0. Rather than generating new equations, let’s recycle the ones we have just used.

Suppose we are already in a desired circular orbit nr_0, and wish to change it back to the elliptical Hohmann transfer orbit with apsides r_0 and nr_0. We have seen that a circular orbit can be distended into an elliptical one. In this case, we simply swap the roles of nr_0 and r_0 in the equation above:

\displaystyle r_0=\bigg(\frac{(1+\alpha_2')^2}{2-(1+\alpha_2')^2}\bigg)nr_0


\displaystyle \alpha_2'=\sqrt{\frac{2}{1+n}}-1

So if we are in a circular orbit of radius nr_0 at velocity v_2 and change the velocity to (1+\alpha_2')v_2, we will be placed on an elliptical path with points of furthest and closest approach nr_0 and r_0 respectively. Now assuming orbits are ‘reversible’, we could supply an impulse in the opposite direction to go from the elliptical orbit to the circular one of radius nr_0. That is, when at apoapsis (for n > 1), the required fractional change in velocity is


Note that this is a fraction of the final velocity v_2, once stable orbit at nr_0 has been achieved. We therefore need to find v_2 in terms of v_0. That is quite easy, since


so analogously


since both orbits are circular. Hence

\displaystyle v_2=\frac{v_0}{\sqrt{n}}

Let’s piece together an expression for \alpha in terms of n. Our first change in velocity \Delta V_1 was given by

\Delta V_1=\alpha_1v_0

The second impulse brings about a change in velocity \Delta V_2 given by

\Delta V_2=\alpha_2v_2

Hence the total change in velocity \Delta V is given by

\Delta V=\alpha_1v_0+\alpha_2v_2

\displaystyle \Delta V=\alpha_1v_0+\alpha_2\frac{v_0}{\sqrt{n}}

\displaystyle \Delta V=\Big(\alpha_1+\frac{\alpha_2}{\sqrt{n}}\Big)v_0

We defined our original quantity \alpha in the following way:

\Delta V=(1+\alpha)v_0-v_0

the total change in velocity of the rocket, given by the difference between its velocity at nr_0 and at r_0.

\Delta V=\alpha v_0

\displaystyle \to\alpha=\alpha_1+\frac{\alpha_2}{\sqrt{n}}

\displaystyle \alpha(n)=\sqrt{\frac{2n}{n+1}}-1+\frac{1}{\sqrt{n}}\bigg(1+\sqrt{\frac{2}{1+n}}\bigg)

Here is a graph said function again n:

hohmann impulse lin

We will pick out some of its main features. First, we see that \alpha=0 for n=1 as we expect: to leave the radius of orbit unchanged, no change in velocity is required. As the radius of orbit goes to 0, the change in velocity actually becomes unboundedly large. The conservation of angular momentum explains this: as the distance between the planet and the rocket decreases, the speed of the rocket increases. In the limit that the radius becomes infinitesimal, the speed required of the rocket becomes infinite.

Finally, as n approaches infinity, the change in velocity required approaches 41.4%. Supplying such an impulse would allow the rocket to escape the planet entirely.

What is interesting is that this figure of 41.4% is not the function’s maximum, far from it in fact. There is a large range of orbits which are, in real terms, ‘more difficult’ to establish than one an infinite distance away. Plotting a log-lin graph of \alpha(n) against n reveals the shape of the graph:

hohmann impulse log

The most expensive orbit and the \Delta V required to achieve it can be found using calculus. The derivative of \alpha(n) is given by

\displaystyle \frac{d\alpha(n)}{dn}=\frac{2}{\sqrt{n}}\sqrt{\frac{n+1}{2}}\frac{1}{(n+1)^2}-\frac{1}{2n^{\frac{3}{2}}}\bigg(1-\sqrt{\frac{2}{n+1}}\bigg)

At its maximum (N,\alpha(N)), the derivative is 0:


Multiplying through by 2N^{\frac{3}{2}}\sqrt{\frac{N+1}{2}}(N+1)^2:

\displaystyle 2N(N+1)-(N+1)^2\sqrt{\frac{N+1}{2}}+(N+1)^2=0

\displaystyle 2N-(N+1)\sqrt{\frac{N+1}{2}}+N+1=0

\displaystyle 3N+1=(N+1)\sqrt{\frac{N+1}{2}}

\displaystyle {\frac{3N+1}{N+1}}^2=\frac{N+1}{2}


The positive root of this equation is N=15.5817 to 4 significant figures. Multiplying your radius of orbit by this figure requires a fractional velocity change of 53.63%. This is the hardest you would ever need to work to carry out a Hohmann transfer.


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