---
title: "Satellites and Planetary Motion"
book: "High School Physics"
subject: physics
language: en
chapter: 27
exercises: 15
source: https://one-course.com/books/physics/2/en/chapter/27-satellites-and-planetary-motion
---

# Chapter 27 — Satellites 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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#ch-g10-universal-gravitation)): the Earth, of mass $M$, pulls a [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) of mass $m$ whose center is a distance $r$ from its own with a [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite). A [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-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).](https://one-course.com/images/onecourse/chapters/physics-2/g12-satellites-kepler/fig-04f2445300da.svg)

*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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) (blue); faster still, it escapes (red).*

## 27.2 The circular orbit: one speed, one period

**Theorem 27.2 (Orbital speed).**

A [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) in [uniform circular](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-ucm) [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) 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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite)’s mass.

**Proof.** Gravity is the only [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force), so [Newton’s second law](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#thm-g12-newtons-laws-second) ([Theorem 25.5](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#thm-g12-newtons-laws-second)) reads $m\vect a = \vect F$; project it on the [Frenet frame](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-frenet) ([Definition 24.13](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-frenet)). The [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-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](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#thm-g12-kinematics-2d-centripetal)). 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.](https://one-course.com/images/onecourse/chapters/physics-2/g12-satellites-kepler/fig-3fe8df61f3fc.svg)

*The whole mechanism of an [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite): velocity tangent, [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) toward the center — gravity never slows the [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite), it only bends its path, forever.*

**Proposition 27.3 (Orbital period).**

The [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) completes its [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) in the [period](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-ucm)

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

**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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) right alongside it. And both fall with $r$: *the higher the [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite), the slower the [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) 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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) 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 $M$):

1. each [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) is an [ellipse](#def-g12-satellites-kepler-ellipse) with the Sun at one focus;
2. the Sun–planet segment sweeps equal areas in equal times;
3. the ratio of squared [period](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-ucm) to cubed [semi-major axis](#def-g12-satellites-kepler-ellipse) is the same for all planets: $T^2/a^3 = 4\pi^2/(G M)$ .

The same laws govern any family of [satellites](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) 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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) the third law is ours: squaring [Proposition 27.3](#prop-g12-satellites-kepler-period) gives $T^2/r^3 = 4\pi^2/(GM)$ — one constant for every [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) of the same center. That [ellipses](#def-g12-satellites-kepler-ellipse), 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](#def-g12-satellites-kepler-ellipse), 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.](https://one-course.com/images/onecourse/chapters/physics-2/g12-satellites-kepler/fig-24bdf63b5568.svg)

*Kepler’s second law: in one month near [perihelion](#def-g12-satellites-kepler-ellipse) $P$ the planet sweeps a short, fat sector; near [aphelion](#def-g12-satellites-kepler-ellipse) $A$, a long, [thin](https://one-course.com/books/physics/2/en/chapter/10-lenses-images-and-the-eye#def-g11-lenses-and-eye-lens) 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.](https://one-course.com/images/onecourse/chapters/physics-2/g12-satellites-kepler/fig-240486924420.svg)

*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](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) ([astronomical units](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-lightyear), years) reads simply $T^2 = a^3$.*

## 27.4 Orbits at work

**Definition 27.9 (Geostationary orbit).**

A [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) is *geostationary* when it hangs over one fixed point of the ground. Its [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) must be circular, equatorial, and of [period](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-ucm) 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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) at $r \approx 2.66 \times 10^{7}\,\mathrm{m}$ (altitude $\approx 20\,200\,\mathrm{km}$), where [Proposition 27.3](#prop-g12-satellites-kepler-period) gives $T \approx
4.32 \times 10^{4}\,\mathrm{s}$: half a [sidereal day](#def-g12-satellites-kepler-geo), so each [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) retraces its ground track daily. Your receiver compares the arrival times of their clock [signals](https://one-course.com/books/physics/2/en/chapter/8-signals-and-waves#def-g10-signals-and-waves-signal) against [orbits](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) 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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) of it (a moon, a probe, a planet), measure its [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) radius $r$ and [period](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-ucm) $T$, and invert the third law:

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

This weighs the *central* body only — the [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite)’s mass cancelled in [Theorem 27.2](#thm-g12-satellites-kepler-speed). Every data-card mass came this way: the Earth from the Moon, the Sun from the Earth’s own [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite).

**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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) is the more *expensive*: climbing against gravity stores potential [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) (the account is settled in [Chapter 29](https://one-course.com/books/physics/2/en/chapter/29-work-and-mechanical-energy#ch-g12-work-and-energy)), and the gain in $E_p$ outweighs the loss in $E_k$. Hence a rocket burns *forward* to reach a higher, slower [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) — and a [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) braked by [thin](https://one-course.com/books/physics/2/en/chapter/10-lenses-images-and-the-eye#def-g11-lenses-and-eye-lens) 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](#def-g12-satellites-kepler-geo) $86\,164\,\mathrm{s}$; Moon’s [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite): $r = 3.84 \times 10^{8}\,\mathrm{m}$, $T =
27.3\,\mathrm{d}$; 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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-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 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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) skimming the Earth’s surface ($r = R$), compute the orbital speed and the [period](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-ucm). Compare with Newton’s cannonball ([Remark 27.1](#rem-g12-satellites-kepler-cannon)).

**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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) *is* Newton’s shot.

**Exercise 27.3 ★.**

A [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) is moved from a circular [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) of radius $r$ to one of radius $2r$. By what factor do its speed and [period](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-ucm) change? Do the answers depend on its mass?

**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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) completed in $24\,\mathrm{h}$.

**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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) per day.

**Exercise 27.5 ★.**

State Kepler’s three laws. Where along its very stretched [ellipse](#def-g12-satellites-kepler-ellipse) is a comet fastest, and which law says so? Does a [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite)’s [period](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-ucm) depend on its own mass?

**Solution of Exercise 27.5.**

[Ellipses](#def-g12-satellites-kepler-ellipse) with the Sun at a focus; equal areas in equal times; $T^2/a^3$ identical for all planets. Fastest at [perihelion](#def-g12-satellites-kepler-ellipse), by the second law (short sweep segment must sweep fast). No: the [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite)’s mass cancels — only the central mass enters.

**Exercise 27.6 ★★.**

Weigh the Earth: from the Moon’s [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) radius and [period](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-ucm) (data card), compute the Earth’s mass and compare with the card’s value.

**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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) hovering permanently above Paris (latitude $49^\circ$ north). From the direction of gravity, explain why every circular [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) is centered on the Earth’s center; conclude the proposal is impossible. Where *can* a [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) hover?

**Solution of Exercise 27.7.**

Gravity points at the Earth’s center, and in circular motion the [net force](https://one-course.com/books/physics/2/en/chapter/16-forces-and-motion#def-g11-forces-and-motion-netforce) 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](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work) 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 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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) 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 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](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#thm-g12-kinematics-2d-centripetal) $a = v^2/r$. Show it equals $g/60^2$, and say why that convinced Newton that one law rules apple and Moon.

**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](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) ([astronomical units](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-lightyear), 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](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-ucm). Halley last passed [perihelion](#def-g12-satellites-kepler-ellipse) in 1986 — when is it due back?

**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](#def-g12-satellites-kepler-ellipse) around $1986 + 75 \approx 2061$.

**Exercise 27.12 ★★★.**

[Thin](https://one-course.com/books/physics/2/en/chapter/10-lenses-images-and-the-eye#def-g11-lenses-and-eye-lens) 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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) slows things down — where does the extra speed come from?

**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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-inventory) does drain [mechanical energy](https://one-course.com/books/physics/2/en/chapter/18-mechanical-energy-and-its-conservation#def-g11-mechanical-energy-em) — the [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) shrinks; but on the way down gravity’s [work](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-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](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) and in Suns. Which mass — star’s or planet’s — *cannot* be obtained this way, and why?

**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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) 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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) radius and the altitude above the Martian surface (data card).

**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](#def-g12-satellites-kepler-ellipse) at $r_p = 8.77 \times 10^{10}\,\mathrm{m}$ moving at $54.5\,\mathrm{km}/\mathrm{s}$; [aphelion](#def-g12-satellites-kepler-ellipse) 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](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) $r_p v_p = r_a v_a$ (equal sweep triangles of area $\tfrac12 r v\,\Delta t$). Deduce the [aphelion](#def-g12-satellites-kepler-ellipse) speed and comment on the ratio.

**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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-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](#def-g12-satellites-kepler-geo) [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite). You are the mission analyst; use the data card. For Part IV: Jupiter’s moon Io [orbits](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) 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.**

1. The [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) must stay over one fixed ground point. What must its [period](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-ucm) be — and why the [sidereal day](#def-g12-satellites-kepler-geo) $86\,164\,\mathrm{s}$ rather than $24\,\mathrm{h}$ ?
2. In circular motion the [net force](https://one-course.com/books/physics/2/en/chapter/16-forces-and-motion#def-g11-forces-and-motion-netforce) points at the circle’s center; here the only [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-force) is gravity. Deduce that the center of any circular [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) is the Earth’s center.
3. 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](#def-g12-satellites-kepler-geo) lie?
4. Write [Newton’s second law](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#thm-g12-newtons-laws-second) in the [Frenet frame](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-frenet) for a circular [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) of radius $r$ and derive $v = \sqrt{GM/r}$ .
5. 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](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-work).**

6. Solve the third law for $r$ and compute the [geostationary](#def-g12-satellites-kepler-geo) radius.
7. Deduce the altitude above the ground.
8. Compute the orbital speed from $v = 2\pi r/T$ .
9. Check it against $v = \sqrt{GM/r}$ .
10. The company’s [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) has mass $4.0 \times 10^{3}\,\mathrm{kg}$ ; a rival’s weighs twice that. Compare their [geostationary](#def-g12-satellites-kepler-geo) radii, and say what the extra mass changes at launch.

**Part III — The station below.**

11. The space station flies at $r = 6.77 \times 10^{6}\,\mathrm{m}$ : compute its speed.
12. Compute its [period](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-ucm) , in seconds and minutes.
13. How many [orbits](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) does it complete in $24\,\mathrm{h}$ — about how many sunrises do the astronauts watch per day?
14. Compute the ratio of radii $r_{\text{geo}}/r_{\text{ISS}}$ , and verify that the ratio of [periods](https://one-course.com/books/physics/2/en/chapter/24-kinematics-in-two-dimensions#def-g12-kinematics-2d-ucm) is that number to the [power](https://one-course.com/books/physics/2/en/chapter/17-work-of-a-force#def-g11-work-of-force-power) $3/2$ , as Kepler’s third law demands.
15. Which [satellite](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) moves faster, and which took more [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) per [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit) to emplace? Reconcile the two in one sentence.

**Part IV — Weighing Jupiter on the side.**

16. From Kepler’s third law applied to Io, express Jupiter’s mass $M_J$ in terms of $G$ , $r$ and $T$ .
17. Compute $M_J$ .
18. How many Earth masses is that? Roughly what fraction of the Sun’s mass?
19. A “jovistationary” relay should hang over one point of Jupiter’s clouds: compute its [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) radius; compare with Io’s.
20. 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 of Problem 27.1.**

**1.** $T = 86\,164\,\mathrm{s}$: the ground turns with the Earth *relative to the stars* once per [sidereal day](#def-g12-satellites-kepler-geo); the $24\,\mathrm{h}$ day adds the degree per day the Earth advances around the Sun.

**2.** The [net force](https://one-course.com/books/physics/2/en/chapter/16-forces-and-motion#def-g11-forces-and-motion-netforce) of circular motion aims at the circle’s center; the only [force](https://one-course.com/books/physics/2/en/chapter/6-forces-and-the-principle-of-inertia#def-g10-inertia-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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) must lie in the *equatorial plane*.

**4.** Normal component of [Newton’s second law](https://one-course.com/books/physics/2/en/chapter/25-newtons-laws#thm-g12-newtons-laws-second): $m v^2/r = GmM/r^2$, 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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite) changes only the launch bill — more fuel for the same [orbit](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite).

**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](https://one-course.com/books/physics/2/en/chapter/4-universal-gravitation-and-weight#def-g10-universal-gravitation-satellite): 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](#def-g12-satellites-kepler-geo) cost more [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-energy) per [kilogram](https://one-course.com/books/physics/2/en/chapter/1-orders-of-magnitude-measuring-the-universe#def-g10-orders-of-magnitude-unit): the climb in potential [energy](https://one-course.com/books/physics/2/en/chapter/9-energy-forms-and-conservation#def-g10-energy-conservation-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.