Tutorial 1

Who pulls harder, what a state vector tells you, and how deep the Sun’s gravity well is

Author

Devendra Ghate

Published

October 16, 2026

How to use this tutorial

The problems below use the material of the \(N\)-body note and the two-body note. Each problem has two hidden parts. Open the Hint only after you have tried for a while; it tells you how to start, not what the answer is. Open the Solution only after you have an answer of your own, and then compare not just the number but the route.

Several solutions end with a remark that goes beyond the question. Those remarks are the point of the tutorial: each problem is chosen because its answer is surprising, or because it exposes a step that is easy to get wrong.

Data

quantity symbol value
Sun’s gravitational parameter \(\mu_S\) \(1.3271\times10^{20}\) m\(^3\)/s\(^2\) \(=1.3271\times10^{11}\) km\(^3\)/s\(^2\)
Earth’s gravitational parameter \(\mu_E\) \(3.9860\times10^{14}\) m\(^3\)/s\(^2\) \(=3.9860\times10^{5}\) km\(^3\)/s\(^2\)
Moon’s gravitational parameter \(\mu_M\) \(4.905\times10^{12}\) m\(^3\)/s\(^2\)
astronomical unit (Sun–Earth \(a\)) \(a_{SE}\) \(1.496\times10^{11}\) m \(=149.6\times10^{6}\) km
Earth–Moon semi-major axis \(a_{EM}\) \(3.844\times10^{8}\) m \(=384\,400\) km
Earth’s radius \(R_E\) \(6378\) km
Sun’s radius \(R_S\) \(6.957\times10^{5}\) km

Problem 1: who pulls the Moon harder?

Compute the gravitational acceleration of the Moon due to the Earth and due to the Sun. Which is larger, and by what factor? If the answer surprises you, explain in what sense the Moon can still be said to orbit the Earth.

Hint A body at distance \(r\) from a mass with gravitational parameter \(\mu\) has acceleration \(\mu/r^2\). The Moon is essentially \(1\) AU from the Sun. For the second part, ask what decides the motion of the Moon relative to the Earth: the Sun’s pull on the Moon, or something else?
Solution

Due to the Earth, with \(r_{EM}=3.844\times10^{8}\) m:

\[ a_{E\to M}=\frac{3.9860\times10^{14}}{(3.844\times10^{8})^2}=\frac{3.9860\times10^{14}}{1.478\times10^{17}}\approx2.70\times10^{-3}\ \text{m/s}^2 . \]

Due to the Sun, with \(r_{SM}\approx1.496\times10^{11}\) m:

\[ a_{S\to M}=\frac{1.3271\times10^{20}}{(1.496\times10^{11})^2}=\frac{1.3271\times10^{20}}{2.238\times10^{22}}\approx5.93\times10^{-3}\ \text{m/s}^2 . \]

\[ \boxed{\;\frac{a_{S\to M}}{a_{E\to M}}\approx2.2\;} \]

The Sun pulls the Moon more than twice as hard as the Earth does.

Why the Earth–Moon two-body picture still works. The relative motion \(\ddot{\mathbf r}_{EM}\) is driven not by the Sun’s pull on the Moon, but by the difference between the Sun’s pulls on the Moon and on the Earth, since the Sun accelerates both. The Earth and Moon are at nearly the same distance from the Sun, so the difference is small. To first order in \(r_{EM}/r_{SE}\) (the tidal acceleration),

\[ a_{\rm tidal}\approx\frac{2\mu_Sr_{EM}}{r_{SE}^3}=\frac{2(1.3271\times10^{20})(3.844\times10^{8})}{(1.496\times10^{11})^3}\approx3.0\times10^{-5}\ \text{m/s}^2, \]

about \(1.1\%\) of \(a_{E\to M}\). Relative to the Earth, the Sun is a small perturbation. (The tidal term is treated properly in the patched-conics note; here it is enough to see that it is the difference of accelerations, not the accelerations themselves, that matters.)

Problem 2: who pulls the Earth harder?

Compute the acceleration of the Earth due to the Sun and due to the Moon, and form their ratio. Contrast it with the ratio of Problem 1 and explain the difference.

Hint The Earth is accelerated by the Moon through the Moon’s gravitational parameter, at the same separation \(r_{EM}\) as in Problem 1. What is the same in the two problems, and what has changed?
Solution

\[ a_{S\to E}=\frac{\mu_S}{r_{SE}^2}\approx5.93\times10^{-3}\ \text{m/s}^2, \qquad a_{M\to E}=\frac{\mu_M}{r_{EM}^2}=\frac{4.905\times10^{12}}{1.478\times10^{17}}\approx3.32\times10^{-5}\ \text{m/s}^2, \]

\[ \boxed{\;\frac{a_{S\to E}}{a_{M\to E}}\approx179\;} \]

against \(2.2\) in Problem 1. The two ratios differ by a factor of about \(81\):

\[ \frac{a_{S\to E}/a_{M\to E}}{a_{S\to M}/a_{E\to M}}=\frac{\mu_E}{\mu_M}=\frac{M_E}{M_M}\approx81.3 , \]

because the Sun’s accelerations of the Earth and of the Moon are essentially equal (both are \(1\) AU away), while the Earth–Moon pair swaps \(\mu_E\) for \(\mu_M\) at the same separation.

Problem 3: a coincidence

What is the ratio of the Sun–Earth semi-major axis to the Earth–Moon semi-major axis? Compare it with the ratio of the diameters of the Sun (\(1.392\times10^{6}\) km) and the Moon (\(3475\) km), and comment on the coincidence.

Hint The angle a body subtends is about (diameter)/(distance).
Solution

\[ \frac{a_{SE}}{a_{EM}}=\frac{149.6\times10^{6}}{384\,400}\approx389, \qquad \frac{D_S}{D_M}=\frac{1.392\times10^{6}}{3475}\approx400 . \]

The two ratios nearly agree, so the Sun and the Moon subtend almost the same angle as seen from the Earth: about \(0.53^\circ\) and \(0.52^\circ\). That near-equality is why total solar eclipses, with the Moon just covering the Sun’s disc and leaving the corona visible, are possible at all. It is an accident of the present epoch: the Moon recedes by about \(3.8\) cm per year, and in some hundreds of millions of years only annular eclipses will remain.

Problem 4: what a state vector tells you

Given the position and velocity vectors \(\mathbf r\), \(\mathbf v\) of a body at one instant and the gravitational parameter \(\mu\), write down an algorithm that returns the specific angular momentum \(h\), the specific energy \(\varepsilon\), the eccentricity \(e\), the semi-latus rectum \(p\), the semi-major axis \(a\), the periapsis distance \(r_p\) and the flight path angle \(\gamma\). State the order of the computation: which quantity needs which.

Hint Start from the three integrals of the two-body problem. Each of them can be computed from \(\mathbf r\) and \(\mathbf v\) directly. Then ask which geometric quantity follows from which integral. For \(\gamma\), split \(\mathbf v\) into a part along \(\mathbf r\) and a part perpendicular to it.
Solution

Every line uses only quantities already computed above it.

\[ \begin{aligned} \mathbf 1.\quad&r=\lVert\mathbf r\rVert,\qquad v=\lVert\mathbf v\rVert\\ \mathbf 2.\quad&\mathbf h=\mathbf r\times\mathbf v,\qquad h=\lVert\mathbf h\rVert\\ \mathbf 3.\quad&\varepsilon=\frac{v^2}2-\frac\mu r\\ \mathbf 4.\quad&\mathbf e=\frac{\mathbf v\times\mathbf h}\mu-\frac{\mathbf r}r,\qquad e=\lVert\mathbf e\rVert\\ \mathbf 5.\quad&p=\frac{h^2}\mu\\ \mathbf 6.\quad&a=-\frac\mu{2\varepsilon}\\ \mathbf 7.\quad&r_p=\frac p{1+e}\\ \mathbf 8.\quad&\gamma=\operatorname{atan2}\left(\mathbf r\cdot\mathbf v,\ h\right) \end{aligned} \]

The dependencies are \((\mathbf r,\mathbf v)\to r,v\to\mathbf h\to\varepsilon\to\mathbf e\to p\to a\to r_p,\gamma\). Steps 1–4 extract the three integrals of motion; steps 5–8 are geometry read off from them.

Five remarks, each worth more than the algorithm itself.

  • Step 3 already classifies the orbit: \(\varepsilon<0\) ellipse, \(\varepsilon=0\) parabola, \(\varepsilon>0\) hyperbola, before any conic is drawn.
  • The eccentricity needs no vectors: \(e=\sqrt{1+2\varepsilon h^2/\mu^2}\) from steps 2 and 3. The vector \(\mathbf e\) is needed only for the direction of periapsis.
  • \(a\) depends only on the magnitudes \(r\) and \(v\): \(a=(2/r-v^2/\mu)^{-1}\) from vis-viva. The direction of \(\mathbf v\) enters only through \(h\), hence only through \(p\), \(e\) and \(\gamma\).
  • Why \(p\) before \(a\), and \(r_p=p/(1+e)\) rather than \(a(1-e)\). The pair \((p,e)\) describes every conic, while \(a=p/(1-e^2)\) blows up at \(e=1\). So \(r_p=p/(1+e)\) works for the ellipse, the parabola (\(r_p=p/2\)) and the hyperbola alike, and in floating point \((p,e)\) is the safer pair near \(e=1\). (The apoapsis \(r_a=p/(1-e)\) exists only for the ellipse.)
  • The flight path angle. With \(v_r=\mathbf r\cdot\mathbf v/r\) and \(v_\perp=h/r\), \(\tan\gamma=v_r/v_\perp=\mathbf r\cdot\mathbf v/h\), \(\sin\gamma=v_r/v\), \(\cos\gamma=h/(rv)\), so \(h=rv\cos\gamma\). The two-argument arctangent keeps the sign: \(\gamma>0\) outbound, \(\gamma<0\) inbound, and \(\gamma=0\) exactly when \(\mathbf r\cdot\mathbf v=0\), i.e. at an apsis. That is the quickest test of whether a state vector sits at periapsis or apoapsis.
Two degenerate cases: if \(h=0\) the motion is rectilinear (\(p=0\), \(e=1\) whatever the energy); if \(e=0\) the orbit is circular and has no periapsis direction.

Problem 5: will the asteroid hit?

A newly discovered asteroid is tracked and, at one instant, its geocentric state is

\[ \mathbf r=(10\,000,\ 0,\ 0)\ \text{km},\qquad\mathbf v=(0,\ 5.15494,\ 0)\ \text{km/s}. \]

With \(\mu_E=398\,600.44\) km\(^3\)/s\(^2\) and \(R_E=6378\) km, find \(h\), \(\varepsilon\), \(a\), \(e\) and the perigee radius \(r_p\). Does the asteroid strike the Earth? If so, where on its orbit, and how fast?

Hint Use the algorithm of Problem 4. Then compare \(r_p\) with \(R_E\). Is the given point perigee, apogee, or neither? For the impact point, solve the orbit equation for the true anomaly at which \(r=R_E\), and think about which of the two solutions the asteroid reaches first.
Solution

Angular momentum. The velocity is perpendicular to \(\mathbf r\), so \(\mathbf h=(0,\ 0,\ 51\,549.4)\) km\(^2\)/s and \(h=51\,549.4\) km\(^2\)/s.

Energy. \(v^2=26.5734\) km\(^2\)/s\(^2\), so \(\varepsilon=13.2867-39.8600=-26.5734\) km\(^2\)/s\(^2<0\): an ellipse.

Size and shape. \(a=-\mu/(2\varepsilon)=7500\) km, and \(e=\sqrt{1+2\varepsilon h^2/\mu^2}=1/3\). Since \(\mathbf r\cdot\mathbf v=0\) the given point is an apsis, and since \(r=10\,000\) km \(>a\) it is the apogee.

Perigee. \(r_p=a(1-e)=5000\) km.

\[ \boxed{\;r_p=5000\ \text{km}<R_E=6378\ \text{km}\;} \]

The orbit dips \(1378\) km below the surface: the asteroid hits the Earth. It never reaches the mathematical perigee.

Where. With \(p=h^2/\mu=6666.7\) km, the orbit reaches \(r=R_E\) where

\[ \cos f=\frac1e\left(\frac p{R_E}-1\right)=3\left(\frac{6666.7}{6378}-1\right)=0.1358, \]

i.e. \(f=\pm82.2^\circ\). The asteroid starts at apogee (\(f=180^\circ\)) and moves on to larger \(f\), falling towards perigee, so it meets the surface first at \(f=360^\circ-82.2^\circ=277.8^\circ\), still descending (flight path angle \(-17.5^\circ\)).

How fast. Vis-viva at \(r=R_E\): \(v=\sqrt{\mu(2/R_E-1/a)}=8.48\) km/s, ignoring the atmosphere.

“Will it hit?” is answered by one scalar, \(r_p\), extracted from one state vector: no propagation in time is needed.

Problem 6: the product of the apsidal speeds

Show that for any elliptic orbit the speeds at periapsis and apoapsis satisfy

\[ v_pv_a=\frac{\mu}{a}. \]

Interpret the result. Can you find a second proof?

Hint Vis-viva at \(r_p=a(1-e)\) and \(r_a=a(1+e)\) gives each speed. For the second proof, use angular momentum at the two apsides, where the velocity is perpendicular to \(\mathbf r\).
Solution

From vis-viva at periapsis,

\[ v_p^2=\mu\left[\frac2{a(1-e)}-\frac1a\right]=\frac\mu a\cdot\frac{2-(1-e)}{1-e}=\frac\mu a\cdot\frac{1+e}{1-e}, \]

and in the same way \(v_a^2=\dfrac\mu a\cdot\dfrac{1-e}{1+e}\). Multiplying, the eccentricity factors are reciprocal and cancel: \(v_p^2v_a^2=\mu^2/a^2\), so \(v_pv_a=\mu/a\).

Interpretation. \(\sqrt{v_pv_a}=\sqrt{\mu/a}\): the geometric mean of the apsidal speeds is the speed of a circular orbit of radius \(a\). Stretching an ellipse at fixed \(a\) speeds up periapsis and slows apoapsis by exactly reciprocal factors \(\sqrt{(1+e)/(1-e)}\).

Second proof. \(h=r_pv_p=r_av_a\), so \(v_pv_a=h^2/(r_pr_a)=\mu a(1-e^2)/\left[a^2(1-e^2)\right]=\mu/a\).

Problem 7: how deep is the Sun’s gravity well?

A spacecraft moves with the Earth on a circular orbit about the Sun at \(1\) AU. Ignore all planetary gravity and treat the Sun as a point mass.

  1. Compute the potential \(\Phi=-\mu_S/r\) at the Sun’s surface and at \(1\) AU, and compare.
  2. Find the escape speed from the Sun’s surface.
  3. Find the escape speed from the Sun at \(1\) AU.
  4. Find the Earth’s orbital speed \(v_E\).
  5. Find the smallest prograde \(\Delta v\) at \(1\) AU that puts the spacecraft on a solar escape trajectory.
  6. Find the retrograde \(\Delta v\) that removes all of its velocity about the Sun.
  7. Explain the surprise: why is it so much harder to fall into the Sun than to leave the solar system?
  8. Suppose instead the speed is cut from \(v_E\) to \(10\) km/s, still along the orbit. Find \(\varepsilon\), \(h\), \(a\), \(e\) and \(r_p\). Does the spacecraft hit the Sun?
Hint (a)–(f) are one-liners with vis-viva and \(v_{\rm esc}=\sqrt{2\mu/r}\). For (g), ask what escaping needs (energy) and what hitting the Sun needs (very low perihelion, i.e. very little angular momentum). For (h), use the algorithm of Problem 4.
Solution

(a) \(\Phi(R_S)=-1.3271\times10^{11}/6.957\times10^{5}=-1.908\times10^{5}\) km\(^2\)/s\(^2\) and \(\Phi(1~\text{AU})=-887.1\) km\(^2\)/s\(^2\). The ratio is \(215=1~\text{AU}/R_S\). At \(1\) AU, \(99.5\%\) of the climb out of the Sun’s well is already done.

(b) \(v_{\rm esc}(R_S)=\sqrt{2\mu_S/R_S}\approx617.7\) km/s.

(c) \(v_{\rm esc}(1~\text{AU})=\sqrt{2(887.1)}\approx42.1\) km/s.

(d) \(v_E=\sqrt{\mu_S/r}=\sqrt{887.1}\approx29.8\) km/s, and \(v_{\rm esc}=\sqrt2\,v_E\), as for any circular orbit.

(e) \(\Delta v=(\sqrt2-1)v_E\approx12.3\) km/s.

(f) \(\Delta v=v_E\approx29.8\) km/s.

(g) Falling into the Sun costs \(2.4\) times as much as leaving the solar system. The two are different problems.

  • Escape is an energy problem, and the circular orbit has paid half the bill. For a circular orbit, \(|\varepsilon|=\mu/(2r)\) equals the kinetic energy already possessed; doubling the kinetic energy is enough, and that costs only a factor \(\sqrt2\) in speed.
  • Hitting the Sun is an angular-momentum problem, and gravity does not help. The Sun is a small target, \(R_S=1~\text{AU}/215\). What keeps the spacecraft out is its transverse speed, and lowering the perihelion to the Sun’s surface means destroying nearly all of it.

Quantitatively, a tangential burn that lowers the perihelion to exactly \(R_S\) needs the aphelion speed of an ellipse with \(r_a=1\) AU and \(r_p=R_S\):

\[ v_a=\sqrt{\frac{2\mu_SR_S}{r_a(r_a+R_S)}}\approx2.87\ \text{km/s},\qquad\Delta v=29.8-2.87\approx26.9\ \text{km/s}, \]

still more than twice the escape budget.

manoeuvre from the Earth’s orbit \(\Delta v\)
escape the solar system \(12.3\) km/s
graze the Sun’s surface (\(r_p=R_S\)) \(26.9\) km/s
fall straight into the Sun \(29.8\) km/s

This is why the Parker Solar Probe did not simply brake towards the Sun: it used seven Venus gravity assists to shed angular momentum over several years.

(h) With \(r=1.496\times10^{8}\) km and \(v=10\) km/s along the orbit:

\[ \varepsilon=50-887.1=-837.1\ \text{km}^2/\text{s}^2,\qquad h=rv=1.496\times10^{9}\ \text{km}^2/\text{s}, \]

\[ a=-\frac{\mu_S}{2\varepsilon}=7.93\times10^{7}\ \text{km}=0.53\ \text{AU},\qquad e=\sqrt{1+\frac{2\varepsilon h^2}{\mu_S^2}}=\sqrt{1-0.2128}=0.887, \]

\[ r_p=a(1-e)=8.94\times10^{6}\ \text{km}=0.060\ \text{AU}\approx12.8\,R_S . \]

(Check: \(p=h^2/\mu_S=1.686\times10^{7}\) km and \(p/(1+e)\) gives the same \(r_p\).) The spacecraft misses the Sun. Removing two thirds of the Earth’s orbital speed, a \(19.8\) km/s burn, still leaves a perihelion almost thirteen solar radii out; the Parker Solar Probe, after all its Venus flybys, reaches about ten.

Problem 8: everything about one ellipse

A satellite has perigee altitude \(z_p=400\) km and apogee altitude \(z_a=4000\) km. With \(\mu=398\,600\) km\(^3\)/s\(^2\) and \(R_E=6378\) km, find

  1. the eccentricity \(e\); (b) the angular momentum \(h\); (c) the perigee speed \(v_p\); (d) the apogee speed \(v_a\); (e) the semi-major axis \(a\); (f) the period \(T\); (g) the radius averaged over true anomaly, \(\bar r_f\); (h) the true anomaly at which \(r=\bar r_f\); (i) the speed there; (j) the flight path angle there; (k) the largest flight path angle \(\gamma_{\max}\), and the true anomaly at which it occurs.
Hint First convert altitudes to radii; every orbital formula measures distance from the Earth’s centre. For (b), write \(p\) in terms of \(r_p\) and \(r_a\) alone. For (g), average \(r=p/(1+e\cos f)\) over \(0\le f<2\pi\), using \(\int_0^{2\pi}df/(1+e\cos f)=2\pi/\sqrt{1-e^2}\). For (k), maximise \(\tan\gamma=e\sin f/(1+e\cos f)\).
Solution

Altitudes are not radii. \(r_p=6378+400=6778\) km and \(r_a=6378+4000=10\,378\) km. Forgetting this is the most common slip in the whole problem.

(a) Adding and subtracting \(r_p=a(1-e)\) and \(r_a=a(1+e)\) eliminates \(a\): \(e=(r_a-r_p)/(r_a+r_p)=3600/17\,156=0.20984\).

(b) The semi-latus rectum is the harmonic mean of the apsidal radii, \(p=a(1-e^2)=2r_pr_a/(r_p+r_a)=8200.29\) km, so \(h=\sqrt{\mu p}=5.7172\times10^{4}\) km\(^2\)/s.

(c), (d) At an apsis the velocity is purely transverse, so \(h=rv\): \(v_p=h/r_p=8.435\) km/s and \(v_a=h/r_a=5.509\) km/s. Check with Problem 6: \(v_pv_a=46.47=\mu/a\).

(e) \(a=(r_p+r_a)/2=8578\) km.

(f) \(T=2\pi\sqrt{a^3/\mu}=7906.6\) s \(=131.78\) min \(=2.196\) h.

(g) Averaging over \(f\),

\[ \bar r_f=\frac1{2\pi}\int_0^{2\pi}\frac{p\,df}{1+e\cos f}=\frac p{\sqrt{1-e^2}}=a\sqrt{1-e^2}=b=8387.02\ \text{km}, \]

the semi-minor axis, which is also the geometric mean \(\sqrt{r_pr_a}\). Its altitude is about \(2009\) km. This is an average over true anomaly, not over time: the satellite lingers near apogee, so the time average is larger, \(\bar r_t=a(1+e^2/2)=8766.9\) km. Always say what you average over.

(h) Inverting the orbit equation, \(\cos f=(p/\bar r_f-1)/e=-0.10610\), so \(f=96.09^\circ\) (outbound) and \(263.91^\circ\) (inbound), mirror images about the line of apsides.

(i) Vis-viva: \(v=\sqrt{\mu(2/\bar r_f-1/a)}=6.970\) km/s, the same at both points, since \(v\) depends only on \(r\).

(j) \(\tan\gamma=e\sin f/(1+e\cos f)=0.21341\), so \(\gamma=+12.05^\circ\) outbound and \(-12.05^\circ\) inbound.

(k) With \(F(f)=e\sin f/(1+e\cos f)\),

\[ \frac{dF}{df}=\frac{e\cos f(1+e\cos f)+e^2\sin^2f}{(1+e\cos f)^2}=\frac{e\cos f+e^2}{(1+e\cos f)^2}, \]

which vanishes at \(\cos f=-e\), i.e. \(f=102.11^\circ\). There \(\sin f=\sqrt{1-e^2}\) and \(\tan\gamma_{\max}=e/\sqrt{1-e^2}\), i.e. \(\sin\gamma_{\max}=e\): the steepest climb depends on \(e\) alone. Here \(\gamma_{\max}=12.11^\circ\) at \(f=102.11^\circ\), and \(-12.11^\circ\) at \(f=257.89^\circ\).

At \(\cos f=-e\) the radius is \(r=p/(1-e^2)=a\): the steepest point is at the ends of the minor axis. Part (j) evaluated \(\gamma\) at \(r=b=8387\) km, close to but not at \(r=a=8578\) km, which is why \(12.05^\circ\) falls just short of \(12.11^\circ\).

What to carry away

  • The size of an acceleration says little about which body “dominates” a motion: what matters is the difference of accelerations across the system (Problems 1, 2).
  • A single state vector gives the whole orbit: \(h\), \(\varepsilon\) and \(\mathbf e\) first, then the geometry (Problem 4). Whether an orbit hits the planet is one number, \(r_p\) (Problem 5).
  • For any ellipse: \(e=\dfrac{r_a-r_p}{r_a+r_p}\), \(a=\dfrac{r_a+r_p}2\), \(p=\dfrac{2r_pr_a}{r_p+r_a}\), \(h=\sqrt{\mu p}\), \(v_{p,a}=h/r_{p,a}\), \(v_pv_a=\mu/a\), \(T=2\pi\sqrt{a^3/\mu}\), and \(\gamma_{\max}=\arcsin e\) at \(r=a\) (Problems 6, 8).
  • Escaping is about energy; falling in is about angular momentum (Problem 7).

All the numbers in the solutions are checked by tutorial-01-verify.py.