High School Physics · Grades 10–12

27Satellites and Planetary Motion

The dish on the balcony is bolted down, yet all day it stares at one fixed point of the sky, $36\,000\,\mathrm{km}$ up; the space station crosses the evening sky in minutes; a constellation of flying clocks tells your phone where you stand. None of these machines has an engine running: they are all simply falling. This chapter computes how fast and how high each must fall — then weighs planets with the same law.

27.1 Falling around the Earth

Everything rests on the law measured earlier (Chapter 4): the Earth, of mass $M$ , pulls a satellite of mass $m$ whose center is a distance $r$ from its own with a force directed toward the Earth’s center, of magnitude

F = G\,\frac{m M}{r^2}, \qquad G = 6.67 \times 10^{-11}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{kg}^{2}.

Mind what $r$ is: the distance between centers. A satellite at altitude $h$ has $r = R + h$ , with $R = 6.37 \times 10^{6}\,\mathrm{m}$ the Earth’s radius — forgetting the $R$ is the classic blunder of this chapter.

Remark 27.1 (Newton’s cannon, revisited)

Fired horizontally, a cannonball lands; faster, it lands farther; at the right speed the ground curves away exactly as fast as the ball falls, and the fall closes on itself: an orbit. A satellite needs no engine — only the right sideways speed.

Newton’s cannon: fired ever faster, the ball lands farther (orange); at about 7.9\, km/ s its fall matches the Earth’s curvature and it orbits (blue); faster still, it escapes (red). — Newton’s cannon: fired ever faster, the ball lands farther (orange); at about $7.9\,\mathrm{km}/\mathrm{s}$ its fall matches the Earth’s curvature and it orbits (blue); faster still, it escapes (red).

27.2 The circular orbit: one speed, one period

Theorem 27.2 (Orbital speed)

A satellite in uniform circular orbit of radius $r$ around a body of mass $M$ travels at the speed

v = \sqrt{\frac{G M}{r}},

which does not depend on the satellite’s mass.

Proof. Gravity is the only force, so Newton’s second law (Theorem 25.5) reads $m\vect a = \vect F$ ; project it on the Frenet frame (Definition 24.13). The force aims at the center: its tangential component is zero, so the speed is constant; its normal component is the whole $GmM/r^2$ , and circular motion has $a_n = v^2/r$ (Theorem 24.12). Hence $m v^2/r = GmM/r^2$ , and — $m$ dividing out — $v^2 = GM/r$ . ∎

The whole mechanism of an orbit: velocity tangent, force toward the center — gravity never slows the satellite, it only bends its path, forever.

Proposition 27.3 (Orbital period)

The satellite completes its orbit in the period

T = \frac{2\pi r}{v} = 2\pi \sqrt{\frac{r^3}{G M}}.

Proof. One lap is the circumference $2\pi r$ at constant speed; insert $v = \sqrt{GM/r}$ : $T = 2\pi r\sqrt{r/(GM)} = 2\pi\sqrt{r^3/(GM)}$ . ∎

Neither $v$ nor $T$ contains $m$ : a screw shaken loose from the space station orbits right alongside it. And both fall with $r$ : the higher the orbit, the slower the satellite, in speed ( $v \propto 1/\sqrt r$ ) and in angle ( $T \propto r^{3/2}$ ).

Example 27.4 (The space station)

The International Space Station flies at altitude $h \approx 400\,\mathrm{km}$ , so $r = 6.37 \times 10^{6} + 4.0 \times 10^{5} = 6.77 \times 10^{6}\,\mathrm{m}$ ; with $GM = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} = 3.98 \times 10^{14}\,\mathrm{m}^{3}/\mathrm{s}^{2}$ , $v = \sqrt{3.98 \times 10^{14}/6.77 \times 10^{6}} \approx 7.7 \times 10^{3}\,\mathrm{m}/\mathrm{s}$ and $T = 2\pi r/v \approx 5.5 \times 10^{3}\,\mathrm{s} \approx 92\,\mathrm{min}$ : a lap of the planet every hour and a half, about sixteen sunrises a day for the astronauts.

Example 27.5 (The Moon falls like the apple)

The Moon, at $r = 3.84 \times 10^{8}\,\mathrm{m}$ , orbits in $T = 27.3\,\mathrm{d} = 2.36 \times 10^{6}\,\mathrm{s}$ , so $v = 2\pi r/T \approx 1.02 \times 10^{3}\,\mathrm{m}/\mathrm{s}$ and $a = v^2/r \approx 2.7 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$ — precisely $g/60^2$ , and the Moon sits $60$ Earth radii away: the pull that drops the apple bends the Moon, weakened by the inverse square. Each second the Moon “falls” about $1.4\,\mathrm{mm}$ toward us, its sideways kilometer carrying it around — the check Newton made in 1666.

27.3 Kepler’s laws

Newton did not guess the inverse square: he extracted it from three regularities that Johannes Kepler distilled, between 1609 and 1619, from Tycho Brahe’s twenty years of naked-eye planet positions — the best data of the pre-telescope world.

Definition 27.6 (Ellipse)

An ellipse is the set of points whose distances to two fixed points, the foci, have a constant sum; half its longest diameter is the semi-major axis $a$ , and a circle is the case where the foci merge. A planet’s orbit points nearest and farthest from the Sun are its perihelion and aphelion.

Theorem 27.7 (Kepler’s laws)

For the planets orbiting the Sun (mass $M$ ):

each orbit is an ellipse with the Sun at one focus;
the Sun–planet segment sweeps equal areas in equal times;
the ratio of squared period to cubed semi-major axis is the same for all planets: $T^2/a^3 = 4\pi^2/(G M)$ .

The same laws govern any family of satellites around any central body, with $M$ the central mass.

Proof. Admitted at this level. ∎

Remark 27.8 (What we can prove this year)

For a circular orbit the third law is ours: squaring Proposition 27.3 gives $T^2/r^3 = 4\pi^2/(GM)$ — one constant for every satellite of the same center. That ellipses, equal areas and the general $a$ follow from the inverse square is shown, with more calculus, in the Year 1 volume. Meanwhile the second law can be read: the planet moves fastest at perihelion, where the short sweep segment must hurry.

Kepler’s second law: in one month near perihelion

P

the planet sweeps a short, fat sector; near aphelion

A

, a long, thin one — equal areas: it hurries when close, dawdles when far.

The eight planets on log–log axes: a straight line of slope 3/2, i.e. T2 a3 — Kepler’s third law, which in these units (astronomical units, years) reads simply T2 = a3. — The eight planets on log–log axes: a straight line of slope $3/2$ , i.e. $T^2 \propto a^3$ — Kepler’s third law, which in these units (astronomical units, years) reads simply $T^2 = a^3$ .

27.4 Orbits at work

Definition 27.9 (Geostationary orbit)

A satellite is geostationary when it hangs over one fixed point of the ground. Its orbit must be circular, equatorial, and of period one sidereal day, $T = 86\,164\,\mathrm{s}$ ( $23\,\mathrm{h}$ $56\,\mathrm{min}$ $4\,\mathrm{s}$ ): the time the Earth takes to turn once relative to the stars. (The $24\,\mathrm{h}$ day is relative to the Sun, toward which the Earth also advances a degree a day.)

Example 27.10 (The geostationary altitude)

Kepler’s third law, read backward, dictates the radius: $r = \left(GMT^2/4\pi^2\right)^{1/3} = \left(3.98 \times 10^{14} \times (86\,164)^2/4\pi^2\right)^{1/3} \approx 4.22 \times 10^{7}\,\mathrm{m}$ — an altitude $h = r - R \approx 3.58 \times 10^{7}\,\mathrm{m} \approx 35\,800\,\mathrm{km}$ , traveled at $v = 2\pi r/T \approx 3.1\,\mathrm{km}/\mathrm{s}$ . Every relay that “hangs still” sits on this one circle over the equator — there is no other.

Example 27.11 (The navigation constellation)

Satellite-navigation systems fly some thirty satellites at $r \approx 2.66 \times 10^{7}\,\mathrm{m}$ (altitude $\approx 20\,200\,\mathrm{km}$ ), where Proposition 27.3 gives $T \approx 4.32 \times 10^{4}\,\mathrm{s}$ : half a sidereal day, so each satellite retraces its ground track daily. Your receiver compares the arrival times of their clock signals against orbits known to meters.

Method 27.12 (Weighing a world from its satellite)

To measure the mass of a planet or a star: find any satellite of it (a moon, a probe, a planet), measure its orbit radius $r$ and period $T$ , and invert the third law:

M = \frac{4\pi^2 r^3}{G\,T^2}.

This weighs the central body only — the satellite’s mass cancelled in Theorem 27.2. Every data-card mass came this way: the Earth from the Moon, the Sun from the Earth’s own orbit.

Remark 27.13 (Higher is slower — but costs more)

Since $v = \sqrt{GM/r}$ , the Moon crawls at $1\,\mathrm{km}/\mathrm{s}$ while the station races at $7.7\,\mathrm{km}/\mathrm{s}$ . Yet the higher orbit is the more expensive: climbing against gravity stores potential energy (the account is settled in Chapter 29), and the gain in $E_p$ outweighs the loss in $E_k$ . Hence a rocket burns forward to reach a higher, slower orbit — and a satellite braked by thin air spirals down and speeds up.

Remark 27.14 (Data card)

Unless stated otherwise: $G = 6.67 \times 10^{-11}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{kg}^{2}$ ; Earth: $M = 5.97 \times 10^{24}\,\mathrm{kg}$ , $R = 6.37 \times 10^{6}\,\mathrm{m}$ , sidereal day $86\,164\,\mathrm{s}$ ; Moon’s orbit: $r = 3.84 \times 10^{8}\,\mathrm{m}$ , $T = 27.3\,\mathrm{d}$ ; Sun: $M = 1.99 \times 10^{30}\,\mathrm{kg}$ ; Earth–Sun distance $1.496 \times 10^{11}\,\mathrm{m}$ ; one year $= 3.156 \times 10^{7}\,\mathrm{s}$ ; Mars: $M = 6.42 \times 10^{23}\,\mathrm{kg}$ , $R = 3.39 \times 10^{6}\,\mathrm{m}$ .

27.5 Exercises

Exercise 27.1 ★

A $1000\,\mathrm{kg}$ satellite sits on the launch pad, then flies at altitude $400\,\mathrm{km}$ . Compute the Earth’s pull in both places: what percentage survives aloft — and why do occupants float?

Solution

Solution of Exercise 27.1.

On the pad: $F = GMm/R^2 = 3.98 \times 10^{17}/4.06 \times 10^{13} \approx 9.8 \times 10^{3}\,\mathrm{N}$ . At $r = 6.77 \times 10^{6}\,\mathrm{m}$ : $F \approx 8.7 \times 10^{3}\,\mathrm{N}$ — still $89\%$ of the ground value. The occupants float because they fall together with the station, not because gravity is gone.

Exercise 27.2 ★

For a (theoretical) satellite skimming the Earth’s surface ( $r = R$ ), compute the orbital speed and the period. Compare with Newton’s cannonball (Remark 27.1).

Solution

Solution of Exercise 27.2.

$v = \sqrt{GM/R} = \sqrt{3.98 \times 10^{14}/6.37 \times 10^{6}} \approx 7.9 \times 10^{3}\,\mathrm{m}/\mathrm{s}$ ; $T = 2\pi R/v \approx 5.1 \times 10^{3}\,\mathrm{s} \approx 84\,\mathrm{min}$ . Exactly the cannonball’s $7.9\,\mathrm{km}/\mathrm{s}$ : a skimming orbit is Newton’s shot.

Exercise 27.3 ★

A satellite is moved from a circular orbit of radius $r$ to one of radius $2r$ . By what factor do its speed and period change? Do the answers depend on its mass?

Solution

Solution of Exercise 27.3.

$v \propto 1/\sqrt{r}$ : divided by $\sqrt 2 \approx 1.41$ . $T \propto r^{3/2}$ : multiplied by $2^{3/2} \approx 2.8$ . Neither contains $m$ : no.

Exercise 27.4 ★

Redo the station’s numbers: from $r = 6.77 \times 10^{6}\,\mathrm{m}$ , compute $v$ , $T$ , and the number of orbits completed in $24\,\mathrm{h}$ .

Solution

Solution of Exercise 27.4.

$v = \sqrt{3.98 \times 10^{14}/6.77 \times 10^{6}} \approx 7.67 \times 10^{3}\,\mathrm{m}/\mathrm{s}$ ; $T = 2\pi r/v \approx 5.55 \times 10^{3}\,\mathrm{s} \approx 92\,\mathrm{min}$ ; $86\,400/5546 \approx 15.6$ orbits per day.

Exercise 27.5 ★

State Kepler’s three laws. Where along its very stretched ellipse is a comet fastest, and which law says so? Does a satellite’s period depend on its own mass?

Solution

Solution of Exercise 27.5.

Ellipses with the Sun at a focus; equal areas in equal times; $T^2/a^3$ identical for all planets. Fastest at perihelion, by the second law (short sweep segment must sweep fast). No: the satellite’s mass cancels — only the central mass enters.

Exercise 27.6 ★★

Weigh the Earth: from the Moon’s orbit radius and period (data card), compute the Earth’s mass and compare with the card’s value.

Solution

Solution of Exercise 27.6.

$T = 2.36 \times 10^{6}\,\mathrm{s}$ : $M = 4\pi^2 r^3/(GT^2) = 2.24 \times 10^{27}/371 \approx 6.0 \times 10^{24}\,\mathrm{kg}$ — within $1\%$ of $5.97 \times 10^{24}\,\mathrm{kg}$ .

Exercise 27.7 ★★

A company proposes a satellite hovering permanently above Paris (latitude $49^\circ$ north). From the direction of gravity, explain why every circular orbit is centered on the Earth’s center; conclude the proposal is impossible. Where can a satellite hover?

Solution

Solution of Exercise 27.7.

Gravity points at the Earth’s center, and in circular motion the net force points at the circle’s center: the two centers coincide. A circle above Paris’s parallel is centered on the axis, not the Earth’s center — impossible. Hovering works only where the parallel is a great circle centered at the center: over the equator.

Exercise 27.8 ★★

Phobos circles Mars at $r = 9.38 \times 10^{6}\,\mathrm{m}$ in $T = 7\,\mathrm{h} 39\,\mathrm{min}$ . Deduce the mass of Mars; compare with the data card.

Solution

Solution of Exercise 27.8.

$T = 2.754 \times 10^{4}\,\mathrm{s}$ : $M = 4\pi^2 r^3/(GT^2) = 3.26 \times 10^{22}/5.06 \times 10^{-2} \approx 6.4 \times 10^{23}\,\mathrm{kg}$ — the data card’s Mars.

Exercise 27.9 ★★

Mars orbits the Sun at $a = 2.279 \times 10^{11}\,\mathrm{m}$ in $687\,\mathrm{d}$ . Compute $T^2/a^3$ for Mars and for the Earth (data card), check Kepler’s third law, and deduce the mass of the Sun.

Solution

Solution of Exercise 27.9.

Earth: $T^2/a^3 = 9.96 \times 10^{14}/3.35 \times 10^{33} = 2.98 \times 10^{-19}\,\mathrm{s}^{2}/\mathrm{m}^{3}$ . Mars: $T = 5.94 \times 10^{7}\,\mathrm{s}$ , $3.52 \times 10^{15}/1.18 \times 10^{34} = 2.98 \times 10^{-19}\,\mathrm{s}^{2}/\mathrm{m}^{3}$ — equal, as the third law demands. Then $M = 4\pi^2/(G \times 2.98 \times 10^{-19}) \approx 1.99 \times 10^{30}\,\mathrm{kg}$ .

Exercise 27.10 ★★

Newton’s Moon test: from the data card, compute the Moon’s speed and its centripetal acceleration $a = v^2/r$ . Show it equals $g/60^2$ , and say why that convinced Newton that one law rules apple and Moon.

Solution

Solution of Exercise 27.10.

$v = 2\pi r/T = 1.02 \times 10^{3}\,\mathrm{m}/\mathrm{s}$ ; $a = v^2/r = 2.72 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$ ; $g/60^2 = 9.81/3600 = 2.73 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$ . The Moon sits $60$ Earth radii out, and its fall is $60^2$ times weaker than the apple’s: exactly the inverse square — one law for both.

Exercise 27.11 ★★

In Sun units (astronomical units, years) Kepler’s third law reads $T^2 = a^3$ : the straight line of the course’s log–log plot. Place on it the asteroid Ceres, $a = 2.77\,\mathrm{au}$ , and Halley’s comet, $a = 17.8\,\mathrm{au}$ : compute both periods. Halley last passed perihelion in 1986 — when is it due back?

Solution

Solution of Exercise 27.11.

Ceres: $T = 2.77^{3/2} = \sqrt{21.3} \approx 4.6\,\mathrm{yr}$ . Halley: $T = 17.8^{3/2} \approx 75\,\mathrm{yr}$ , so next perihelion around $1986 + 75 \approx 2061$ .

Exercise 27.12 ★★★

Thin air brushes the station, yet its speed increases as it loses altitude. Compute $v$ at altitudes $400\,\mathrm{km}$ and $350\,\mathrm{km}$ , then resolve the paradox: friction slows things down — where does the extra speed come from?

Solution

Solution of Exercise 27.12.

At $400\,\mathrm{km}$ : $v \approx 7.67 \times 10^{3}\,\mathrm{m}/\mathrm{s}$ ; at $350\,\mathrm{km}$ ( $r = 6.72 \times 10^{6}\,\mathrm{m}$ ): $v \approx 7.70 \times 10^{3}\,\mathrm{m}/\mathrm{s}$ . Friction does drain mechanical energy — the orbit shrinks; but on the way down gravity’s work more than repays the loss of speed to drag: $E_m$ falls while $E_k$ rises.

Exercise 27.13 ★★★

A telescope finds an exoplanet circling its star at $r = 7.48 \times 10^{9}\,\mathrm{m}$ every $3.50\,\mathrm{d}$ . Weigh the star, in kilograms and in Suns. Which mass — star’s or planet’s — cannot be obtained this way, and why?

Solution

Solution of Exercise 27.13.

$T = 3.024 \times 10^{5}\,\mathrm{s}$ : $M = 4\pi^2 r^3/(GT^2) = 1.65 \times 10^{31}/6.10 \approx 2.7 \times 10^{30}\,\mathrm{kg} \approx 1.4$ Suns. The planet’s mass is out of reach: it cancelled from the orbit equations.

Exercise 27.14 ★★★

Mars settlers want an “areostationary” relay over their base. Mars turns relative to the stars in $24\,\mathrm{h}$ $37\,\mathrm{min}$ : compute the orbit radius and the altitude above the Martian surface (data card).

Solution

Solution of Exercise 27.14.

$T = 8.862 \times 10^{4}\,\mathrm{s}$ , $GM = 4.28 \times 10^{13}\,\mathrm{m}^{3}/\mathrm{s}^{2}$ : $r^3 = GMT^2/(4\pi^2) = 8.52 \times 10^{21}$ , so $r \approx 2.04 \times 10^{7}\,\mathrm{m}$ and $h = r - R \approx 1.7 \times 10^{7}\,\mathrm{m} \approx 17\,000\,\mathrm{km}$ above the Martian ground.

Exercise 27.15 ★★★

Halley’s comet passes perihelion at $r_p = 8.77 \times 10^{10}\,\mathrm{m}$ moving at $54.5\,\mathrm{km}/\mathrm{s}$ ; aphelion is at $r_a = 5.25 \times 10^{12}\,\mathrm{m}$ . At both extremes the velocity is perpendicular to the Sun–comet segment, and Kepler’s second law then forces $r_p v_p = r_a v_a$ (equal sweep triangles of area $\tfrac12 r v\,\Delta t$ ). Deduce the aphelion speed and comment on the ratio.

Solution

Solution of Exercise 27.15.

$v_a = r_p v_p / r_a = 8.77 \times 10^{10} \times 5.45 \times 10^{4} / 5.25 \times 10^{12} \approx 9.1 \times 10^{2}\,\mathrm{m}/\mathrm{s}$ . Since $r_a/r_p \approx 60$ , the comet is $60$ times farther and $60$ times slower: it spends decades dawdling in the dark and weeks sprinting past the Sun.

27.6 Problem: A Seat Over the Equator

Problem 27.1

Weekend problem — putting a satellite over the equator: why the only parking spot in the sky is a circle $35\,800\,\mathrm{km}$ up, how it compares with the station racing below, and how the same law weighs Jupiter on the way

A telecom company wants a relay that its ground dishes, bolted down once, never have to chase: a geostationary satellite. You are the mission analyst; use the data card. For Part IV: Jupiter’s moon Io orbits at $r = 4.22 \times 10^{8}\,\mathrm{m}$ in $T = 1.77\,\mathrm{d}$ , and Jupiter turns relative to the stars in $9\,\mathrm{h}$ $56\,\mathrm{min}$ .

Part I — The rules of hanging still.

The satellite must stay over one fixed ground point. What must its period be — and why the sidereal day $86\,164\,\mathrm{s}$ rather than $24\,\mathrm{h}$ ?
In circular motion the net force points at the circle’s center; here the only force is gravity. Deduce that the center of any circular orbit is the Earth’s center.
A circle hovering above Paris’s latitude would be centered on the Earth’s axis, north of the center. Conclude: in which plane must a geostationary orbit lie?
Write Newton’s second law in the Frenet frame for a circular orbit of radius $r$ and derive $v = \sqrt{GM/r}$ .
Deduce $T = 2\pi\sqrt{r^3/(GM)}$ and rewrite it as Kepler’s third law $T^2/r^3 = 4\pi^2/(GM)$ .

Part II — The one circle that works.

Solve the third law for $r$ and compute the geostationary radius.
Deduce the altitude above the ground.
Compute the orbital speed from $v = 2\pi r/T$ .
Check it against $v = \sqrt{GM/r}$ .
The company’s satellite has mass $4.0 \times 10^{3}\,\mathrm{kg}$ ; a rival’s weighs twice that. Compare their geostationary radii, and say what the extra mass changes at launch.

Part III — The station below.

The space station flies at $r = 6.77 \times 10^{6}\,\mathrm{m}$ : compute its speed.
Compute its period, in seconds and minutes.
How many orbits does it complete in $24\,\mathrm{h}$ — about how many sunrises do the astronauts watch per day?
Compute the ratio of radii $r_{\text{geo}}/r_{\text{ISS}}$ , and verify that the ratio of periods is that number to the power $3/2$ , as Kepler’s third law demands.
Which satellite moves faster, and which took more energy per kilogram to emplace? Reconcile the two in one sentence.

Part IV — Weighing Jupiter on the side.

From Kepler’s third law applied to Io, express Jupiter’s mass $M_J$ in terms of $G$ , $r$ and $T$ .
Compute $M_J$ .
How many Earth masses is that? Roughly what fraction of the Sun’s mass?
A “jovistationary” relay should hang over one point of Jupiter’s clouds: compute its orbit radius; compare with Io’s.
Report to the board, two sentences: the altitude where the company must park its relay, and the mass of Jupiter that the very same law delivered for free.

Solution

Solution of Problem 27.1.

1. $T = 86\,164\,\mathrm{s}$ : the ground turns with the Earth relative to the stars once per sidereal day; the $24\,\mathrm{h}$ day adds the degree per day the Earth advances around the Sun.

2. The net force of circular motion aims at the circle’s center; the only force, gravity, aims at the Earth’s center: the two centers are the same point.

3. A Paris-hovering circle would be centered on the axis north of the center — excluded by 2. The orbit must lie in the equatorial plane.

4. Normal component of Newton’s second law: $m v^2/r = GmM/r^2$ , so $v = \sqrt{GM/r}$ ( $m$ cancels).

5. $T = 2\pi r/v = 2\pi\sqrt{r^3/(GM)}$ ; squaring, $T^2/r^3 = 4\pi^2/(GM)$ .

6. $r = \left(GMT^2/4\pi^2\right)^{1/3} = (7.49 \times 10^{22})^{1/3} \approx 4.22 \times 10^{7}\,\mathrm{m}$ .

7. $h = r - R = 4.22 \times 10^{7} - 6.37 \times 10^{6} \approx 3.58 \times 10^{7}\,\mathrm{m} \approx 35\,800\,\mathrm{km}$ .

8. $v = 2\pi \times 4.22 \times 10^{7}/86\,164 \approx 3.07 \times 10^{3}\,\mathrm{m}/\mathrm{s}$ .

9. $\sqrt{3.98 \times 10^{14}/4.22 \times 10^{7}} \approx 3.07 \times 10^{3}\,\mathrm{m}/\mathrm{s}$ — consistent.

10. Identical radii: the mass cancelled in 4. The heavier satellite changes only the launch bill — more fuel for the same orbit.

11. $v = \sqrt{3.98 \times 10^{14}/6.77 \times 10^{6}} \approx 7.67 \times 10^{3}\,\mathrm{m}/\mathrm{s}$ .

12. $T = 2\pi r/v \approx 5.55 \times 10^{3}\,\mathrm{s} \approx 92.4\,\mathrm{min}$ .

13. $86\,400/5546 \approx 15.6$ orbits: about $16$ sunrises a day.

14. $r_{\text{geo}}/r_{\text{ISS}} = 4.22 \times 10^{7}/6.77 \times 10^{6} = 6.23$ ; $6.23^{3/2} \approx 15.5 = 86\,164/5546$ — Kepler’s third law, verified on the spot.

15. The station is faster ( $7.67\,\mathrm{km}/\mathrm{s}$ against $3.07\,\mathrm{km}/\mathrm{s}$ ), yet the geostationary cost more energy per kilogram: the climb in potential energy outweighs the loss of speed.

16. $M_J = 4\pi^2 r^3/(G T^2)$ .

17. $T = 1.529 \times 10^{5}\,\mathrm{s}$ : $M_J = 2.97 \times 10^{27}/1.56 \approx 1.90 \times 10^{27}\,\mathrm{kg}$ .

18. $1.90 \times 10^{27}/5.97 \times 10^{24} \approx 318$ Earths — and about $1/1000$ of the Sun.

19. $T_J = 3.576 \times 10^{4}\,\mathrm{s}$ : $r^3 = GM_J T_J^2/(4\pi^2) = 4.11 \times 10^{24}$ , $r \approx 1.60 \times 10^{8}\,\mathrm{m}$ — Io, at $4.22 \times 10^{8}\,\mathrm{m}$ , flies $2.6$ times higher and drifts across Jupiter’s sky.

20. Park the relay on the equatorial circle of altitude $35\,800\,\mathrm{km}$ , where it rides at $3.07\,\mathrm{km}/\mathrm{s}$ and never leaves the dish’s aim; and the same law, fed Io’s month, weighed Jupiter at $1.90 \times 10^{27}\,\mathrm{kg}$ — $318$ Earths — for free.