Capture at Mars: an Elliptical Mars-Centred Orbit
A worked companion to the hyperbolic-flyby problem — showing \(\varepsilon<0\) and \(e<1\)
1 Why this problem is different from the flyby problem
In the flyby problem the spacecraft arrives from far outside Mars’s gravitational reach, so its Mars-centred energy is evaluated in the limit \(r\rightarrow\infty\), where \(\varepsilon=v_\infty^2/2>0\) always. Within a pure Mars-only two-body model, energy is conserved, so a spacecraft that starts outside can never end up bound: unpowered capture on a patched conic is impossible.
To obtain a bound Mars-centred orbit the state must be evaluated at a finite radius \(r\), with the Mars-relative speed below the local escape speed,
\[ v<v_\mathrm{esc}(r)=\sqrt{\frac{2\mu_M}{r}} \qquad\Longleftrightarrow\qquad \varepsilon=\frac{v^2}{2}-\frac{\mu_M}{r}<0 , \]
and something must have put it there: a Mars-orbit-insertion (MOI) burn, aerobraking, or three-body ballistic capture. The example below uses an MOI burn, and gives the heliocentric velocity before and after it at the same point, so that the same arithmetic yields a hyperbola in one case and an ellipse in the other.
2 Problem statement
A spacecraft approaches Mars on a slow, nearly velocity-matched trajectory. Use the same heliocentric frame as in the flyby problem: at the encounter epoch \(+x\) points radially outward from the Sun through Mars, and \(+y\) lies along the instantaneous direction of motion of Mars.
| Symbol | Value | Meaning |
|---|---|---|
| \(\mathbf r\) | \(\begin{bmatrix}-1.0\times10^{5} & 0\end{bmatrix}^{\mathsf T}\) km | position of the spacecraft relative to Mars at the epoch |
| \(\mathbf V_M\) | \(\begin{bmatrix}0 & 24.13\end{bmatrix}^{\mathsf T}\) km/s | heliocentric velocity of Mars |
| \(\mathbf V_0\) | \(\begin{bmatrix}0.585 & 23.35\end{bmatrix}^{\mathsf T}\) km/s | heliocentric velocity of the spacecraft before the MOI burn |
| \(\mathbf V_1\) | \(\begin{bmatrix}0.36 & 23.65\end{bmatrix}^{\mathsf T}\) km/s | heliocentric velocity of the spacecraft after the MOI burn |
| \(\mu_M\) | \(4.2828\times10^{4}\ \mathrm{km^3/s^2}\) | gravitational parameter of Mars |
| \(R_M\) | \(3396\) km | radius of Mars |
| \(r_\mathrm{SOI}\) | \(5.77\times10^{5}\) km | radius of Mars’s sphere of influence |
The burn is instantaneous and occurs at the position \(\mathbf r\) given above, so the position is the same before and after.
Tasks.
- Transform both heliocentric velocities into the Mars-centred frame and compare each Mars-relative speed with the local escape speed \(v_\mathrm{esc}(r)\).
- For each case compute the specific energy \(\varepsilon=v^2/2-\mu_M/r\) and the specific angular momentum \(h=xv_y-yv_x\), and hence the eccentricity \(e=\sqrt{1+2\varepsilon h^2/\mu_M^2}\).
- Show that the pre-burn state is hyperbolic (\(\varepsilon>0\), \(e>1\)) and the post-burn state is elliptical (\(\varepsilon<0\), \(e<1\)): the spacecraft has been captured.
- For the captured orbit find \(a\), \(p\), \(r_p\), \(r_a\), the periapsis altitude, the orbital period, and the true anomaly at the epoch.
- Check that the capture orbit lies entirely inside the sphere of influence, so that the Mars-only two-body model is self-consistent.
- Find the burn magnitude \(|\Delta\mathbf V|\), and comment on what happened to the heliocentric speed.
3 Solution
3.1 Mars-relative velocities
Subtracting the velocity of Mars, \(\mathbf v=\mathbf V-\mathbf V_M\), gives
\[ \mathbf v_0=\begin{bmatrix}0.585\\ 23.35-24.13\end{bmatrix} =\begin{bmatrix}0.585\\ -0.780\end{bmatrix}\ \mathrm{km/s}, \qquad \mathbf v_1=\begin{bmatrix}0.36\\ 23.65-24.13\end{bmatrix} =\begin{bmatrix}0.36\\ -0.48\end{bmatrix}\ \mathrm{km/s}. \]
Both are multiples of the same unit vector \(\hat{\mathbf u}=\begin{bmatrix}0.6&-0.8\end{bmatrix}^{\mathsf T}\), so the burn is purely retrograde relative to Mars:
\[ v_0=|\mathbf v_0|=0.975\ \mathrm{km/s}, \qquad v_1=|\mathbf v_1|=\sqrt{0.36^2+0.48^2}=\sqrt{0.36}=0.600\ \mathrm{km/s}. \]
Note that both heliocentric speeds, \(|\mathbf V_0|=23.357\) km/s and \(|\mathbf V_1|=23.653\) km/s, are below the Mars speed of \(24.13\) km/s: the spacecraft is being overtaken by Mars, which is exactly the slow-approach geometry that makes capture affordable.
3.2 The escape-speed test at \(r=10^{5}\) km
With \(r=|\mathbf r|=1.0\times10^{5}\) km,
\[ v_\mathrm{esc}(r)=\sqrt{\frac{2\mu_M}{r}}=\sqrt{\frac{2(42828)}{10^{5}}}=\sqrt{0.85656} =\boxed{0.9255\ \mathrm{km/s}} . \]
Therefore \(v_0=0.975>v_\mathrm{esc}\) (unbound) while \(v_1=0.600<v_\mathrm{esc}\) (bound). Everything that follows is just this statement expressed through \(\varepsilon\) and \(e\).
3.3 Specific energy
\[ \varepsilon_0=\frac{v_0^2}{2}-\frac{\mu_M}{r} =\frac{0.975^2}{2}-\frac{42828}{10^{5}} =0.475313-0.428280 =\boxed{+0.047032\ \mathrm{km^2/s^2}} , \]
\[ \varepsilon_1=\frac{v_1^2}{2}-\frac{\mu_M}{r} =\frac{0.600^2}{2}-\frac{42828}{10^{5}} =0.180000-0.428280 =\boxed{-0.248280\ \mathrm{km^2/s^2}} . \]
The post-burn energy is negative: the kinetic term \(v^2/2\) no longer covers the potential well depth \(\mu_M/r\) at this radius. The corresponding semi-major axes are
\[ a_0=-\frac{\mu_M}{2\varepsilon_0}=-\frac{42828}{0.094065}=-455302\ \mathrm{km}<0 \quad\text{(hyperbola)}, \]
\[ a_1=-\frac{\mu_M}{2\varepsilon_1}=+\frac{42828}{0.496560}=\boxed{86249\ \mathrm{km}>0} \quad\text{(ellipse)} . \]
3.4 Specific angular momentum
With \(\mathbf r=\begin{bmatrix}-10^{5}&0\end{bmatrix}^{\mathsf T}\) and \(h=xv_y-yv_x\),
\[ h_0=(-10^{5})(-0.780)-0=7.8000\times10^{4}\ \mathrm{km^2/s}, \qquad h_1=(-10^{5})(-0.480)-0=\boxed{4.8000\times10^{4}\ \mathrm{km^2/s}} . \]
Both are positive, so the motion is counter-clockwise in this frame; the retro burn removed angular momentum in the same proportion as speed, because it did not change the velocity direction.
3.5 Eccentricity
Using \(e=\sqrt{1+2\varepsilon h^2/\mu_M^2}\) with \(\mu_M^2=1.834238\times10^{9}\ \mathrm{km^6/s^4}\):
\[ e_0=\sqrt{1+\frac{2(0.047032)(7.8\times10^{4})^2}{1.834238\times10^{9}}} =\sqrt{1+0.312006}=\boxed{1.1454>1}\quad\text{(hyperbola)}, \]
\[ e_1=\sqrt{1-\frac{2(0.248280)(4.8\times10^{4})^2}{1.834238\times10^{9}}} =\sqrt{1-0.623677}=\sqrt{0.376323}=\boxed{0.6134<1}\quad\text{(ellipse)} . \]
The sign of \(\varepsilon\) and the value of \(e\) carry the same information, since \(2\varepsilon h^2/\mu_M^2=e^2-1\): \(\varepsilon<0\Leftrightarrow e<1\). The pre-burn orbit would have escaped with \(v_\infty=\sqrt{2\varepsilon_0}=0.3067\) km/s; the post-burn orbit has no \(v_\infty\) at all.
3.6 Elements of the capture orbit
\[ p=\frac{h_1^2}{\mu_M}=\frac{(4.8\times10^{4})^2}{42828}=53797\ \mathrm{km}, \]
\[ r_p=a_1(1-e_1)=86249(0.386594)=\boxed{33343\ \mathrm{km}}, \qquad r_a=a_1(1+e_1)=86249(1.613406)=\boxed{139155\ \mathrm{km}}, \]
\[ h_p=r_p-R_M=33343-3396=29947\ \mathrm{km}\ \text{(periapsis altitude)} . \]
The periapsis and apoapsis speeds follow from \(h=rv\) at the apsides:
\[ v_p=\frac{h_1}{r_p}=\frac{48000}{33343}=1.4396\ \mathrm{km/s}, \qquad v_a=\frac{h_1}{r_a}=\frac{48000}{139155}=0.3449\ \mathrm{km/s}. \]
The period is
\[ T=2\pi\sqrt{\frac{a_1^3}{\mu_M}} =2\pi\sqrt{\frac{(86249)^3}{42828}} =7.690\times10^{5}\ \mathrm{s}=\boxed{8.90\ \mathrm{days}} . \]
The true anomaly at the epoch comes from the conic equation, \(1+e\cos f=p/r\):
\[ \cos f=\frac{1}{e_1}\left(\frac{p}{r}-1\right) =\frac{0.537966-1}{0.613406}=-0.75323 \quad\Longrightarrow\quad f=138.87^\circ\ \text{or}\ 221.13^\circ . \]
The radial velocity is \(v_r=\hat{\mathbf r}\cdot\mathbf v_1=(-1)(0.36)=-0.36\) km/s \(<0\), so the spacecraft is falling inward and the correct branch is \(\boxed{f=221.13^\circ}\), i.e. \(-138.87^\circ\): periapsis lies ahead, about \(139^\circ\) of true anomaly away.
3.7 Consistency check against the sphere of influence
\[ \frac{r_a}{r_\mathrm{SOI}}=\frac{139155}{5.77\times10^{5}}=0.24 . \]
The whole capture ellipse stays within a quarter of the sphere of influence, so treating Mars as the only gravitating body is self-consistent — this orbit really is closed and the spacecraft really is trapped.
The same check also exposes the point made at the start: a closed orbit with \(r_a=139155\) km never reaches the SOI boundary, so the spacecraft cannot have arrived on it from interplanetary space. The negative energy is itself the proof that a burn (or some non-two-body effect) occurred.
3.8 The burn, and the heliocentric bookkeeping
\[ \Delta\mathbf V=\mathbf v_1-\mathbf v_0 =\begin{bmatrix}0.36-0.585\\ -0.48+0.78\end{bmatrix} =\begin{bmatrix}-0.225\\ 0.300\end{bmatrix}\ \mathrm{km/s}, \qquad |\Delta\mathbf V|=0.975-0.600=\boxed{0.375\ \mathrm{km/s}}, \]
directed exactly opposite to \(\mathbf v_0\). Note the frame-dependence, echoing the flyby lesson in reverse: this is a braking burn in the Mars frame (\(0.975\rightarrow0.600\) km/s), yet in the heliocentric frame the spacecraft speeds up, from \(23.357\) km/s to \(23.653\) km/s, because the retro-burn direction relative to Mars happens to point partly along the heliocentric motion. Only the Mars-relative energy decides whether the spacecraft is captured; the heliocentric speed says nothing about it.
4 Summary
| Quantity | Before MOI | After MOI |
|---|---|---|
| heliocentric \(\mathbf V\) (km/s) | \(\begin{bmatrix}0.585&23.35\end{bmatrix}^{\mathsf T}\) | \(\begin{bmatrix}0.36&23.65\end{bmatrix}^{\mathsf T}\) |
| Mars-relative \(\mathbf v\) (km/s) | \(\begin{bmatrix}0.585&-0.780\end{bmatrix}^{\mathsf T}\) | \(\begin{bmatrix}0.36&-0.48\end{bmatrix}^{\mathsf T}\) |
| speed \(v\) vs. \(v_\mathrm{esc}=0.9255\) km/s | \(0.975>v_\mathrm{esc}\) | \(0.600<v_\mathrm{esc}\) |
| \(\varepsilon\ (\mathrm{km^2/s^2})\) | \(+0.0470\) | \(\mathbf{-0.2483}\) |
| \(h\ (\mathrm{km^2/s})\) | \(7.80\times10^{4}\) | \(4.80\times10^{4}\) |
| \(e\) | \(1.1454\) | \(\mathbf{0.6134}\) |
| \(a\) (km) | \(-455302\) | \(+86249\) |
| conic | hyperbola (escapes, \(v_\infty=0.307\) km/s) | ellipse — captured |
| \(r_p,\ r_a\) (km) | — | \(33343,\ 139155\) |
| period | — | \(8.90\) days |
5 Teaching notes
- Capture is a statement about a finite radius. Students who try to evaluate \(\varepsilon\) “at infinity” for the captured orbit will find no \(v_\infty\) exists; the orbit simply does not extend that far. The escape-speed comparison \(v\lessgtr\sqrt{2\mu_M/r}\) is the most physical way to introduce this.
- Velocity matching is what makes capture cheap. Here \(|\mathbf V_0-\mathbf V_M|=0.975\) km/s, so only \(0.375\) km/s of \(\Delta V\) is needed. In the flyby problem \(v_\infty=3.00\) km/s, and capture at the same radius would demand a burn of \(\sqrt{9+0.85656}-0.6\approx2.54\) km/s.
- The threshold at the SOI boundary. To be even marginally bound at \(r_\mathrm{SOI}\) requires \(v<\sqrt{2\mu_M/r_\mathrm{SOI}}=0.385\) km/s — the heliocentric velocities of spacecraft and planet must agree to within about \(0.4\) km/s. This is the quantitative reason that unpowered capture is a three-body (weak-stability-boundary) phenomenon and not a patched-conic one.
- A good follow-up exercise. Ask for the burn that circularises the orbit at \(r_p\): \(v_\mathrm{circ}=\sqrt{\mu_M/r_p}=1.1334\) km/s against \(v_p=1.4396\) km/s, so a further \(0.306\) km/s at periapsis produces a circular \(33343\) km orbit — the second half of a standard two-burn insertion.