University Physics — Year 1 · Bachelor Year 1

16Central 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 — 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 if its magnitude depends only on $r = OM$ : $\vect F = F(r)\,\vect e_r$ , which derives from a potential energy $E_p(r)$ 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 force:

keeps its angular momentum $\vect L_O$ ; its motion is planar and, in polar coordinates in that plane, $r^2\dot\theta = C$ is constant (law of areas, Corollary 15.5);
keeps its mechanical energy $E_m = \tfrac12 m(\dot r^2 + r^2\dot\theta^2) + E_p(r)$ , 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 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. The force is 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 of the angular motion, acts as a repulsive centrifugal barrier in the radial problem, to which the one-dimensional analysis of Chapter 13 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).

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 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 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. — The effective potential of Newtonian gravitation: the attraction $-GMm/r$ plus the centrifugal barrier $L^2/2mr^2$ . 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.

Theorem 16.5 (Circular orbits)

On a circular orbit 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 about the same center (Kepler’s third law).

Proof. Uniform circular motion needs the centripetal force $mv^2/r = GMm/r^2$ ; $T = 2\pi r/v$ ; $E_k = \tfrac12 mv^2 = GMm/2r$ 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 $v^2/r = 2.7 \times 10^{-3}\,\mathrm{m}/\mathrm{s}^{2}$ 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 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 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 ( $a = r$ , $e = 0$ ) every statement reduces to Theorem 16.5, 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. — 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 of radius $r_1$ to a coplanar circular orbit 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. ∎

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. — 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.

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) the effective potential has no well: every trajectory 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).

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

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 from the sidereal day $86\,164\,\mathrm{s}$ , and the orbital speed.

Solution

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 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

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. A body at $30\,\mathrm{AU}$ (Neptune)?

Solution

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

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

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 from the ISS orbit ( $6770\,\mathrm{km}$ ) to the geostationary orbit ( $42\,200\,\mathrm{km}$ ): both burns, total, and duration.

Solution

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 $\sqrt{GM/r_c^3}$ — the orbital frequency itself. What does this say about a slightly disturbed circular orbit?

Solution

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

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 from $r = 3.84 \times 10^{8}\,\mathrm{m}$ and $T = 27.3\,\mathrm{d}$ , and compare with $g(R/r)^2$ . Conclude.

Solution

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

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 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 becomes heat?

Solution

Solution of Exercise 16.12.

$E_m = -GMm/2r$ decreases $\Rightarrow$ $r$ decreases $\Rightarrow$ $v = \sqrt{GM/r}$ increases. The kinetic energy $GMm/2r$ rises by $\abs{\Delta E}$ while the potential energy $-GMm/r$ falls by $2\abs{\Delta E}$ : half of the potential energy 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. — 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.

Part I — The orbit.

Compute the orbital radius and the altitude.
Compute the orbital speed.
Compute the satellite’s acceleration and compare it with $g$ at the surface.
Compute the mechanical energy per kilogram, and the kinetic and potential parts.
Why choose half a sidereal day rather than half a solar day? What does a ground observer see repeat?
Seen from a satellite, what angle does the Earth’s disk subtend?
Compare the orbit’s radius with the geostationary one; which is higher, and by what factor?

Part II — Getting there.

Energy per kilogram to go from rest at the surface to the final orbit (ignore the air and the Earth’s spin).
The launcher first reaches a low circular orbit at $400\,\mathrm{km}$ . Speed there?
Hohmann transfer from that orbit to the final one: the two burns and their sum.
Duration of the transfer.
Compare the sum of the burns with the escape velocity from low orbit: how far from “leaving the Earth” is this mission?
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.

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.
With $24$ satellites spread uniformly, how many are visible on average from one point? Is the minimum of $4$ needed for a fix plausible?
Light travel time from a satellite at the zenith; at the horizon.
To locate the receiver to $3\,\mathrm{m}$ , to what precision must the signals be timed?
The receiver’s own clock is a cheap quartz, wrong by milliseconds: why are four satellites needed instead of three?
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 $GM(1/R - 1/r)/c^2$ .

Compute the slowing due to speed, in microseconds per day.
Compute the speeding-up due to altitude, in microseconds per day.
Net effect, and the position error it would cause per day if uncorrected.
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.
The orbits are not exactly Keplerian: name three perturbations and say how the system copes.
Summarize the chain: two conservation laws, one orbit, one constellation, and the one correction nobody would have guessed.

Solution

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 fix a circular orbit 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.

Terms defined in this chapter

See all 393 terms in the glossary