# 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.

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

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

$GM=r_0v_0^2$

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

$R=r_0$

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:

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

yielding

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

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

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

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

$GM=r_0v_0^2$

so analogously

$GM=nr_0v_2^2$

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 $n$:

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:

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 $0$:

$\displaystyle\frac{2}{\sqrt{N}}\sqrt{\frac{N+1}{2}}\frac{1}{(N+1)^2}-\frac{1}{2N^{\frac{3}{2}}}\bigg(1-\sqrt{\frac{2}{N+1}}\bigg)=0$

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

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

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

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

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

$N^3-15N^2-9N-1=0$

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.