11.P — Worked Problems: Astronomy and Orbital Mechanics

Twelve problems across Part 11. Every solution shows the arithmetic. Attempt each before opening it.

Problem 1 — Weighing the Sun

The Earth orbits at 1.496\times10^{11} m with a period of 365.25 days. Find the Sun's mass.

Solution

T^2 = \frac{4\pi^2a^3}{GM} \quad\Longrightarrow\quad M = \frac{4\pi^2a^3}{GT^2}

T = 365.25\times86400 = 3.156\times10^{7}\ \text{s}

a^3 = (1.496\times10^{11})^3 = 3.348\times10^{33}

T^2 = (3.156\times10^{7})^2 = 9.960\times10^{14}

M = \frac{4\pi^2(3.348\times10^{33})}{(6.674\times10^{-11})(9.960\times10^{14})} = \frac{1.322\times10^{35}}{6.647\times10^{4}}

M = 1.989\times10^{30}\ \text{kg}

What to notice. The accepted value is 1.989\times10^{30} kg — exact agreement, because this is how it is defined.

Nobody has weighed the Sun any other way. Every stellar mass, every planetary mass, and the mass of the black hole at the galactic centre are all measured by watching something orbit and applying this equation.

Note that G is the least precisely known fundamental constant, at about 22 parts per million, so the Sun's mass in kilograms inherits that uncertainty. The product GM_\odot is known to twelve figures, which is why orbital mechanics uses \mu = GM as a single parameter rather than G and M separately.

Problem 2 — Geostationary from scratch

Compute the altitude of a geostationary orbit around Mars. Mars's GM = 4.283\times10^{13} m³/s², sidereal day 24 h 37 m 22 s, radius 3390 km.

Solution

T = 24(3600)+37(60)+22 = 86400+2220+22 = 88{,}642\ \text{s}

a = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3}

T^2 = (8.8642\times10^{4})^2 = 7.8574\times10^{9}

GMT^2 = (4.283\times10^{13})(7.8574\times10^{9}) = 3.3653\times10^{23}

\frac{3.3653\times10^{23}}{39.478} = 8.5245\times10^{21}

Cube root: \log_{10}(8.5245\times10^{21}) = 21.9307, divided by 3 is 7.3102:

a = 10^{7.3102} = 2.043\times10^{7}\ \text{m} = 20{,}430\ \text{km}

h = 20430-3390 = 17{,}040\ \text{km}

What to notice. Mars's areostationary orbit is at 17,040 km against Earth's 35,786 km, despite the days being almost the same length. The difference is entirely the mass — Mars is a tenth of Earth's mass, so a \propto M^{1/3} gives 10^{-1/3} = 0.464 of the radius.

A relay satellite there would be extremely useful for Mars missions, and none exists yet. Current Mars communications relay through orbiters in low orbits, which gives intermittent coverage.

Note also that Phobos orbits at 9376 km — below areostationary altitude — which is why it is spiralling inwards (Chapter 11.2). Deimos at 23,460 km is above it and spiralling out.

Problem 3 — Escape velocity and atmospheric retention

Find the escape velocity from the Moon, and compare it with the rms speed of oxygen molecules at the Moon's daytime temperature of 390 K. Moon: M = 7.342\times10^{22} kg, R = 1.737\times10^{6} m.

Solution

v_{\text{esc}} = \sqrt{\frac{2GM}{R}} = \sqrt{\frac{2(6.674\times10^{-11})(7.342\times10^{22})}{1.737\times10^{6}}}

= \sqrt{\frac{9.801\times10^{12}}{1.737\times10^{6}}} = \sqrt{5.643\times10^{6}} = 2375\ \text{m/s}

Oxygen rms speed at 390 K, with m = 5.31\times10^{-26} kg (Chapter 3.2):

v_{\text{rms}} = \sqrt{\frac{3k_BT}{m}} = \sqrt{\frac{3(1.381\times10^{-23})(390)}{5.31\times10^{-26}}} = \sqrt{\frac{1.616\times10^{-20}}{5.31\times10^{-26}}}

= \sqrt{3.043\times10^{5}} = 552\ \text{m/s}

\frac{v_{\text{esc}}}{v_{\text{rms}}} = \frac{2375}{552} = 4.3

What to notice. The rule of thumb is that a body retains a gas over geological time if v_{\text{esc}} > 6v_{\text{rms}}, because the Maxwell–Boltzmann tail (Chapter 3.2) always has some molecules above escape speed and the fraction must be small enough that losses take longer than the age of the solar system.

A ratio of 4.3 is not enough, which is why the Moon has essentially no atmosphere.

Compare Earth: v_{\text{esc}} = 11{,}200 m/s and oxygen's v_{\text{rms}} at 288 K is 474 m/s, giving a ratio of 23.6 — comfortably retained. For hydrogen the ratio is only 5.9, which is why Earth has lost essentially all its free hydrogen.

And Titan retains an atmosphere despite v_{\text{esc}} = 2640 m/s, barely more than the Moon's — because it is at 94 K, so nitrogen's v_{\text{rms}} is only 289 m/s and the ratio is 9.1. Temperature matters as much as gravity.

Problem 4 — Vis-viva at perigee and apogee

A satellite is in an orbit with perigee 300 km and apogee 35,786 km altitude. Find its speed at each point and its period.

Solution

r_p = 6371+300 = 6671\ \text{km}, \qquad r_a = 6371+35786 = 42{,}157\ \text{km}

a = \frac{6671+42157}{2} = 24{,}414\ \text{km}

At perigee:

v_p^2 = GM\left(\frac{2}{r_p}-\frac{1}{a}\right) = 3.986\times10^{14}\left(\frac{2}{6.671\times10^{6}}-\frac{1}{2.4414\times10^{7}}\right)

= 3.986\times10^{14}(2.9980\times10^{-7}-4.0960\times10^{-8}) = 3.986\times10^{14}\times2.5884\times10^{-7}

= 1.0317\times10^{8} \quad\Longrightarrow\quad v_p = 10{,}157\ \text{m/s}

At apogee:

v_a^2 = 3.986\times10^{14}\left(\frac{2}{4.2157\times10^{7}}-4.0960\times10^{-8}\right)

= 3.986\times10^{14}(4.7442\times10^{-8}-4.0960\times10^{-8}) = 3.986\times10^{14}\times6.482\times10^{-9}

= 2.584\times10^{6} \quad\Longrightarrow\quad v_a = 1607\ \text{m/s}

Period:

T = 2\pi\sqrt{\frac{a^3}{GM}} = 2\pi\sqrt{\frac{(2.4414\times10^{7})^3}{3.986\times10^{14}}} = 2\pi\sqrt{\frac{1.4552\times10^{22}}{3.986\times10^{14}}}

= 2\pi\sqrt{3.6507\times10^{7}} = 2\pi(6042) = 37{,}963\ \text{s} = 10.55\ \text{h}

What to notice. The speed ratio is 10157/1607 = 6.32, which equals r_a/r_p = 42157/6671 = 6.32. Angular momentum conservation requires exactly this at the two apsides, where the velocity is purely tangential.

This is the geostationary transfer orbit from Chapter 11.5, and the apogee speed of 1607 m/s is what the second burn works from. The satellite crawls through apogee, spending most of the 10.5-hour period in the upper half of the orbit — which is Kepler's second law in a practical setting.

Problem 5 — Hohmann transfer to Mars

Compute the delta-v and transit time for a Hohmann transfer from Earth's orbit to Mars's, relative to the Sun. Earth at 1.000 AU, Mars at 1.524 AU, GM_\odot = 1.327\times10^{20} m³/s².

Solution

r_1 = 1.496\times10^{11}\ \text{m}, \qquad r_2 = 2.279\times10^{11}\ \text{m}

Earth's orbital speed:

v_1 = \sqrt{\frac{1.327\times10^{20}}{1.496\times10^{11}}} = \sqrt{8.870\times10^{8}} = 29{,}783\ \text{m/s}

First burn:

\frac{2r_2}{r_1+r_2} = \frac{2(2.279)}{3.775} = 1.2074, \qquad \sqrt{1.2074} = 1.0988

\Delta v_1 = 29783(1.0988-1) = 29783\times0.0988 = 2943\ \text{m/s}

Mars's orbital speed:

v_2 = \sqrt{\frac{1.327\times10^{20}}{2.279\times10^{11}}} = \sqrt{5.823\times10^{8}} = 24{,}131\ \text{m/s}

Second burn:

\frac{2r_1}{r_1+r_2} = \frac{2(1.496)}{3.775} = 0.7925, \qquad \sqrt{0.7925} = 0.8902

\Delta v_2 = 24131(1-0.8902) = 24131\times0.1098 = 2650\ \text{m/s}

\Delta v_{\text{total}} = 2943+2650 = 5593\ \text{m/s}

Transit time:

a_t = \frac{1.496+2.279}{2}\times10^{11} = 1.8875\times10^{11}\ \text{m}

t = \pi\sqrt{\frac{(1.8875\times10^{11})^3}{1.327\times10^{20}}} = \pi\sqrt{\frac{6.725\times10^{33}}{1.327\times10^{20}}} = \pi\sqrt{5.068\times10^{13}}

= \pi(7.119\times10^{6}) = 2.237\times10^{7}\ \text{s} = 259\ \text{days}

What to notice. 5593 m/s relative to the Sun — and this is not the delta-v a spacecraft needs, because it starts in Earth orbit rather than free in space.

The Oberth effect helps enormously. Departing from a 300 km parking orbit, the required hyperbolic excess is 2943 m/s, and the burn from low Earth orbit is:

\Delta v = \sqrt{v_{\text{esc}}^2+v_\infty^2}-v_{\text{circ}} = \sqrt{(10{,}930)^2+(2943)^2}-7726 = 11{,}319-7726 = 3593\ \text{m/s}

Only 3593 m/s, because burning deep in Earth's gravity well is far more effective (Chapter 11.5).

And 259 days is why crewed Mars missions are hard. With a 26-month synodic period, a minimum-energy round trip means about 26 months away — 9 months out, 500 days waiting for the return window, 9 months back.

Problem 6 — Plane change cost

A satellite is in a 500 km circular orbit at 51.6° inclination (the ISS orbit) and must move to 0° for a geostationary mission. Find the delta-v (a) done in low orbit, (b) done at geostationary altitude.

Solution

(a) In low orbit:

v = \sqrt{\frac{3.986\times10^{14}}{6.871\times10^{6}}} = \sqrt{5.801\times10^{7}} = 7616\ \text{m/s}

\Delta v = 2v\sin\frac{\theta}{2} = 2(7616)\sin(25.8°) = 2(7616)(0.4352) = 6629\ \text{m/s}

(b) At geostationary altitude, where v = 3075 m/s:

\Delta v = 2(3075)(0.4352) = 2676\ \text{m/s}

What to notice. Doing it low costs 6629 m/s — more than reaching orbit in the first place, and completely prohibitive.

Doing it high costs 2676 m/s, still enormous, and the trick from Chapter 11.5 is to combine it with the circularisation burn, which brings the effective cost down further.

This is why launch inclination is chosen at the pad. A satellite launched from Baikonur at 45.6°N cannot economically reach geostationary orbit, which is why Russia uses Molniya orbits and why Sea Launch operated from an equatorial platform.

And it is why the ISS is at 51.6°. That is Baikonur's latitude, and the orbit was chosen so that both Russian and American vehicles could reach it — with the Americans paying a plane-change penalty from Florida's 28.5°.

Problem 7 — Lagrange point distance

Compute the Sun–Earth L2 distance using the Hill radius approximation, and find the orbital period of a spacecraft there.

Solution

r = R\left(\frac{M_E}{3M_\odot}\right)^{1/3} = 1.496\times10^{11}\times(1.0008\times10^{-6})^{1/3}

From Chapter 11.6, the cube root is 1.0003\times10^{-2}:

r = 1.496\times10^{9}\ \text{m} = 1.50\ \text{million km}

The period is by construction one year, because that is the defining property — the spacecraft orbits the Sun with the Earth's period despite being further out.

Verify why that needs the Earth's help. A body at R+r = 1.5110\times10^{11} m orbiting the Sun alone would have:

T = 2\pi\sqrt{\frac{(1.5110\times10^{11})^3}{1.327\times10^{20}}} = 2\pi\sqrt{\frac{3.4500\times10^{33}}{1.327\times10^{20}}} = 2\pi\sqrt{2.6000\times10^{13}}

= 2\pi(5.099\times10^{6}) = 3.204\times10^{7}\ \text{s} = 370.8\ \text{days}

Nearly six days too long. The Earth's gravity supplies the extra centripetal force that shortens it to exactly 365.25.

What to notice. JWST sits here, and the six-day discrepancy is what its station-keeping fights. L2 is unstable with a growth timescale of about 23 days, so the telescope needs a small burn every three weeks.

And it can only push away from the Sun, because its thrusters are on the sunward side and firing towards the Sun would contaminate the cold optics. So the halo orbit is designed to always require an outward correction, and if the propellant runs out the telescope drifts away rather than towards the Earth. Its 20-year propellant margin, better than planned because of an accurate launch, is what sets its lifetime.

Problem 8 — Gravity assist gain

A spacecraft approaches Jupiter at 10 km/s relative to the planet and is turned through 90°. Jupiter's orbital speed is 13.1 km/s. Find the change in the spacecraft's heliocentric speed.

Solution

Work in the Sun's frame. Let Jupiter move along +x at 13.1 km/s.

Before: the spacecraft's velocity relative to Jupiter is \vec{v}_\infty. Take it as arriving along -x:

\vec{v}_{\text{in}} = (-10, 0)+(13.1, 0) = (3.1, 0), \qquad |v_{\text{in}}| = 3.1\ \text{km/s}

After a 90° turn, the relative velocity points along +y:

\vec{v}_{\text{out}} = (0, 10)+(13.1, 0) = (13.1, 10)

|v_{\text{out}}| = \sqrt{13.1^2+10^2} = \sqrt{171.6+100} = \sqrt{271.6} = 16.48\ \text{km/s}

\Delta v = 16.48-3.1 = 13.4\ \text{km/s}

What to notice. The spacecraft gained 13.4 km/s with no fuel. In Jupiter's frame it arrived at 10 km/s and left at 10 km/s — only the direction changed.

The maximum possible gain is a 180° turn, giving \Delta v = 2v_p = 26.2 km/s, and that requires passing extremely close, which is limited by the planet's radius and radiation environment.

Where the energy came from: Jupiter, which slowed by about 10^{-21} m/s (Chapter 11.6). Momentum and energy are conserved exactly; the asymmetry is entirely the mass ratio.

Problem 9 — Rocket equation and staging

A single-stage rocket has 90 t propellant, 10 t dry mass and no payload, with v_e = 3200 m/s. Compare its delta-v with a two-stage version: stage 1 with 70 t propellant and 7 t dry, stage 2 with 20 t propellant and 3 t dry.

Solution

Single stage:

\Delta v = 3200\ln\frac{100}{10} = 3200\times2.3026 = 7368\ \text{m/s}

Two stage, stage 1:

m_0 = 100\ \text{t}, \qquad m_f = 100-70 = 30\ \text{t}

\Delta v_1 = 3200\ln\frac{100}{30} = 3200\times1.2040 = 3853\ \text{m/s}

Stage 2, after dropping stage 1's 7 t dry mass:

m_0 = 23\ \text{t}, \qquad m_f = 23-20 = 3\ \text{t}

\Delta v_2 = 3200\ln\frac{23}{3} = 3200\times2.0369 = 6518\ \text{m/s}

\Delta v_{\text{total}} = 3853+6518 = 10{,}371\ \text{m/s}

A gain of 3003 m/s, from the same total mass and the same propellant.

What to notice. The single-stage version falls 2000 m/s short of orbit; the two-stage version has 1000 m/s of margin. The difference is entirely that the second stage does not have to accelerate 7 tonnes of empty first-stage tank.

And the gain diminishes with more stages. A three-stage version of the same vehicle would add perhaps another 1000 m/s, and each staging event costs separation hardware and reliability. Two or three is the practical optimum, which is why essentially every launcher ever built has used exactly that.

Problem 10 — Suicide burn

A booster is falling at 250 m/s at 1500 m altitude, with a mass of 28 t and one engine producing 780 kN. Find the ignition altitude and burn duration, ignoring mass change. Then find the sensitivity to a 0.5 s timing error.

Solution

a_{\text{thrust}} = \frac{780{,}000}{28{,}000} = 27.86\ \text{m/s}^2

a_{\text{net}} = 27.86-9.81 = 18.05\ \text{m/s}^2

Required burn altitude:

h = \frac{v^2}{2a_{\text{net}}} = \frac{(250)^2}{2(18.05)} = \frac{62{,}500}{36.10} = 1731\ \text{m}

The vehicle is at 1500 m and needs 1731. It is already too late.

Take the vehicle at 2000 m falling at 250 m/s instead. It must fall until it reaches the ignition point, which given continued acceleration at g:

v^2 = 250^2+2(9.81)\Delta h

Solving \frac{v^2}{2(18.05)} = 2000-\Delta h together with the above gives \Delta h = 216 m, v = 258 m/s, ignition at 1784 m.

t = \frac{258}{18.05} = 14.3\ \text{s}

A 0.5 s late ignition. The vehicle falls a further 258(0.5)+\frac{1}{2}(9.81)(0.25) = 130.2 m, arriving at 1654 m with v = 263 m/s, needing:

h_{\text{needed}} = \frac{(263)^2}{36.10} = 1916\ \text{m}

Short by 262 m. Impact speed:

v = \sqrt{263^2-2(18.05)(1654)} = \sqrt{69{,}169-59{,}709} = \sqrt{9460} = 97\ \text{m/s}

What to notice. Half a second late gives a 97 m/s impact. This is why the guidance runs at 10 Hz or faster and re-solves continuously (Chapter 11.8).

Including the mass loss softens it slightly. At 780 kN and I_{sp} = 282 s the flow rate is 282 kg/s, so 14.3 seconds burns 4.0 t, taking the mass from 28 t to 24 t and raising the final acceleration to 22.7 m/s². The average net acceleration is about 20.2 rather than 18.05, shortening the required burn altitude by about 11 %.

Problem 11 — Cepheid distance

A Cepheid variable has a period of 30 days and an apparent magnitude of 24.0 in a distant galaxy. Find the distance in megaparsecs.

Solution

M_V = -2.81\log_{10}P-1.43 = -2.81\log_{10}(30)-1.43

\log_{10}(30) = 1.4771

M_V = -2.81(1.4771)-1.43 = -4.151-1.43 = -5.58

Distance modulus:

m-M = 24.0-(-5.58) = 29.58

d = 10^{(m-M+5)/5} = 10^{(29.58+5)/5} = 10^{6.916}\ \text{pc}

d = 8.24\times10^{6}\ \text{pc} = 8.24\ \text{Mpc} = 26.9\ \text{million light years}

What to notice. A single star, individually resolved and measured, at 27 million light years. This is what Hubble Space Telescope Cepheid observations do, and it is the second rung of the distance ladder.

The uncertainty matters enormously. A 0.1 magnitude error in M_V gives a 4.7 % distance error, and that propagates directly into the Hubble constant. The 5-sigma Hubble tension (Chapter 11.9) turns on exactly this kind of calibration.

And crowding is the main worry. At 27 million light years, one Hubble pixel covers several parsecs, so other stars can blend into the Cepheid's measurement and make it look brighter and therefore closer. JWST's better resolution has tested this and found the effect small, which strengthens rather than resolves the tension.

Problem 12 — Resolving power and the Event Horizon Telescope

Sagittarius A* has a mass of 4.3\times10^{6}\,M_\odot and is 8.1 kpc away. Find the angular size of its event horizon, and the baseline needed to resolve it at 1.3 mm.

Solution

Schwarzschild radius (Chapter 6.9):

r_s = 2.95\ \text{km}\times4.3\times10^{6} = 1.269\times10^{10}\ \text{m}

Distance:

d = 8.1\times10^{3}\times3.086\times10^{16} = 2.500\times10^{20}\ \text{m}

Angular size of the horizon diameter:

\theta = \frac{2r_s}{d} = \frac{2.537\times10^{10}}{2.500\times10^{20}} = 1.015\times10^{-10}\ \text{rad}

Converting: 1.015\times10^{-10}\times206265 = 2.09\times10^{-5} arcseconds $= 21\ \mu$as.

The observed shadow is larger — about \sqrt{27} \approx 5.2 Schwarzschild radii in diameter, from light bending (Chapter 6.9) — giving about 52 μas.

Baseline needed:

B = \frac{\lambda}{\theta} = \frac{1.3\times10^{-3}}{1.015\times10^{-10}} = 1.28\times10^{7}\ \text{m} = 12{,}800\ \text{km}

What to notice. 12,800 km is almost exactly the Earth's diameter. The Event Horizon Telescope is at the limit of what a planet-sized instrument can do, and resolving the horizon itself rather than the larger shadow required using the shortest wavelength the atmosphere permits.

Going further requires either a shorter wavelength — blocked by the atmosphere — or a longer baseline, which means going to space. Both are being studied.

And Sagittarius A* is harder than M87* despite being 2000 times closer, because it is 1500 times less massive, so the angular sizes are similar — and because it varies on timescales of minutes rather than days, so the image smears during the observation. M87* was imaged first for exactly that reason.