---
title: "Central Forces: Planets and Satellites"
book: "University Physics — Year 1"
subject: physics
language: en
chapter: 16
exercises: 12
source: https://one-course.com/books/physics/3/en/chapter/16-central-forces-planets-and-satellites
---

# Chapter 16 — Central Forces: Planets and Satellites

Thirty-one satellites circle the Earth twice a day, each one a clock and a radio, and a phone that hears four of them knows where it is to a few meters. Halley’s comet falls toward the Sun for thirty-eight years, swings round it in a few weeks, and climbs away again for thirty-eight more. A probe bound for Mars leaves Earth at exactly the speed that will make its path just kiss the orbit of Mars. All of this is the motion of a body attracted toward a fixed point by a force that falls as the inverse square of the distance, and all of it follows from two conserved quantities — energy and [angular momentum](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#def-b1-angular-momentum-L) — plus a single new idea, the effective potential. This chapter derives the orbits’ shapes and speeds, Kepler’s three laws, and the arithmetic of putting a satellite where one wants it.

## 16.1 Central conservative forces

**Definition 16.1 (Central force; central conservative force).**

A force is *central* with center $O$ if it is always directed along $\vect{OM}$: $\vect F = F(M)\,\vect e_r$. It is *central [conservative](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative)* if its magnitude depends only on $r = OM$: $\vect F =
F(r)\,\vect e_r$, which derives from a [potential energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative) $E_p(r)$ with $F(r) = -\dd E_p/\dd r$.

**Theorem 16.2 (Motion under a central conservative force).**

A particle of mass $m$ subject only to a [central conservative](#def-b1-central-forces-central) force:

1. keeps its [angular momentum](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#def-b1-angular-momentum-L) $\vect L_O$ ; its motion is planar and, in [polar coordinates](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#def-b1-point-kinematics-cylindrical) in that plane, $r^2\dot\theta = C$ is constant (law of areas, [Corollary 15.5](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#cor-b1-angular-momentum-central) );
2. keeps its [mechanical energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#thm-b1-work-and-energy-em) $E_m = \tfrac12 m(\dot r^2 + r^2\dot\theta^2)  + E_p(r)$, which can be written $$E_m = \tfrac12 m\dot r^2 + E_{p,\mathrm{eff}}(r), \qquad  E_{p,\mathrm{eff}}(r) = E_p(r) + \frac{mC^2}{2r^2} = E_p(r) + \frac{L^2}{2mr^2} .$$

The radial motion is thus that of a one-dimensional particle in the *effective potential* $E_{p,\mathrm{eff}}$: the radial [turning points](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-turning) are the roots of $E_{p,\mathrm{eff}}(r) = E_m$, the motion is bound if it is trapped between two of them, and circular if $r$ sits at a minimum of $E_{p,\mathrm{eff}}$.

**Proof.** Part 1 is [Corollary 15.5](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#cor-b1-angular-momentum-central). The force is [conservative](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative) and no other force acts, so $E_m$ is conserved; in the plane, $v^2 = \dot r^2 + r^2\dot\theta^2$ and $r^2\dot\theta^2 = C^2/r^2$. The term $mC^2/2r^2$, the [kinetic energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#thm-b1-work-and-energy-ke) of the angular motion, acts as a repulsive *[centrifugal barrier](#prop-b1-central-forces-effective)* in the radial problem, to which the one-dimensional analysis of [Chapter 13](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#ch-b1-work-and-energy) applies. ∎

## 16.2 Newtonian gravitation

**Definition 16.3 (Universal gravitation).**

A point mass $M$ at $O$ attracts a point mass $m$ at $M$ with

$$
\vect F = -\frac{GMm}{r^2}\,\vect e_r, \qquad E_p = -\frac{GMm}{r} \quad (E_p \to 0 \text{ at infinity}),
$$

$G = 6.67 \times 10^{-11}\,\mathrm{N}\,\mathrm{m}^{2}/\mathrm{kg}^{2}$. A spherically symmetric body attracts outside itself as if its mass were at its center (admitted; a consequence of Gauss’s theorem, [Chapter 26](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#ch-b1-electrostatics-gauss)).

**Proposition 16.4 (The effective potential of gravitation).**

For $E_p = -GMm/r$,

$$
E_{p,\mathrm{eff}}(r) = -\frac{GMm}{r} + \frac{L^2}{2mr^2}
$$

tends to $+\infty$ at $r \to 0$ (for $L \neq 0$), to $0^-$ at infinity, and has one minimum at $r_c = L^2/GMm^2$, of value $-G^2M^2m^3/2L^2$. Hence: $E_m < 0$ — [bound motion](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-turning) between a minimum and a maximum radius (perigee and apogee); $E_m \geq 0$ — unbound, the body comes from and returns to infinity; $E_m$ equal to the minimum — [circular orbit](#prop-b1-central-forces-effective) of radius $r_c$.

**Proof.** Limits by inspection; $E_{p,\mathrm{eff}}' = GMm/r^2 - L^2/mr^3 = 0$ at $r_c$; substitute. The circular case is $\dot r = 0$ forever, possible only at the minimum. ∎

![The effective potential of Newtonian gravitation: the attraction -GMm/r plus the centrifugal barrier L2/2mr2. Below zero energy the radial motion is trapped between two turning points (perigee, apogee); at the minimum the orbit is circular; at or above zero the body escapes.](https://one-course.com/images/onecourse/chapters/physics-3/b1-central-forces/fig-4f6732b45192.svg)

*The effective potential of Newtonian [gravitation](#def-b1-central-forces-gravitation): the attraction $-GMm/r$ plus the [centrifugal barrier](#prop-b1-central-forces-effective) $L^2/2mr^2$. Below zero energy the radial motion is trapped between two [turning points](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-turning) (perigee, apogee); at the minimum the orbit is circular; at or above zero the body escapes.*

**Theorem 16.5 (Circular orbits).**

On a [circular orbit](#prop-b1-central-forces-effective) of radius $r$ about a mass $M$:

$$
v = \sqrt{\frac{GM}{r}}, \qquad
T = 2\pi\sqrt{\frac{r^3}{GM}}, \qquad
E_m = -\frac{GMm}{2r} = -E_k = \tfrac12 E_p .
$$

In particular $T^2/r^3 = 4\pi^2/GM$ is the same for all [circular orbits](#prop-b1-central-forces-effective) about the same center ([Kepler’s third law](#thm-b1-central-forces-circular)).

**Proof.** Uniform [circular motion](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#cor-b1-point-kinematics-circular) needs the [centripetal force](https://one-course.com/books/physics/3/en/chapter/12-newtons-laws-of-dynamics#prop-b1-newton-dynamics-centripetal) $mv^2/r = GMm/r^2$; $T = 2\pi r/v$; $E_k = \tfrac12 mv^2 = GMm/2r$ and $E_p = -GMm/r$. ∎

**Example 16.6 (Three orbits).**

$GM_\oplus = 3.99 \times 10^{14}\,\mathrm{m}^{3}/\mathrm{s}^{2}$. Low orbit ($r = 6770\,\mathrm{km}$, $400\,\mathrm{km}$ up): $v = 7.7\,\mathrm{km}/\mathrm{s}$, $T = 92\,\mathrm{min}$. Geostationary ($T = 86\,164\,\mathrm{s}$, one sidereal day): $r = (GMT^2/4\pi^2)^{1/3} =
42\,200\,\mathrm{km}$, $v = 3.1\,\mathrm{km}/\mathrm{s}$. The Moon ($r = 384\,000\,\mathrm{km}$): $T = 27.4\,\mathrm{d}$ — and its [centripetal acceleration](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#cor-b1-point-kinematics-circular) $v^2/r =
2.7 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$ is $g/3600$ at sixty Earth radii: the inverse square, checked by Newton himself.

## 16.3 Kepler’s laws and the shape of orbits

**Theorem 16.7 (Trajectories in a Newtonian field).**

The [trajectory](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#def-b1-point-kinematics-frame) of a body in the field of a fixed mass $M$ is a conic with $O$ at one focus: an ellipse if $E_m < 0$, a parabola if $E_m = 0$, a hyperbola if $E_m > 0$. For the ellipse of semi-major axis $a$ and eccentricity $e$:

- (Kepler I) the center of force is at a focus; perigee and apogee are at $r_p = a(1 - e)$ and $r_a = a(1 + e)$ ;
- (Kepler II) the [areal velocity](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#cor-b1-angular-momentum-central) is constant (the law of areas);
- (Kepler III) the period obeys $T^2 = 4\pi^2a^3/GM$ ;
- the energy depends on $a$ alone, $E_m = -GMm/2a$, so that the speed at distance $r$ is given by $$v^2 = GM\left(\frac{2}{r} - \frac{1}{a}\right) .$$

**Proof.** *Admitted at this level.* ∎

**Remark 16.8 (What is proved and what is admitted).**

Kepler II and the energy classification were proved above. That the bound orbits are exactly ellipses with $O$ at a focus, with Kepler III and $E_m = -GMm/2a$ in general, needs the integration of the radial equation (the Year 2 volume does it); for [circular orbits](#prop-b1-central-forces-effective) ($a = r$, $e = 0$) every statement reduces to [Theorem 16.5](#thm-b1-central-forces-circular), which is the check we can make here. The relation $v^2 = GM(2/r - 1/a)$ follows from $E_m = \tfrac12 mv^2 - GMm/r = -GMm/2a$ once that energy formula is granted.

![Left: an elliptical orbit about the focus O — semi-major axis a, semi-minor b, focal distance c = ea; the body is fastest at the perigee P and slowest at the apogee A. Right: a hyperbola (positive energy) and a parabola (zero energy) about the same focus — the unbound paths of comets and probes.](https://one-course.com/images/onecourse/chapters/physics-3/b1-central-forces/fig-2c1aadf4ebcf.svg)

*Left: an elliptical orbit about the focus $O$ — semi-major axis $a$, semi-minor $b$, focal distance $c = ea$; the body is fastest at the perigee $P$ and slowest at the apogee $A$. Right: a hyperbola (positive energy) and a parabola (zero energy) about the same focus — the unbound paths of comets and probes.*

**Example 16.9 (Halley’s comet).**

Perihelion $0.586\,\mathrm{AU}$, aphelion $35.1\,\mathrm{AU}$: $a = 17.8\,\mathrm{AU}$, $e = 0.967$, $T = \sqrt{17.8^3} = 75\,\mathrm{years}$ (Kepler III in astronomical units and years, where $T^2 = a^3$ for the Sun). With $GM_\odot = 1.33 \times 10^{20}\,\mathrm{m}^{3}/\mathrm{s}^{2}$, the speed at perihelion is $\sqrt{GM(2/r_p - 1/a)} = 54\,\mathrm{km}/\mathrm{s}$ and at aphelion $0.9\,\mathrm{km}/\mathrm{s}$ — sixty times slower, as the law of areas demands ($r_pv_p = r_av_a$).

## 16.4 Satellites

**Proposition 16.10 (Escape velocity; geostationary orbit).**

From the surface of a body of mass $M$ and radius $R$, the minimum launch speed to reach infinity is $v_e = \sqrt{2GM/R} = \sqrt2\,v_c$, where $v_c = \sqrt{GM/R}$ is the circular speed at ground level: $11.2\,\mathrm{km}/\mathrm{s}$ for the Earth. A *geostationary* satellite orbits in the equatorial plane with the Earth’s sidereal period, at $r = 42\,200\,\mathrm{km}$ ($35\,800\,\mathrm{km}$ above ground).

**Proof.** $E_m \geq 0$: $\tfrac12 mv_e^2 - GMm/R = 0$. Kepler III with $T = 86\,164\,\mathrm{s}$. ∎

**Proposition 16.11 (Hohmann transfer).**

To move a satellite from a [circular orbit](#prop-b1-central-forces-effective) of radius $r_1$ to a coplanar [circular orbit](#prop-b1-central-forces-effective) of radius $r_2 > r_1$ with two brief burns, fire at $r_1$ to enter the *transfer ellipse* of perigee $r_1$ and apogee $r_2$ ($a = (r_1 + r_2)/2$), coast half a period, and fire again at $r_2$ to circularize:

$$
\Delta v_1 = \sqrt{\frac{GM}{r_1}}\left(\sqrt{\frac{2r_2}{r_1 + r_2}} - 1\right),
\qquad
\Delta v_2 = \sqrt{\frac{GM}{r_2}}\left(1 - \sqrt{\frac{2r_1}{r_1 + r_2}}\right),
\qquad
t = \pi\sqrt{\frac{(r_1 + r_2)^3}{8GM}} .
$$

**Proof.** Speeds on the transfer ellipse from $v^2 = GM(2/r - 1/a)$ at $r_1$ and $r_2$; subtract the circular speeds; half the period of [Theorem 16.7](#thm-b1-central-forces-conics). ∎

![A Hohmann transfer between two circular orbits: a first burn at the inner orbit raises the apogee to r_2; half an ellipse later a second burn raises the perigee — the least-fuel two-burn route, and the one that takes a probe to Mars.](https://one-course.com/images/onecourse/chapters/physics-3/b1-central-forces/fig-1c4f63cefac0.svg)

*A [Hohmann transfer](#prop-b1-central-forces-hohmann) between two [circular orbits](#prop-b1-central-forces-effective): a first burn at the inner orbit raises the apogee to $r_2$; half an ellipse later a second burn raises the perigee — the least-fuel two-burn route, and the one that takes a probe to Mars.*

**Example 16.12 (Low orbit to geostationary).**

$r_1 = 6770\,\mathrm{km}$, $r_2 = 42\,200\,\mathrm{km}$: $\Delta v_1 = 2.4\,\mathrm{km}/\mathrm{s}$, $\Delta v_2 = 1.5\,\mathrm{km}/\mathrm{s}$, coast $5.3\,\mathrm{h}$. The total, $3.9\,\mathrm{km}/\mathrm{s}$, is half what the launch to low orbit already cost: the “last mile” to a high orbit is expensive, and satellites carry their own engine for it.

**Remark 16.13 (Repulsive centers).**

For a repulsive inverse-square force (two like charges, [Chapter 26](https://one-course.com/books/physics/3/en/chapter/26-electrostatics-field-and-gausss-law#ch-b1-electrostatics-gauss)) the effective potential has no well: every [trajectory](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#def-b1-point-kinematics-frame) is a hyperbola, the particle approaches, is deflected and leaves. Aiming an $\alpha$ particle straight at a gold nucleus, its distance of closest approach follows from energy alone, $E_k =
2Ze^2/4\pi\varepsilon_0 r_{\min}$ — Rutherford’s way of bounding the size of the nucleus ([Exercise 16.11](#exo-b1-central-forces-11)).

## 16.5 Exercises

**Exercise 16.1 ★.**

The International Space Station orbits at $400\,\mathrm{km}$ altitude ($R_\oplus = 6370\,\mathrm{km}$, $GM_\oplus = 3.99 \times 10^{14}\,\mathrm{m}^{3}/\mathrm{s}^{2}$). Speed, period, number of sunrises its crew sees per day.

**Solution of Exercise 16.1.**

$r = 6770\,\mathrm{km}$: $v = \sqrt{GM/r} = 7.7\,\mathrm{km}/\mathrm{s}$; $T = 2\pi r/v
= 5.5 \times 10^{3}\,\mathrm{s} = 92\,\mathrm{min}$; $1440/92 \approx 16$ sunrises a day.

**Exercise 16.2 ★.**

Find the radius and altitude of the [geostationary orbit](#prop-b1-central-forces-escape) from the sidereal day $86\,164\,\mathrm{s}$, and the [orbital speed](#thm-b1-central-forces-circular).

**Solution of Exercise 16.2.**

$r = (GMT^2/4\pi^2)^{1/3} = (3.99\times10^{14} \times 7.42\times10^9/39.5)^{1/3}
= 4.22 \times 10^{7}\,\mathrm{m}$: altitude $35\,800\,\mathrm{km}$; $v = 2\pi r/T = 3.1\,\mathrm{km}/\mathrm{s}$.

**Exercise 16.3 ★.**

[Escape velocity](#prop-b1-central-forces-escape) from the Earth and from the Moon ($GM = 4.90 \times 10^{12}\,\mathrm{m}^{3}/\mathrm{s}^{2}$, $R = 1740\,\mathrm{km}$). Why does the Moon keep no atmosphere?

**Solution of Exercise 16.3.**

$v_e = \sqrt{2GM/R}$: Earth $11.2\,\mathrm{km}/\mathrm{s}$; Moon $\sqrt{2 \times 4.90\times
10^{12}/1.74\times10^6} = 2.4\,\mathrm{km}/\mathrm{s}$. Gas molecules at a few hundred meters per second have a tail of their speed distribution above $2.4\,\mathrm{km}/\mathrm{s}$; over geological times the Moon’s gas leaked away.

**Exercise 16.4 ★.**

Mars has $a = 1.524\,\mathrm{AU}$, Jupiter $5.20\,\mathrm{AU}$: their years, from [Kepler’s third law](#thm-b1-central-forces-circular). A body at $30\,\mathrm{AU}$ (Neptune)?

**Solution of Exercise 16.4.**

$T = a^{3/2}$ (years, AU): Mars $1.88\,\mathrm{yr}$; Jupiter $11.9\,\mathrm{yr}$; $30\,\mathrm{AU}$: $164\,\mathrm{yr}$.

**Exercise 16.5 ★★.**

Energy needed to place $1000\,\mathrm{kg}$ on the ISS orbit starting from rest at the surface (ignore the Earth’s rotation and the air). Compare with the chemical energy of gasoline ($46\,\mathrm{MJ}/\mathrm{kg}$).

**Solution of Exercise 16.5.**

$\Delta E = \tfrac12 mv^2 + GMm(1/R - 1/r) = 2.94\times10^{10} + 3.99\times
10^{17} \times (1.570 - 1.477)\times10^{-7} = 2.94\times10^{10} + 0.37\times
10^{10} = 3.3 \times 10^{10}\,\mathrm{J}$ — $33\,\mathrm{MJ}/\mathrm{kg}$, the energy of $0.7\,\mathrm{kg}$ of gasoline per kilogram placed in orbit (rockets need far more, having to lift their own fuel).

**Exercise 16.6 ★★.**

Halley’s comet: perihelion $0.586\,\mathrm{AU}$, aphelion $35.1\,\mathrm{AU}$. Compute $a$, $e$, the period, and the speeds at perihelion and aphelion ($GM_\odot = 1.33 \times 10^{20}\,\mathrm{m}^{3}/\mathrm{s}^{2}$, $1\,\mathrm{AU} = 1.496 \times 10^{11}\,\mathrm{m}$).

**Solution of Exercise 16.6.**

$a = (0.586 + 35.1)/2 = 17.8\,\mathrm{AU}$; $e = (35.1 - 0.586)/35.7 = 0.967$; $T = 17.8^{3/2} = 75\,\mathrm{yr}$. $v_p^2 = GM(2/r_p - 1/a)$ with $r_p =
8.77\times10^{10}$ m, $a = 2.67\times10^{12}$ m: $v_p = 54\,\mathrm{km}/\mathrm{s}$; $v_a = v_pr_p/r_a = 0.91\,\mathrm{km}/\mathrm{s}$.

**Exercise 16.7 ★★.**

[Hohmann transfer](#prop-b1-central-forces-hohmann) from the ISS orbit ($6770\,\mathrm{km}$) to the [geostationary orbit](#prop-b1-central-forces-escape) ($42\,200\,\mathrm{km}$): both burns, total, and duration.

**Solution of Exercise 16.7.**

$a = 24\,490\,\mathrm{km}$. $v_1 = 7.67\,\mathrm{km}/\mathrm{s}$; on the ellipse at $r_1$: $\sqrt{GM(2/r_1 - 1/a)} = 10.07\,\mathrm{km}/\mathrm{s}$: $\Delta v_1 = 2.4\,\mathrm{km}/\mathrm{s}$. At $r_2$: $\sqrt{GM(2/r_2 - 1/a)} = 1.62\,\mathrm{km}/\mathrm{s}$, circular $3.07\,\mathrm{km}/\mathrm{s}$: $\Delta v_2 = 1.45\,\mathrm{km}/\mathrm{s}$. Total $3.85\,\mathrm{km}/\mathrm{s}$; duration $\pi\sqrt{a^3/GM} = 1.9 \times 10^{4}\,\mathrm{s} = 5.3\,\mathrm{h}$.

**Exercise 16.8 ★★.**

Show that the minimum of $E_{p,\mathrm{eff}}$ sits at $r_c = L^2/GMm^2$, and that small radial oscillations about it have the [angular frequency](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) $\sqrt{GM/r_c^3}$ — the orbital frequency itself. What does this say about a slightly disturbed [circular orbit](#prop-b1-central-forces-effective)?

**Solution of Exercise 16.8.**

$E_{p,\mathrm{eff}}' = GMm/r^2 - L^2/mr^3 = 0$ at $r_c = L^2/GMm^2$. $E_{p,\mathrm{eff}}'' = -2GMm/r^3 + 3L^2/mr^4$; at $r_c$, $L^2 = GMm^2r_c$ gives $E'' = GMm/r_c^3$, so $\omega_r^2 = E''/m = GM/r_c^3 = \omega_{\text{orb}}^2$. A small radial disturbance oscillates exactly once per revolution: the orbit closes on itself — a slightly eccentric ellipse, not a rosette.

**Exercise 16.9 ★★.**

Astronauts aboard the ISS float. Explain with the forces; what is the value of $g$ at their altitude, and why does it not contradict the floating?

**Solution of Exercise 16.9.**

Only gravity acts on station and crew alike; both fall on the same orbit, so no contact force is needed between them — no floor pushes, hence “weightlessness”. $g = GM/r^2 = 3.99\times10^{14}/(6.77\times10^6)^2
= 8.7\,\mathrm{m}/\mathrm{s}^{2}$: gravity is almost full strength; weight is not absent, it is entirely spent bending the path.

**Exercise 16.10 ★★★.**

Newton’s “Moon test”: compute the Moon’s [centripetal acceleration](https://one-course.com/books/physics/3/en/chapter/11-kinematics-of-a-point#cor-b1-point-kinematics-circular) from $r = 3.84 \times 10^{8}\,\mathrm{m}$ and $T = 27.3\,\mathrm{d}$, and compare with $g(R/r)^2$. Conclude.

**Solution of Exercise 16.10.**

$a = 4\pi^2r/T^2 = 39.5 \times 3.84\times10^8/(2.36\times10^6)^2 =
2.7 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$; $g(R/r)^2 = 9.81/60.3^2 = 2.7 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$: the same force law that drops an apple holds the Moon, diluted as $1/r^2$.

**Exercise 16.11 ★★★.**

An $\alpha$ particle ($q = 2e$, $E_k = 5.0\,\mathrm{MeV}$) is fired straight at a gold nucleus ($Z = 79$), taken fixed. Using energy conservation with $E_p = 2Ze^2/4\pi\varepsilon_0r$ ($1/4\pi\varepsilon_0 = 8.99 \times 10^{9}\,\mathrm{SI}$), find the distance of closest approach; compare with the nuclear radius, about $7\,\mathrm{fm}$. What energy would be needed to touch the nucleus?

**Solution of Exercise 16.11.**

$E_k = 2Ze^2/4\pi\varepsilon_0r_{\min}$: $r_{\min} = 8.99\times10^9 \times 2 \times 79
\times (1.60\times10^{-19})^2/8.0\times10^{-13} = 4.5 \times 10^{-14}\,\mathrm{m} = 45\,\mathrm{fm}$, six times the nuclear radius: the $\alpha$ turns back before touching. To reach $7\,\mathrm{fm}$: $E_k \approx 32\,\mathrm{MeV}$.

**Exercise 16.12 ★★★.**

A satellite in low orbit is slowed by the residual atmosphere, so its [mechanical energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#thm-b1-work-and-energy-em) decreases. Show with $E_m = -GMm/2r$ and $v = \sqrt{GM/r}$ that it then moves *faster*. Where does the energy go, and what fraction of the lost [potential energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative) becomes heat?

**Solution of Exercise 16.12.**

$E_m = -GMm/2r$ decreases $\Rightarrow$ $r$ decreases $\Rightarrow$ $v =
\sqrt{GM/r}$ increases. The [kinetic energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#thm-b1-work-and-energy-ke) $GMm/2r$ rises by $\abs{\Delta E}$ while the [potential energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative) $-GMm/r$ falls by $2\abs{\Delta E}$: half of the [potential energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#def-b1-work-and-energy-conservative) released becomes kinetic, half is dissipated as heat by the drag.

![A GPS satellite (NASA artist’s view): one of the thirty-odd in the 20\,200\, km constellation of the weekend problem, each a clock in a Keplerian orbit.](https://one-course.com/images/onecourse/chapters/physics-3/b1-central-forces/img-cc30b677620c.jpg)

*A GPS satellite (NASA artist’s view): one of the thirty-odd in the $20\,200\,\mathrm{km}$ constellation of the weekend problem, each a clock in a Keplerian orbit.*

## 16.6 Problem: The satellites that tell you where you are

**Problem 16.1.**

Weekend problem — a navigation constellation twenty thousand kilometers up: its orbit, the fuel to get there, how many satellites a phone can see, and why their clocks must be told to run slow

Data: $GM_\oplus = 3.986 \times 10^{14}\,\mathrm{m}^{3}/\mathrm{s}^{2}$, $R_\oplus = 6371\,\mathrm{km}$, sidereal day $86\,164\,\mathrm{s}$, $c = 2.998 \times 10^{8}\,\mathrm{m}/\mathrm{s}$. The satellites circle the Earth in exactly half a sidereal day on [circular orbits](#prop-b1-central-forces-effective).

**Part I — The orbit.**

1. Compute the orbital radius and the altitude.
2. Compute the [orbital speed](#thm-b1-central-forces-circular) .
3. Compute the satellite’s acceleration and compare it with $g$ at the surface.
4. Compute the [mechanical energy](https://one-course.com/books/physics/3/en/chapter/13-work-energy-and-potential-energy#thm-b1-work-and-energy-em) per kilogram, and the kinetic and potential parts.
5. Why choose half a sidereal day rather than half a solar day? What does a ground observer see repeat?
6. Seen from a satellite, what angle does the Earth’s disk subtend?
7. Compare the orbit’s radius with the geostationary one; which is higher, and by what factor?

**Part II — Getting there.**

8. Energy per kilogram to go from rest at the surface to the final orbit (ignore the air and the Earth’s spin).
9. The launcher first reaches a low [circular orbit](#prop-b1-central-forces-effective) at $400\,\mathrm{km}$ . Speed there?
10. [Hohmann transfer](#prop-b1-central-forces-hohmann) from that orbit to the final one: the two burns and their sum.
11. Duration of the transfer.
12. Compare the sum of the burns with the [escape velocity](#prop-b1-central-forces-escape) from low orbit: how far from “leaving the Earth” is this mission?
13. The upper stage and satellite mass $2000\,\mathrm{kg}$ before the transfer; the engine’s exhaust speed is $3.0\,\mathrm{km}/\mathrm{s}$ and the rocket equation gives $\Delta v = u\ln(m_0/m_1)$ . Propellant needed for the two burns?

**Part III — Coverage and timing.**

14. A satellite is visible from every point of the Earth that lies within the cone of half-angle $\theta$ with $\cos\theta = R/r$ about the Earth–satellite line. Compute $\theta$ and the fraction $(1 - \cos\theta)/2$ of the Earth’s surface it covers.
15. With $24$ satellites spread uniformly, how many are visible on average from one point? Is the minimum of $4$ needed for a fix plausible?
16. Light travel time from a satellite at the zenith; at the horizon.
17. To locate the receiver to $3\,\mathrm{m}$ , to what precision must the [signals](https://one-course.com/books/physics/3/en/chapter/5-signal-propagation-travelling-and-standing-waves#def-b1-wave-propagation-signal) be timed?
18. The receiver’s own clock is a cheap quartz, wrong by milliseconds: why are four satellites needed instead of three?
19. The satellites’ clocks are atomic. Give the relative stability needed for a $3\,\mathrm{m}$ error after one day.

**Part IV — Relativity, briefly.** Two effects shift the satellite clocks relative to ground clocks (formulas admitted, from the Year 3 volume): motion slows a clock by the fraction $v^2/2c^2$; height speeds it up by $GM(1/R - 1/r)/c^2$.

20. Compute the slowing due to speed, in microseconds per day.
21. Compute the speeding-up due to altitude, in microseconds per day.
22. Net effect, and the position error it would cause per day if uncorrected.
23. The satellite clocks are set before launch to $10.229\,999\,995\,43\,\mathrm{MHz}$ instead of $10.23\,\mathrm{MHz}$ . Check that this compensates the net effect.
24. The orbits are not exactly Keplerian: name three perturbations and say how the system copes.
25. Summarize the chain: two conservation laws, one orbit, one constellation, and the one correction nobody would have guessed.

**Solution of Problem 16.1.**

**1.** $T = 43\,082\,\mathrm{s}$: $r = (GMT^2/4\pi^2)^{1/3} = (3.986\times10^{14}
\times 1.856\times10^9/39.48)^{1/3} = 2.656 \times 10^{7}\,\mathrm{m}$; altitude $20\,190\,\mathrm{km}$.

**2.** $v = \sqrt{GM/r} = 3.87\,\mathrm{km}/\mathrm{s}$.

**3.** $a = v^2/r = GM/r^2 = 0.565\,\mathrm{m}/\mathrm{s}^{2} = g(R/r)^2$, seventeen times weaker than at the surface — but not zero: the satellite falls continuously around the Earth.

**4.** $E_m/m = -GM/2r = -7.5\,\mathrm{MJ}/\mathrm{kg}$; $E_k/m = +7.5\,\mathrm{MJ}/\mathrm{kg}$, $E_p/m = -15.0\,\mathrm{MJ}/\mathrm{kg}$.

**5.** The Earth turns once per sidereal day relative to the stars (the frame of the orbit); after two orbits the satellite is back over the same ground point: the ground tracks repeat every sidereal day.

**6.** $\sin\beta = R/r = 0.240$: $\beta = 13.9^\circ$, a disk $28^\circ$ wide.

**7.** Geostationary $42\,200\,\mathrm{km}$: $1.59$ times higher; the navigation orbit is below it.

**8.** $GM(1/R - 1/2r) = 3.986\times10^{14}(1.570\times10^{-7} - 1.88\times
10^{-8}) = 5.5 \times 10^{7}\,\mathrm{J}/\mathrm{kg} = 55\,\mathrm{MJ}/\mathrm{kg}$.

**9.** $r_1 = 6771\,\mathrm{km}$: $v = 7.67\,\mathrm{km}/\mathrm{s}$.

**10.** $a = 16\,665\,\mathrm{km}$; at $r_1$: $\sqrt{GM(2/r_1 - 1/a)} =
9.69\,\mathrm{km}/\mathrm{s}$, $\Delta v_1 = 2.0\,\mathrm{km}/\mathrm{s}$; at $r_2$: $2.47\,\mathrm{km}/\mathrm{s}$ against circular $3.87\,\mathrm{km}/\mathrm{s}$, $\Delta v_2 = 1.4\,\mathrm{km}/\mathrm{s}$; total $3.4\,\mathrm{km}/\mathrm{s}$.

**11.** $\pi\sqrt{a^3/GM} = 1.07 \times 10^{4}\,\mathrm{s} \approx 3.0\,\mathrm{h}$.

**12.** Escape from low orbit: $(\sqrt2 - 1)\times 7.67 = 3.2\,\mathrm{km}/\mathrm{s}$: the mission’s $3.4\,\mathrm{km}/\mathrm{s}$ exceeds it — reaching this orbit costs about as much as leaving the Earth altogether.

**13.** $m_0/m_1 = \eu^{3.4/3.0} = 3.1$: $m_1 = 640\,\mathrm{kg}$, propellant $1360\,\mathrm{kg}$ — two thirds of the mass.

**14.** $\cos\theta = 0.240$, $\theta = 76^\circ$; fraction $(1 - 0.240)/2
= 0.38$.

**15.** $24 \times 0.38 = 9$ on average; four is comfortably exceeded almost always.

**16.** Zenith: $2.02\times10^7/c = 67\,\mathrm{ms}$; horizon: distance $\sqrt{r^2 - R^2} = 2.58 \times 10^{7}\,\mathrm{m}$, $86\,\mathrm{ms}$.

**17.** $3/c = 10\,\mathrm{ns}$.

**18.** The receiver’s clock offset is a fourth unknown besides its three coordinates: four distances, four equations.

**19.** $10^{-8}\ \mathrm{s}/86\,400\,\mathrm{s} \approx 1 \times 10^{-13}$.

**20.** $v^2/2c^2 = (3874)^2/(2 \times 8.99\times10^{16}) = 8.3 \times 10^{-11}$: $8.3\times10^{-11} \times 86400 = 7.2\,\text{µ}\mathrm{s}$ per day, slow.

**21.** $GM(1/R - 1/r)/c^2 = 3.986\times10^{14} \times 1.19\times10^{-7}/
8.99\times10^{16} = 5.3 \times 10^{-10}$: $45.7\,\text{µ}\mathrm{s}$ per day, fast.

**22.** Net $+38.5\,\text{µ}\mathrm{s}$ per day; $c \times 38.5\times10^{-6}
= 11.5\,\mathrm{km}$ of error per day.

**23.** $(10.23 - 10.22999999543)/10.23 = 4.47 \times 10^{-10}$, matching the net $4.5\times10^{-10}$ ($38.5\,\text{µ}\mathrm{s}$ over $86\,400\,\mathrm{s}$).

**24.** The Earth’s equatorial bulge, the Moon’s and Sun’s attraction, solar radiation pressure (and small thruster firings): the ground segment measures the orbits continuously and uploads fresh ephemerides, which the satellites broadcast.

**25.** Energy and [angular momentum](https://one-course.com/books/physics/3/en/chapter/15-angular-momentum#def-b1-angular-momentum-L) fix a [circular orbit](#prop-b1-central-forces-effective) from its period; Kepler III places it; geometry fixes coverage and light times; and the clocks — the heart of the system — must be corrected for relativity by $38\,\text{µ}\mathrm{s}$ a day, or the fix drifts by ten kilometers by nightfall.