Tutorial 1
Problems
Attempt each problem before opening its solution.
1. Compute the gravitational acceleration of the Moon due to the Earth and due to the Sun. Which of the two is larger, and by what factor? If your answer surprises you, explain in what sense the Moon can still be said to orbit the Earth.
Answer
For a body at distance \(r\) from a mass with gravitational parameter \(\mu = GM\),
\[ a = \frac{\mu}{r^{2}} . \]
Due to the Earth. With \(r_{EM} = 3.844\times10^{8}\ \mathrm{m}\) and \(\mu_E = 3.9860\times10^{14}\ \mathrm{m^3/s^2}\),
\[ a_{E\to M} = \frac{3.9860\times10^{14}}{(3.844\times10^{8})^{2}} = \frac{3.9860\times10^{14}}{1.478\times10^{17}} \approx 2.70\times10^{-3}\ \mathrm{m/s^2}. \]
Due to the Sun. The Moon is essentially \(1\ \mathrm{AU}\) from the Sun, so with \(r_{SM} \approx 1.496\times10^{11}\ \mathrm{m}\) and \(\mu_S = 1.3271\times10^{20}\ \mathrm{m^3/s^2}\),
\[ a_{S\to M} = \frac{1.3271\times10^{20}}{(1.496\times10^{11})^{2}} = \frac{1.3271\times10^{20}}{2.238\times10^{22}} \approx 5.93\times10^{-3}\ \mathrm{m/s^2}. \]
| Source | Acceleration of the Moon |
|---|---|
| Earth | \(2.70\times10^{-3}\ \mathrm{m/s^2}\) |
| Sun | \(5.93\times10^{-3}\ \mathrm{m/s^2}\) |
| Sun / Earth | \(\approx 2.20\) |
\[ \boxed{\ \frac{a_{S\to M}}{a_{E\to M}} \approx 2.2\ } \]
The Sun pulls the Moon more than twice as hard as the Earth does.
Why we can still predict the path of Moon using E-M two-body system? What governs the relative motion \(\ddot{\mathbf r}_{EM}\) is not the Sun’s absolute pull on the Moon, but the difference between the Sun’s pull on the Moon and on the Earth — the solar tidal term. To first order in \(r_{EM}/r_{SE}\),
\[ a_{\text{tidal}} \approx \frac{2\mu_S\, r_{EM}}{r_{SE}^{3}} = \frac{2(1.3271\times10^{20})(3.844\times10^{8})}{(1.496\times10^{11})^{3}} \approx 3.0\times10^{-5}\ \mathrm{m/s^2}, \]
which is only about \(1.1\%\) of \(a_{E\to M}\). In the Earth-centred frame the Sun is a small perturbation, and the two-body picture is excellent.
We have not covered the tidal terms yet in this course. We will try to cover tidal terms in the later part of the course. However, this problem should show you the subtle arguments behind justifying two body problem setup.
2. Compute the acceleration of the Earth due to the Sun and due to the Moon, and form their ratio. Contrast this ratio with the one you obtained in Problem 1 and explain the difference.
Answer
Earth accelerated by the Sun.
\[ a_{S\to E} = \frac{\mu_S}{r_{SE}^{2}} = \frac{1.3271\times10^{20}}{(1.496\times10^{11})^{2}} \approx 5.93\times10^{-3}\ \mathrm{m/s^2}. \]
Earth accelerated by the Moon. Here the Moon’s gravitational parameter must be used:
\[ a_{M\to E} = \frac{\mu_M}{r_{EM}^{2}} = \frac{4.905\times10^{12}}{(3.844\times10^{8})^{2}} = \frac{4.905\times10^{12}}{1.478\times10^{17}} \approx 3.32\times10^{-5}\ \mathrm{m/s^2}. \]
Hence
\[ \boxed{\ \frac{a_{S\to E}}{a_{M\to E}} = \frac{5.93\times10^{-3}}{3.32\times10^{-5}} \approx 179 \ } \]
The contrast. From Problem 1, \(a_{S\to M}/a_{E\to M} \approx 2.2\), whereas here \(a_{S\to E}/a_{M\to E} \approx 179\). The two differ by a factor of about \(81\), which is precisely the Earth/Moon mass ratio:
\[ \frac{a_{S\to E}/a_{M\to E}}{a_{S\to M}/a_{E\to M}} = \frac{\mu_E}{\mu_M} = \frac{M_E}{M_M} \approx 81.3 , \]
because the Sun’s accelerations of the Earth and of the Moon are essentially equal (both bodies are \(1\ \mathrm{AU}\) away), while the Earth–Moon pair swaps \(\mu_E\) for \(\mu_M\) at the same separation \(r_{EM}\).
3. What is the ratio of the Sun–Earth semi-major axis to the Earth–Moon semi-major axis? Comment on the numerical coincidence you find.
Answer
\[ \frac{a_{SE}}{a_{EM}} = \frac{149.6\times10^{6}\ \mathrm{km}}{384\,400\ \mathrm{km}} \qquad\Longrightarrow\qquad \boxed{\ \frac{a_{SE}}{a_{EM}} \approx 389 \ } \]
that is, \(a_{SE} : a_{EM} \approx 389 : 1\).
The coincidence. The ratio of the Sun’s diameter to the Moon’s diameter is
\[ \frac{D_S}{D_M} \approx \frac{1.392\times10^{6}}{3475} \approx 400 , \]
remarkably close to \(389\). Since angular size is \(\theta \approx D/r\), the two objects subtend nearly the same angle (\(\approx 0.5^{\circ}\)) as seen from the Earth. That near-equality is what makes total solar eclipses — with their thin ring of corona — possible at all, and it is a genuine accident of the present epoch: the Moon recedes by about \(3.8\ \mathrm{cm/year}\), so in a few hundred million years only annular eclipses will remain.
4. Given the instantaneous position and velocity vectors \(\mathbf r\), \(\mathbf v\) of a body and the gravitational parameter \(\mu\), write down an algorithm that returns the specific angular momentum \(h\), the specific energy \(\varepsilon\), the eccentricity \(e\), semi-latus rectum \(p\), semi-major axis \(a\), periapsis distance \(r_p\) and flight path angle \(\gamma\). State the order of the computation.
Answer
Algorithm. Everything is built out of \(\mathbf r\), \(\mathbf v\) and \(\mu\) alone, in the following order. Each line uses only quantities already computed above it.
\[ \begin{aligned} \mathbf{1.}\quad & r = \lVert\mathbf r\rVert, \qquad v = \lVert\mathbf v\rVert \\[3pt] \mathbf{2.}\quad & \mathbf h = \mathbf r\times\mathbf v, \qquad h = \lVert\mathbf h\rVert \\[3pt] \mathbf{3.}\quad & \varepsilon = \frac{v^{2}}{2}-\frac{\mu}{r} \\[3pt] \mathbf{4.}\quad & \mathbf e = \frac{\mathbf v\times\mathbf h}{\mu}-\frac{\mathbf r}{r}, \qquad e = \lVert\mathbf e\rVert \\[3pt] \mathbf{5.}\quad & p = \frac{h^{2}}{\mu} \\[3pt] \mathbf{6.}\quad & a = -\frac{\mu}{2\varepsilon} \\[3pt] \mathbf{7.}\quad & r_p = \frac{p}{1+e} \\[3pt] \mathbf{8.}\quad & \gamma = \operatorname{atan2}\!\left(\mathbf r\cdot\mathbf v,\ h\right) \end{aligned} \]
Order of computation. The dependency chain is
\[ (\mathbf r,\mathbf v)\ \longrightarrow\ r,v\ \longrightarrow\ \mathbf h,h\ \longrightarrow\ \varepsilon\ \longrightarrow\ \mathbf e,e\ \longrightarrow\ p\ \longrightarrow\ a\ \longrightarrow\ r_p,\ \gamma . \]
| Step | Output | Formula | Requires |
|---|---|---|---|
| 1 | \(r,\ v\) | \(\lVert\mathbf r\rVert,\ \lVert\mathbf v\rVert\) | \(\mathbf r,\mathbf v\) |
| 2 | \(\mathbf h,\ h\) | \(\mathbf r\times\mathbf v\) | \(\mathbf r,\mathbf v\) |
| 3 | \(\varepsilon\) | \(v^{2}/2-\mu/r\) | \(r,v\) |
| 4 | \(\mathbf e,\ e\) | \((\mathbf v\times\mathbf h)/\mu-\mathbf r/r\) | \(\mathbf h,r\) |
| 5 | \(p\) | \(h^{2}/\mu\) | \(h\) |
| 6 | \(a\) | \(-\mu/2\varepsilon\) | \(\varepsilon\) |
| 7 | \(r_p\) | \(p/(1+e)\) | \(p,e\) |
| 8 | \(\gamma\) | \(\operatorname{atan2}(\mathbf r\cdot\mathbf v,\ h)\) | \(\mathbf r,\mathbf v,h\) |
Note that \(\varepsilon\), \(h\) and \(e\) are the three integrals of motion: steps 1–4 extract them, and steps 5–8 are pure geometry read off from them.
Step 3 already classifies the orbit, before any conic is drawn:
\[ \varepsilon<0 \Rightarrow \text{ellipse},\qquad \varepsilon=0 \Rightarrow \text{parabola},\qquad \varepsilon>0 \Rightarrow \text{hyperbola}. \]
Eccentricity without vectors. Steps 2 and 3 already determine the magnitude \(e\) on their own,
\[ e = \sqrt{1+\frac{2\varepsilon h^{2}}{\mu^{2}}}, \]
so the vector \(\mathbf e\) is needed only when the direction of periapsis is wanted — \(\mathbf e\) points from the focus towards periapsis, and its magnitude is the eccentricity.
Semi-major axis. Equivalently, straight from vis-viva \(v^{2}=\mu\left(\dfrac{2}{r}-\dfrac{1}{a}\right)\),
\[ a = \left(\frac{2}{r}-\frac{v^{2}}{\mu}\right)^{-1}, \]
so \(a\) depends only on the magnitudes \(r\) and \(v\) — not on the direction of \(\mathbf v\). The direction enters only through \(h\), hence only through \(p\), \(e\) and \(\gamma\).
Why \(p\) before \(a\), and \(r_p = p/(1+e)\). The pair \((p,e)\) describes every conic, whereas \(a\) blows up at \(e=1\):
\[ p = a\left(1-e^{2}\right) \quad\Longleftrightarrow\quad a = \frac{p}{1-e^{2}} . \]
For a parabola \(\varepsilon = 0\), so \(a\to\infty\) while \(p\) stays perfectly finite. Writing the periapsis distance as \(r_p = p/(1+e)\) — rather than \(r_p = a(1-e)\) — therefore works for the ellipse, the parabola (\(r_p = p/2\)) and the hyperbola alike. In floating-point work one should prefer \((p,e)\) to \((a,e)\) near \(e=1\) for the same reason. (Apoapsis exists only for the ellipse: \(r_a = p/(1-e) = a(1+e)\).)
Flight path angle. \(\gamma\) is the angle between \(\mathbf v\) and the local horizon, i.e. the amount by which the velocity tilts away from being purely transverse. Split \(\mathbf v\) into its radial and transverse parts:
\[ v_r = \dot r = \frac{\mathbf r\cdot\mathbf v}{r}, \qquad v_\perp = \frac{h}{r}, \qquad v^{2}=v_r^{2}+v_\perp^{2}, \]
so that
\[ \boxed{\ \tan\gamma = \frac{v_r}{v_\perp} = \frac{\mathbf r\cdot\mathbf v}{h}\ }, \qquad \sin\gamma = \frac{v_r}{v}, \qquad \cos\gamma = \frac{h}{r v} . \]
The last identity is worth remembering on its own:
\[ h = r\,v\cos\gamma . \]
Use atan2 rather than arctan so that the sign is carried automatically: \(\gamma>0\) outbound (rising from periapsis to apoapsis), \(\gamma<0\) inbound, and
\[ \gamma = 0 \iff \mathbf r\cdot\mathbf v = 0 \iff \text{the body is at an apse}, \]
which is the quickest check that a given state vector sits at periapsis or apoapsis. In terms of the true anomaly, \(\tan\gamma = \dfrac{e\sin f}{1+e\cos f}\); for a circle \(\gamma\equiv 0\).
Two degenerate cases to watch. If \(h=0\) the motion is rectilinear: \(p=0\), \(e=1\) whatever the energy, and \(\gamma=\pm90^{\circ}\) — the shape parameters stop meaning anything. If \(e=0\) the orbit is circular and \(\mathbf e = \mathbf 0\) gives no periapsis direction, so periapsis is undefined (any point will do).
5. A newly discovered asteroid is tracked and, at one instant, its geocentric state vector is
\[ \mathbf r = \begin{bmatrix}10\,000\\0\\0\end{bmatrix}\ \mathrm{km}, \qquad \mathbf v = \begin{bmatrix}0\\5.15494\\0\end{bmatrix}\ \mathrm{km/s}. \]
Using \(\mu_E = 398\,600.44\ \mathrm{km^3/s^2}\) and \(R_E = 6378\ \mathrm{km}\), determine \(h\), \(\varepsilon\), \(a\), \(e\) and the perigee radius \(r_p\). Does the asteroid strike the Earth?
Answer
Apply the algorithm of Problem 4.
Angular momentum. The velocity is purely transverse, so
\[ \mathbf h = \mathbf r\times\mathbf v = \begin{bmatrix}10\,000\\0\\0\end{bmatrix}\times \begin{bmatrix}0\\5.15494\\0\end{bmatrix} = \begin{bmatrix}0\\0\\51\,549.4\end{bmatrix}\ \mathrm{km^2/s}, \qquad h = 51\,549.4\ \mathrm{km^2/s}. \]
Energy. With \(v^{2} = (5.15494)^{2} = 26.5734\ \mathrm{km^2/s^2}\),
\[ \varepsilon = \frac{26.5734}{2}-\frac{398\,600.44}{10\,000} = 13.2867-39.8600 = -26.5734\ \mathrm{km^2/s^2} < 0 , \]
so the orbit is an ellipse.
Semi-major axis.
\[ a = -\frac{\mu_E}{2\varepsilon} = \frac{398\,600.44}{2(26.5734)} = 7500\ \mathrm{km}. \]
Eccentricity.
\[ e = \sqrt{1+\frac{2\varepsilon h^{2}}{\mu_E^{2}}} = \frac13 = 0.3333 . \]
Since the velocity is perpendicular to \(\mathbf r\), the point given is an apse; as \(r = 10\,000\ \mathrm{km} > a\), it is the apogee.
Perigee.
\[ r_p = a(1-e) = 7500\left(1-\tfrac13\right) = 5000\ \mathrm{km}. \]
Verdict.
\[ \boxed{\ r_p = 5000\ \mathrm{km} \;<\; R_E = 6378\ \mathrm{km}\ } \]
The Keplerian trajectory dips \(1378\ \mathrm{km}\) below the Earth’s surface, so the asteroid impacts the Earth — it never reaches the mathematical perigee. Impact occurs where \(r = R_E\), at a true anomaly measured from perigee of
\[ \cos f = \frac{1}{e}\left(\frac{p}{R_E}-1\right), \qquad p = \frac{h^{2}}{\mu_E} = 6666.7\ \mathrm{km}, \]
giving \(f \approx \pm 122^{\circ}\); the body is still descending there, and the impact speed follows from vis-viva at \(r = R_E\): \(v = \sqrt{\mu_E\left(2/R_E - 1/a\right)} \approx 9.7\ \mathrm{km/s}\) (ignoring the atmosphere).
The moral is that “will it hit?” is answered by a single scalar, \(r_p\), extracted from one state vector — no propagation in time is needed.
6. Show that for any elliptical orbit the speeds at periapsis and apoapsis satisfy
\[ v_p\,v_a = \frac{\mu}{a}. \]
Interpret the result.
Answer
At the apses, \(r_p = a(1-e)\) and \(r_a = a(1+e)\). From vis-viva, \(v^{2}=\mu\left(\dfrac{2}{r}-\dfrac{1}{a}\right)\), at periapsis
\[ v_p^{2} = \mu\left[\frac{2}{a(1-e)}-\frac1a\right] = \frac{\mu}{a}\left[\frac{2}{1-e}-1\right] = \frac{\mu}{a}\,\frac{2-(1-e)}{1-e}, \]
that is
\[ \boxed{\ v_p^{2}=\frac{\mu}{a}\,\frac{1+e}{1-e}\ }, \qquad\text{and identically}\qquad \boxed{\ v_a^{2}=\frac{\mu}{a}\,\frac{1-e}{1+e}\ }. \]
Multiplying, the eccentricity factors are reciprocals and cancel:
\[ v_p^{2}v_a^{2} = \frac{\mu^{2}}{a^{2}} \qquad\Longrightarrow\qquad \boxed{\ v_p v_a = \frac{\mu}{a}\ } \]
(speeds being positive).
Interpretation. Writing it as
\[ \sqrt{v_p v_a} = \sqrt{\frac{\mu}{a}} = v_c(a), \]
the geometric mean of the apsidal speeds equals the speed of a circular orbit of radius \(a\). The eccentricity cancels entirely: stretching an ellipse at fixed \(a\) speeds up periapsis and slows down apoapsis by exactly reciprocal factors, \(\sqrt{(1+e)/(1-e)}\).
A second route makes this obvious. Angular momentum gives \(h = r_p v_p = r_a v_a\), so
\[ v_p v_a = \frac{h^{2}}{r_p r_a} = \frac{\mu\,a(1-e^{2})}{a^{2}(1-e^{2})} = \frac{\mu}{a}, \]
using \(h^{2} = \mu p = \mu a (1-e^{2})\) and \(r_p r_a = b^{2} = a^{2}(1-e^{2})\).
7. A spacecraft moves with the Earth on a circular heliocentric orbit at \(1\ \mathrm{AU}\). Neglect all planetary gravity and treat the Sun as a point mass, with \(\mu_S = 1.3271\times10^{11}\ \mathrm{km^3/s^2}\), \(1\ \mathrm{AU} = 1.496\times10^{8}\ \mathrm{km}\) and \(R_S = 6.957\times10^{5}\ \mathrm{km}\).
- Compute \(\Phi = -\mu_S/r\) at the Sun’s surface and at \(1\ \mathrm{AU}\), and compare their magnitudes.
- Find the escape speed from the Sun’s surface.
- Find the escape speed from the Sun at \(1\ \mathrm{AU}\).
- Find the Earth’s heliocentric orbital speed \(v_E\).
- Find the minimum prograde \(\Delta v\) at \(1\ \mathrm{AU}\) that puts the spacecraft on a solar escape trajectory.
- Find the retrograde \(\Delta v\) that removes all of its heliocentric tangential velocity.
- Explain the surprise: why is it so much harder to fall into the Sun than to leave the Solar System?
- Suppose instead the speed is reduced instantaneously from \(v_E\) to \(10\ \mathrm{km/s}\), still purely tangential. Compute \(\varepsilon\), \(h\), \(e\), \(a\) and \(r_p = a(1-e)\). Does the orbit intersect the Sun?
Answer
(a)
\[ \begin{aligned} \Phi(R_S) &= -\frac{1.3271\times10^{11}}{6.957\times10^{5}} = -1.908\times10^{5}\ \mathrm{km^2/s^2},\\[4pt] \Phi(1\,\mathrm{AU}) &= -\frac{1.3271\times10^{11}}{1.496\times10^{8}} = -887.1\ \mathrm{km^2/s^2}. \end{aligned} \]
The ratio is \(215\) — exactly \(1\,\mathrm{AU}/R_S\), since \(\Phi\propto 1/r\). The Earth sits on the shallow lip of the well: \(99.5\%\) of the climb out of the Sun’s potential is already done by the time one reaches \(1\ \mathrm{AU}\).
(b) Escape from the surface.
\[ v_{\text{esc}}(R_S)=\sqrt{\frac{2\mu_S}{R_S}}=\sqrt{2(1.908\times10^{5})} \approx 617.7\ \mathrm{km/s}. \]
(c) Escape at \(1\ \mathrm{AU}\).
\[ v_{\text{esc}}(1\,\mathrm{AU})=\sqrt{2(887.1)} \approx 42.1\ \mathrm{km/s}. \]
(d) Earth’s orbital speed.
\[ v_E=\sqrt{\frac{\mu_S}{r}}=\sqrt{887.1}\approx 29.8\ \mathrm{km/s}. \]
Note \(v_{\text{esc}} = \sqrt{2}\,v_E\), as it must be for a circular orbit.
(e) Prograde \(\Delta v\) to escape.
\[ \Delta v = \left(\sqrt2-1\right)v_E = 0.4142\times 29.8 \approx \boxed{12.3\ \mathrm{km/s}}. \]
(f) Retrograde \(\Delta v\) to kill the tangential velocity.
\[ \Delta v = v_E \approx \boxed{29.8\ \mathrm{km/s}}. \]
(g) The surprise, explained. Falling into the Sun costs \(2.4\) times the \(\Delta v\) of leaving the Solar System altogether. The reason is that these are two different problems:
- Escaping is an energy problem, and Earth’s orbit has already paid half the bill. For a circular orbit \(\varepsilon = -\mu_S/2r\), i.e. \(\lvert\varepsilon\rvert\) equals the kinetic energy already possessed. Doubling the kinetic energy suffices, and doubling energy costs only a factor \(\sqrt2\) in speed.
- Falling in is an angular-momentum problem, and gravity does not help. The Sun subtends a target of radius \(R_S = a_{SE}/215\). What keeps the spacecraft out is not energy but its own transverse motion, \(h = r v_\theta\), and the only way to lower perihelion onto the Sun is to destroy nearly all of \(v_\theta\).
Quantitatively, a single tangential burn that lowers perihelion to exactly \(R_S\) needs the apoapsis speed of an ellipse with \(r_a = 1\ \mathrm{AU}\), \(r_p = R_S\):
\[ v_a=\sqrt{\frac{2\mu_S R_S}{r_a\left(r_a+R_S\right)}} \approx 2.87\ \mathrm{km/s}, \qquad \Delta v = 29.8-2.87 \approx 26.9\ \mathrm{km/s}, \]
still more than twice the escape budget. This is why 
did not simply brake towards the Sun: it used seven Venus gravity assists to shed angular momentum over seven years.
| Manoeuvre from Earth’s orbit | \(\Delta v\) |
|---|---|
| Escape the Solar System | \(12.3\ \mathrm{km/s}\) |
| Graze the Sun’s surface (\(r_p = R_S\)) | \(26.9\ \mathrm{km/s}\) |
| Fall radially into the Sun (\(v_\theta = 0\)) | \(29.8\ \mathrm{km/s}\) |
(h) Slowing to \(10\ \mathrm{km/s}\). With \(\mathbf r = [1.496\times10^{8},0,0]^{T}\ \mathrm{km}\) and \(\mathbf v = [0,10,0]^{T}\ \mathrm{km/s}\):
\[ \varepsilon = \frac{10^{2}}{2}-\frac{1.3271\times10^{11}}{1.496\times10^{8}} = 50 - 887.1 = -837.1\ \mathrm{km^2/s^2}, \]
\[ h = r v = (1.496\times10^{8})(10) = 1.496\times10^{9}\ \mathrm{km^2/s}, \]
\[ a = -\frac{\mu_S}{2\varepsilon} = \frac{1.3271\times10^{11}}{1674.2} = 7.93\times10^{7}\ \mathrm{km} = 0.53\ \mathrm{AU}, \]
\[ e = \sqrt{1+\frac{2\varepsilon h^{2}}{\mu_S^{2}}} = \sqrt{1-0.2128} = 0.887 , \]
\[ r_p = a(1-e) = (7.93\times10^{7})(0.1127) = 8.94\times10^{6}\ \mathrm{km} = 0.060\ \mathrm{AU}. \]
(Check: \(r_p = p/(1+e)\) with \(p = h^{2}/\mu_S = 1.686\times10^{7}\ \mathrm{km}\) gives the same \(8.94\times10^{6}\ \mathrm{km}\).)
\[ \boxed{\ r_p = 8.94\times10^{6}\ \mathrm{km} \approx 12.8\,R_S \;>\; R_S\ } \]
The spacecraft misses the Sun. Removing two-thirds of the Earth’s orbital speed — a colossal \(19.8\ \mathrm{km/s}\) burn — still leaves a perihelion almost thirteen solar radii out. For comparison, Parker Solar Probe reaches about \(9.9\,R_S\). That single number is the most vivid statement of how deep and how steep the Sun’s gravity well really is.
8. An Earth satellite is in an elliptical orbit with perigee altitude \(z_p = 400\ \mathrm{km}\) and apogee altitude \(z_a = 4000\ \mathrm{km}\). Taking \(\mu = 398\,600\ \mathrm{km^3/s^2}\) and \(R_E = 6378\ \mathrm{km}\), determine
- the eccentricity \(e\);
- the specific angular momentum \(h\);
- the perigee speed \(v_p\);
- the apogee speed \(v_a\);
- the semi-major axis \(a\);
- the orbital period \(T\);
- the true-anomaly-averaged radius \(\bar r_f\);
- the true anomaly at which \(r = \bar r_f\);
- the speed at \(r = \bar r_f\);
- the flight path angle \(\gamma\) at \(r = \bar r_f\);
- the maximum flight path angle \(\gamma_{\max}\) and the true anomaly at which it occurs.
Answer
Setup: altitudes are not radii. The data are altitudes, measured from the surface; every orbital formula needs distances from the focus, i.e. from the Earth’s centre:
\[ r_p = R_E+z_p = 6378+400 = 6778\ \mathrm{km}, \qquad r_a = R_E+z_a = 6378+4000 = 10\,378\ \mathrm{km}. \]
This is the single most common slip in the whole problem.
(a) Eccentricity. From \(r_p = a(1-e)\) and \(r_a = a(1+e)\), adding gives \(r_p+r_a = 2a\) and subtracting gives \(r_a-r_p = 2ae\), so their ratio eliminates \(a\) entirely:
\[ e = \frac{r_a-r_p}{r_a+r_p} = \frac{3600}{17\,156} \qquad\Longrightarrow\qquad \boxed{\ e = 0.20984\ } \]
(b) Angular momentum. The semi-latus rectum can be written without \(a\) as the harmonic mean of the apsidal radii,
\[ p = a\left(1-e^{2}\right) = \frac{2r_pr_a}{r_p+r_a} = \frac{2(6778)(10\,378)}{17\,156} = 8200.29\ \mathrm{km}, \]
and since \(p = h^{2}/\mu\),
\[ h = \sqrt{\mu p} = \sqrt{(398\,600)(8200.29)} \qquad\Longrightarrow\qquad \boxed{\ h = 5.7172\times10^{4}\ \mathrm{km^2/s}\ } \]
(c), (d) Apsidal speeds. At an apse the radial velocity vanishes, so the velocity is purely transverse and \(h = rv\):
\[ v_p = \frac{h}{r_p} = \frac{57\,172}{6778} = \boxed{8.435\ \mathrm{km/s}}, \qquad v_a = \frac{h}{r_a} = \frac{57\,172}{10\,378} = \boxed{5.509\ \mathrm{km/s}}. \]
As required by \(r_pv_p = r_av_a = h\), the satellite is faster at perigee. This also checks the identity of Problem 6:
\[ v_pv_a = (8.435)(5.509) = 46.47 = \frac{\mu}{a} = \frac{398\,600}{8578}. \quad\checkmark \]
(e) Semi-major axis.
\[ a = \frac{r_p+r_a}{2} = \frac{6778+10\,378}{2} = \boxed{8578\ \mathrm{km}}. \]
(f) Period. From Kepler’s third law,
\[ T = 2\pi\sqrt{\frac{a^{3}}{\mu}} = 2\pi\sqrt{\frac{8578^{3}}{398\,600}} = 7906.6\ \mathrm{s} \qquad\Longrightarrow\qquad \boxed{\ T = 131.78\ \mathrm{min} = 2.196\ \mathrm{h}\ } \]
(g) True-anomaly-averaged radius. Averaging the orbit equation \(r = p/(1+e\cos f)\) uniformly in \(f\),
\[ \bar r_f = \frac{1}{2\pi}\int_0^{2\pi}\frac{p\,\mathrm df}{1+e\cos f} = \frac{p}{\sqrt{1-e^{2}}}, \]
using the standard integral \(\displaystyle\int_0^{2\pi}\frac{\mathrm df}{1+e\cos f} = \frac{2\pi}{\sqrt{1-e^{2}}}\). With \(p = a(1-e^{2})\) this collapses to a remarkably clean result:
\[ \boxed{\ \bar r_f = a\sqrt{1-e^{2}} = b\ } = 8578\sqrt{1-0.209839^{2}} = 8387.02\ \mathrm{km}, \]
the semi-minor axis — also the geometric mean \(\sqrt{r_pr_a}\) of the apsidal radii. The corresponding mean altitude is \(\bar z_f = 8387.02-6378 \approx 2009\ \mathrm{km}\).
This is an average with respect to true anomaly, not with respect to time. The satellite loiters near apogee, so the time average is larger: \(\bar r_t = a\left(1+e^{2}/2\right) = 8766.9\ \mathrm{km}\). Averaging over a different variable gives a different answer — always say which one you mean.
(h) True anomaly at \(r = \bar r_f\). Inverting the orbit equation,
\[ \cos f = \frac{1}{e}\left(\frac{p}{\bar r_f}-1\right) = \frac{8200.29/8387.02-1}{0.209839} = -0.10610, \]
so
\[ \boxed{\ f = 96.09^{\circ}\ \ (\text{outbound}) \qquad\text{and}\qquad f = 263.91^{\circ}\ \ (\text{inbound})\ } \]
the two being mirror images about the apse line.
(i) Speed there. From vis-viva,
\[ v = \sqrt{\mu\left(\frac{2}{\bar r_f}-\frac1a\right)} = \sqrt{398\,600\left(\frac{2}{8387.02}-\frac{1}{8578}\right)} = \boxed{6.970\ \mathrm{km/s}}, \]
identical at both true anomalies, since \(v\) depends on position only through \(r\).
(j) Flight path angle there. Using \(\tan\gamma = \dfrac{e\sin f}{1+e\cos f}\) from Problem 4,
\[ \tan\gamma = \frac{0.209839\sin 96.09^{\circ}}{1+0.209839\cos 96.09^{\circ}} = 0.21341, \]
so
\[ \boxed{\ \gamma = +12.05^{\circ}\ \text{(outbound)}, \qquad -12.05^{\circ}\ \text{(inbound)}\ } \]
The sign follows the sign of \(\sin f\): climbing away from perigee gives \(\gamma>0\), falling back gives \(\gamma<0\).
(k) Maximum flight path angle. Since \(\lvert\gamma\rvert<90^{\circ}\), maximising \(\gamma\) is the same as maximising \(\tan\gamma\). With \(F(f) = \dfrac{e\sin f}{1+e\cos f}\),
\[ \frac{\mathrm dF}{\mathrm df} = \frac{e\cos f\left(1+e\cos f\right)+e^{2}\sin^{2}f}{\left(1+e\cos f\right)^{2}} = \frac{e\cos f+e^{2}}{\left(1+e\cos f\right)^{2}}, \]
where \(\sin^{2}f+\cos^{2}f=1\) has collapsed the numerator. Setting this to zero and dividing by \(e\neq0\),
\[ \boxed{\ \cos f_{\gamma_{\max}} = -e\ } \qquad\Longrightarrow\qquad f_{\gamma_{\max}} = \cos^{-1}(-0.209839) = \boxed{102.11^{\circ}}. \]
At that point \(\sin f = \sqrt{1-e^{2}}\), so
\[ \tan\gamma_{\max} = \frac{e\sqrt{1-e^{2}}}{1-e^{2}} = \frac{e}{\sqrt{1-e^{2}}} \qquad\Longrightarrow\qquad \boxed{\ \sin\gamma_{\max} = e\ }, \]
a result of striking simplicity: the maximum flight path angle depends on the eccentricity alone, not on \(a\) or \(\mu\). Here
\[ \gamma_{\max} = \sin^{-1}(0.209839) = \boxed{12.11^{\circ}}, \]
reached outbound at \(f = 102.11^{\circ}\), and \(\gamma = -12.11^{\circ}\) inbound at \(f = 257.89^{\circ}\).
Note where this happens: \(\cos f = -e\) gives \(r = \dfrac{p}{1-e^{2}} = a\). The velocity tilts most steeply exactly where \(r = a\), i.e. at the ends of the minor axis. Part (j) evaluated \(\gamma\) at \(r = b = 8387\ \mathrm{km}\), close to but not at that point, which is why \(12.05^{\circ}\) falls just short of \(12.11^{\circ}\).
Summary.
| Quantity | Result |
|---|---|
| Perigee / apogee radius \(r_p,\ r_a\) | \(6778\ \mathrm{km}\), \(10\,378\ \mathrm{km}\) |
| Eccentricity \(e\) | \(0.20984\) |
| Angular momentum \(h\) | \(57\,172\ \mathrm{km^2/s}\) |
| Perigee / apogee speed \(v_p,\ v_a\) | \(8.435\ \mathrm{km/s}\), \(5.509\ \mathrm{km/s}\) |
| Semi-major axis \(a\) | \(8578\ \mathrm{km}\) |
| Period \(T\) | \(7906.6\ \mathrm{s} = 131.78\ \mathrm{min}\) |
| Averaged radius \(\bar r_f = b\) | \(8387.02\ \mathrm{km}\) |
| \(f\) at \(r = \bar r_f\) | \(96.09^{\circ},\ 263.91^{\circ}\) |
| Speed at \(r = \bar r_f\) | \(6.970\ \mathrm{km/s}\) |
| \(\gamma\) at \(r = \bar r_f\) | \(\pm12.05^{\circ}\) |
| \(\gamma_{\max}\), at \(f\) | \(12.11^{\circ}\), at \(102.11^{\circ}\) |
Results worth carrying away. For any ellipse,
\[ e=\frac{r_a-r_p}{r_a+r_p},\qquad a=\frac{r_a+r_p}{2},\qquad p=a\left(1-e^{2}\right)=\frac{2r_pr_a}{r_p+r_a},\qquad h=\sqrt{\mu p}, \]
\[ v_p=\frac{h}{r_p},\qquad v_a=\frac{h}{r_a},\qquad T=2\pi\sqrt{\frac{a^{3}}{\mu}},\qquad \bar r_f=a\sqrt{1-e^{2}}=b, \]
\[ \tan\gamma=\frac{e\sin f}{1+e\cos f},\qquad \gamma_{\max}=\sin^{-1}e \ \ \text{at}\ \ \cos f=-e \ \ (r=a). \]
