Corrected exercises: universal gravitation (203-SN1-RE)

$g=\dfrac{GM}{R^{2}}=\dfrac{6.674\times 10^{-11}\times 5.972\times 10^{24}}{\left(6.371\times 10^{6}\right)^{2}}=9.82$ m/s². The difference from 9.80 comes from two effects ignored here: the Earth is not perfectly spherical, it is flattened at the poles, and its rotation subtracts a small centrifugal component, largest at the equator. The measured value therefore varies from 9.78 at the equator to 9.83 at the poles.

$F=\dfrac{6.674\times 10^{-11}\times 70.0\times 70.0}{1.00^{2}}=3.27\times 10^{-7}$ N, about 0.33 micronewton. That is the weight of a speck of dust: completely imperceptible. Gravitation is by far the WEAKEST of the fundamental interactions; it dominates at large scales only because it is always attractive and can therefore never cancel out, unlike the electric force, where positive and negative charges offset each other.

We want $\dfrac{GM}{r^{2}}=\dfrac{1}{2}\dfrac{GM}{R^{2}}$, so $r^{2}=2R^{2}$ and $r=R\sqrt{2}=9.010\times 10^{6}$ m. The altitude is $r-R=2.639\times 10^{6}$ m, about 2640 km. You have to rise by more than a third of an Earth radius for gravity to drop by half: it decreases much more slowly than intuition suggests.

$g'=\dfrac{G\left(8M\right)}{\left(2R\right)^{2}}=\dfrac{8}{4}\cdot\dfrac{GM}{R^{2}}=2g$. The surface field there would therefore be DOUBLE that of the Earth. This example shows that $g$ depends on the ratio $\dfrac{M}{R^{2}}$ and not on the mass alone: a much more massive planet can perfectly well have a modest surface field if it is also much larger, which is the case for the gas giants.

Gravitation is the ONLY force, and it is directed toward the centre: it therefore plays exactly the role of the centripetal force. Hence $\dfrac{GMm}{r^{2}}=\dfrac{mv^{2}}{r}$. The mass $m$ of the satellite appears on both sides and CANCELS, leaving $v=\sqrt{\dfrac{GM}{r}}$. A bolt and a 400-tonne station therefore follow the same orbit at the same speed, which is the orbital version of the universality of free fall.

$r=6.371\times 10^{6}+4.00\times 10^{5}=6.771\times 10^{6}$ m. Then $g=\dfrac{GM}{r^{2}}=8.69$ m/s², that is $\dfrac{8.69}{9.82}=88.5$ % of the value at the ground.

$v=\sqrt{\dfrac{6.674\times 10^{-11}\times 5.972\times 10^{24}}{6.771\times 10^{6}}}=7.67\times 10^{3}$ m/s, that is 7.67 km/s or about 27,600 km/h. The period is $T=\dfrac{2\pi r}{v}=\dfrac{2\pi\times 6.771\times 10^{6}}{7672}=5545$ s, that is 92.4 minutes. The Station therefore circles the Earth about 16 times a day, and its crew sees 16 sunrises.

Because the weightlessness they feel is not the absence of gravity, but the absence of a CONTACT force. On the ground, what gives us the sensation of weight is the floor pushing us upward. In orbit, the station and the astronaut undergo the same gravitational acceleration and fall together: the floor no longer presses on the astronaut, there is no contact force left, hence the sensation of floating. The exact term is FREE FALL, or microgravity. The same sensation can be had for a second or two in an elevator whose cable snaps, or in an aircraft flying a parabolic trajectory.

Nothing spectacular: it stays on the SAME orbit, at the same speed, right next to the Station, since its motion does not depend on its mass. It seems motionless relative to the crew and drifts very slowly, because of the tiny relative speed given to it when it was released and of the residual atmospheric drag, slightly different depending on its shape. That is why tools are tethered: a lost object stays in orbit for months and becomes dangerous debris, crossing the paths of other satellites at several kilometres per second.

First part: a homogeneous spherical shell exerts NO net gravitational force on a body located inside it. The attractions of all its elements cancel exactly, the small nearby portion being offset by the large distant portion. Second part: seen from outside, that same shell acts exactly as if all its mass were concentrated at its centre, which justifies the use of $r$ between centres since the start of the chapter. Consequence for a miner at depth $d$: all the matter ABOVE them, that is, the shell of inner radius $r=R-d$, stops counting. Only the inner sphere of radius $r$ still attracts them.

The inner mass is $M_{in}=\rho\frac{4}{3}\pi r^{3}$ while the total mass is $M=\rho\frac{4}{3}\pi R^{3}$, so $M_{in}=M\frac{r^{3}}{R^{3}}$. The field is then $g(r)=\frac{GM_{in}}{r^{2}}=\frac{GM}{R^{3}}r$. Recognizing $g_{surface}=\frac{GM}{R^{2}}$, we get $g(r)=g_{surface}\frac{r}{R}$: the field decreases LINEARLY toward the centre, where it vanishes. With $g_{surface}=\frac{\left(6.674\times 10^{-11}\right)\left(5.972\times 10^{24}\right)}{\left(6.371\times 10^{6}\right)^{2}}\approx 9.82$ m/s$^{2}$ and $r=6.371\times 10^{6}-3.00\times 10^{6}=3.371\times 10^{6}$ m: $g\approx 9.82\left(0.5291\right)\approx 5.20$ m/s$^{2}$. The miner would weigh almost half as much there.

From $r=0$ to $r=R$, the field rises linearly from zero to $9.82$ m/s$^{2}$. Beyond $R$, it decreases as $\frac{1}{r^{2}}$. The field is therefore MAXIMUM at the surface, exactly at $r=R$, and it decreases whether you go up or down. At $r=2R$: $g=\frac{9.82}{4}\approx 2.45$ m/s$^{2}$. The complete curve is a rising straight line joined to a decreasing hyperbola, with a sharp peak at the surface. In the real Earth, whose core is much denser than its mantle, $g$ even increases slightly at the start of the descent before falling toward zero.

In the tunnel, locating the position by $x$ measured from the centre, the only force is that of the inner sphere: $F=-mg(x)=-\frac{GMm}{R^{3}}x$, directed toward the centre on either side. This is exactly the form $F=-kx$ of a spring, with $k=\frac{GMm}{R^{3}}$: the motion is therefore simple harmonic, with amplitude $R$. Angular frequency: $\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{GM}{R^{3}}}=\sqrt{\frac{g_{surface}}{R}}\approx 1.241\times 10^{-3}$ rad/s. Period: $T=\frac{2\pi}{\omega}\approx 5.06\times 10^{3}$ s, that is $84.4$ minutes for the complete round trip, and therefore about $42$ minutes from one end of the globe to the other. Speed at the centre, where it is greatest: $v_{max}=\omega R\approx 7.91\times 10^{3}$ m/s.

Satellite in a grazing orbit, $r=R$: $v=\sqrt{\frac{GM}{R}}\approx 7.91\times 10^{3}$ m/s and $T=\frac{2\pi R}{v}\approx 5.06\times 10^{3}$ s. Both results are EXACTLY identical to those of the tunnel. This is not a numerical coincidence: both motions are governed by the same quantity $\frac{GM}{R^{3}}$, since $\omega_{orbit}=\sqrt{\frac{GM}{R^{3}}}$ is exactly the angular frequency found in d). Geometrically, the motion in the tunnel is the PROJECTION onto a diameter of the uniform circular motion of the grazing satellite: an object released in the tunnel stays permanently directly beneath the satellite. It is the same statement that makes simple harmonic motion the projection of uniform circular motion, and it will be taken up again unchanged in the chapter on SHM.

First law: the planets describe ELLIPSES with the Sun at one focus. Second law, the law of areas: the segment joining the planet to the Sun sweeps out equal areas in equal times, which amounts to saying that the planet moves faster at perihelion. Third law: the square of the period is proportional to the cube of the semi-major axis, $\dfrac{T^{2}}{a^{3}}$ being the same constant for all bodies orbiting the same celestial body.

From $\dfrac{GMm}{r^{2}}=\dfrac{mv^{2}}{r}$ we get $v^{2}=\dfrac{GM}{r}$. Now $v=\dfrac{2\pi r}{T}$, so $\dfrac{4\pi^{2}r^{2}}{T^{2}}=\dfrac{GM}{r}$, which gives $T^{2}=\dfrac{4\pi^{2}}{GM}r^{3}$. The constant $\dfrac{4\pi^{2}}{GM}$ depends ONLY on the mass of the central body: this is why the law holds for all the planets of the Solar System with the same value, and for all the satellites of the Earth with another value.

$T=2\pi\sqrt{\dfrac{r^{3}}{GM}}=2\pi\sqrt{\dfrac{\left(3.844\times 10^{8}\right)^{3}}{6.674\times 10^{-11}\times 5.972\times 10^{24}}}=2.372\times 10^{6}$ s, that is 27.5 days. The observed value is 27.3 days: the agreement is excellent, the difference of less than 1 % coming from the fact that the lunar orbit is slightly elliptical and perturbed by the Sun.

A geostationary satellite must have exactly the rotation period of the Earth, that is 86,164 s. From $T^{2}=\dfrac{4\pi^{2}}{GM}r^{3}$ we get $r=\sqrt[3]{\dfrac{GMT^{2}}{4\pi^{2}}}=4.216\times 10^{7}$ m. The altitude is $r-R=4.216\times 10^{7}-6.371\times 10^{6}=3.579\times 10^{7}$ m, about 35,800 km.

Because this altitude is the ONLY one for which the period is 24 sidereal hours: the third law sets up a one-to-one correspondence between radius and period, so there is no choice. A lower satellite would go round too fast, a higher one too slowly, and in both cases it would drift across the sky, forcing the antennas to track it. As for the equator: the plane of the orbit must contain the CENTRE of the Earth, since the gravitational force is directed toward it. An orbit that stayed permanently above Montreal would be a circle whose centre is not the centre of the Earth, which is dynamically impossible. Only an equatorial orbit can therefore appear fixed, and that is why satellite dishes in Montreal all point south, quite low on the horizon.

The reference is taken at infinity, where $U=0$: two bodies infinitely far apart no longer interact. To bring them closer, the force of attraction does POSITIVE work, so the potential energy DECREASES and becomes negative. A negative value therefore means that energy would have to be supplied to separate the bodies: it is the signature of a BOUND state. A body with negative mechanical energy cannot escape.

With $r=6.771\times 10^{6}$ m: $U=-\dfrac{6.674\times 10^{-11}\times 5.972\times 10^{24}\times 100}{6.771\times 10^{6}}=-5.886\times 10^{9}$ J. The orbital speed being 7672 m/s, $K=\frac{1}{2}\times 100\times 7672^{2}=2.943\times 10^{9}$ J. The mechanical energy is $E=K+U=-2.943\times 10^{9}$ J. We do find $E=\frac{1}{2}U$ and $E=-K$: this is the virial theorem for a circular orbit.

At the ground, the satellite at rest has $U_{ground}=-\dfrac{GMm}{R}=-6.256\times 10^{9}$ J and $K=0$, so $E_{ground}=-6.256\times 10^{9}$ J. The energy to supply is $\Delta E=-2.943\times 10^{9}-\left(-6.256\times 10^{9}\right)=3.31\times 10^{9}$ J, that is 3.31 GJ for 100 kg. This is the equivalent of about 80 kg of gasoline in pure energy, and a real rocket burns far more because of its efficiency and drag.

To escape means to reach infinity with zero speed, and therefore to have zero mechanical energy: $\frac{1}{2}mv^{2}-\dfrac{GMm}{R}=0$, so $v_{esc}=\sqrt{\dfrac{2GM}{R}}$. Numerically, $v_{esc}=1.119\times 10^{4}$ m/s, that is 11.2 km/s. Note that $v_{esc}=\sqrt{2}\,v_{orbit}$ at the same distance: you only need to multiply the orbital speed by 1.41 to escape.

Its speed INCREASES, which is the paradox. Friction removes mechanical energy, so $E$ becomes more negative, so $r$ decreases according to $E=-\dfrac{GMm}{2r}$. Now $v=\sqrt{\dfrac{GM}{r}}$ grows when $r$ decreases: the satellite descends and SPEEDS UP. The resolution of the paradox is that the potential energy drops twice as fast as the kinetic energy increases, so the overall balance is indeed a loss. This is why the International Space Station has to be reboosted regularly, and why atmospheric re-entry takes place at very high speed.

The major axis is the sum of the two extreme distances, so $a=\frac{r_{p}+r_{a}}{2}=\frac{7.00\times 10^{6}+4.20\times 10^{7}}{2}=2.45\times 10^{7}$ m. Eccentricity: $e=\frac{r_{a}-r_{p}}{r_{a}+r_{p}}=\frac{3.50\times 10^{7}}{4.90\times 10^{7}}\approx 0.714$, a very elongated ellipse. Kepler's third law, valid for an ellipse provided the radius is replaced by the SEMI-MAJOR AXIS: $T=2\pi\sqrt{\frac{a^{3}}{GM}}=2\pi\sqrt{\frac{\left(2.45\times 10^{7}\right)^{3}}{3.986\times 10^{14}}}\approx 3.82\times 10^{4}$ s, that is $10.6$ hours.

The gravitational force is central: it passes permanently through the centre of the Earth, so its torque about that point is zero and the angular momentum of the satellite is conserved. At perigee as at apogee, the velocity is perpendicular to the radius, so $L=mv_{p}r_{p}=mv_{a}r_{a}$, hence $\frac{v_{p}}{v_{a}}=\frac{r_{a}}{r_{p}}=6.00$. Conservation of energy provides the second relation: $\frac{1}{2}v_{p}^{2}-\frac{GM}{r_{p}}=\frac{1}{2}v_{a}^{2}-\frac{GM}{r_{a}}$. Substituting $v_{a}=v_{p}\frac{r_{p}}{r_{a}}$ gives $v_{p}=\sqrt{\frac{2GMr_{a}}{r_{p}\left(r_{p}+r_{a}\right)}}\approx 9.88\times 10^{3}$ m/s, and therefore $v_{a}=\frac{9880}{6.00}\approx 1.65\times 10^{3}$ m/s. The satellite passes perigee six times faster than apogee.

During an interval $dt$, the radius vector sweeps out a thin triangle with base $v\,dt$ and height equal to the lever arm. Its area is $dA=\frac{1}{2}r^{2}d\theta$, so the areal velocity is $\frac{dA}{dt}=\frac{1}{2}r^{2}\frac{d\theta}{dt}=\frac{1}{2}rv_{\perp}$. Now the angular momentum is $L=mrv_{\perp}$, hence $\frac{dA}{dt}=\frac{L}{2m}$. The law of areas states that this areal velocity is constant: that is exactly saying that $L$ is constant. Kepler had therefore discovered empirically, as early as 1609, the conservation of angular momentum, without having the concept.

At perigee: $E=\frac{1}{2}mv_{p}^{2}-\frac{GMm}{r_{p}}=\frac{1}{2}(1{,}500)\left(9880.1\right)^{2}-\frac{\left(3.986\times 10^{14}\right)(1{,}500)}{7.00\times 10^{6}}=7.32\times 10^{10}-8.54\times 10^{10}\approx -1.22\times 10^{10}$ J. Check: $-\frac{GMm}{2a}=-\frac{\left(3.986\times 10^{14}\right)(1{,}500)}{2\left(2.45\times 10^{7}\right)}\approx -1.22\times 10^{10}$ J, identical. This relation is remarkable: the energy of an orbit depends ONLY on the semi-major axis, not on the eccentricity. Two orbits with the same $a$, one nearly circular and the other very elongated, cost exactly the same energy, and have exactly the same period.

You must act at APOGEE, where the speed is lowest. The circular speed at that distance is $v_{c}=\sqrt{\frac{GM}{r_{a}}}=\sqrt{\frac{3.986\times 10^{14}}{4.20\times 10^{7}}}\approx 3.08\times 10^{3}$ m/s, whereas the satellite is moving there at only $1647$ m/s. You must therefore add $\Delta v=3081-1647\approx 1.43\times 10^{3}$ m/s in the direction of motion. Corresponding energy: the new circular orbit has $a'=r_{a}$, so $E'=-\frac{GMm}{2r_{a}}\approx -7.12\times 10^{9}$ J, against $-1.22\times 10^{10}$ J before, an input of $\Delta E\approx 5.08\times 10^{9}$ J. You act at apogee because a thrust mainly changes the DIAMETRICALLY OPPOSITE point of the orbit: speeding up at apogee raises the perigee, which is precisely what is needed to round off the ellipse. This is the second burn of a Hohmann transfer, the standard manoeuvre for placing a satellite in geostationary orbit.

We look for $d$ such that $\dfrac{GM_{E}}{d^{2}}=\dfrac{GM_{Moon}}{\left(D-d\right)^{2}}$, with $D=3.844\times 10^{8}$ m. Cancelling $G$ and taking the root: $\dfrac{D-d}{d}=\sqrt{\dfrac{M_{Moon}}{M_{E}}}=\sqrt{\dfrac{7.348\times 10^{22}}{5.972\times 10^{24}}}=0.1109$. So $D=d\left(1+0.1109\right)$ and $d=\dfrac{3.844\times 10^{8}}{1.1109}=3.460\times 10^{8}$ m.

The neutral point is at $\dfrac{3.460}{3.844}=90.0$ % of the way, so much closer to the Moon than to the Earth. This is consistent with the Earth being 81 times more massive: you have to go a long way from it for its field, decreasing as $\frac{1}{r^{2}}$, to fall to the level of the Moon's. It is also the point a probe must cross to “fall” toward the Moon rather than back to the Earth.

The lunar field is stronger on the side of the Earth nearest the Moon, and weaker on the opposite side. In the frame of the Earth, which is itself falling toward the Moon with the acceleration of its CENTRE, the water on the near side is attracted MORE than the centre, so it rises toward the Moon; the water on the far side is attracted LESS than the centre, so it is left behind, which moves it away from the centre and raises it as well. TWO diametrically opposite bulges thus form, and as the Earth rotates on itself, a point on the globe passes through both in a little over 24 hours (one lunar day): hence two high tides a day. It is the DIFFERENCES in field, called tidal forces, that matter, not the field itself.

This is the EQUIVALENCE PRINCIPLE: no local experiment can distinguish a frame in free fall in a gravitational field from an inertial frame without gravity, nor a uniformly accelerated frame from one at rest in a gravitational field. It is the starting point of general relativity, in which Einstein deduces that gravitation is not a force but a curvature of spacetime.

The GRAVITATIONAL mass, the one in the law of gravitation, measures how a body creates and undergoes gravitation: it is a kind of “gravitational charge”, analogous to electric charge. The INERTIAL mass, the one in $F=ma$, measures the resistance of the body to any acceleration, whatever its cause. Nothing in the principles requires them to be equal, any more than the electric charge of a body is equal to its mass. Yet their equality has been verified experimentally to better than $10^{-15}$. Its immediate consequence is that $m$ cancels in $mg=ma$: all bodies fall with the same acceleration, independently of their mass and composition. It is this coincidence, unexplained in Newton's work, that Einstein raised to the status of a principle.

$F=G\frac{m_{1}m_{2}}{r^{2}}=\frac{\left(6.674\times 10^{-11}\right)(158)(0.730)}{(0.230)^{2}}\approx 1.46\times 10^{-7}$ N. The weight of the small sphere is $0.730(9.82)\approx 7.17$ N, nearly $50$ million times more. Having to measure a force $5\times 10^{7}$ times weaker than the weight of the object that undergoes it explains the whole Cavendish apparatus: a TORSION balance, in which the force to be measured acts perpendicular to gravity and the deflection is amplified by a light beam reflected off a mirror. The apparatus also had to be shielded from air currents, which Cavendish did by observing it from another room through a telescope.

The Earth is not weighed, it is DEDUCED. A body of mass $m$ at the surface undergoes $mg=\frac{GMm}{R^{2}}$, where $m$ cancels, so $M=\frac{gR^{2}}{G}$. Every quantity on the right-hand side can be measured in the laboratory: $g$ with a pendulum, $R$ by geodesy, and $G$ by Cavendish. This is why his experiment was billed as a weighing of the Earth. Numerically: $M=\frac{9.82\left(6.371\times 10^{6}\right)^{2}}{6.674\times 10^{-11}}\approx 5.97\times 10^{24}$ kg.

The Earth is in a nearly circular orbit, so $\frac{GM_{S}m}{r^{2}}=\frac{4\pi^{2}mr}{T^{2}}$, where the mass of the Earth cancels. This gives $M_{S}=\frac{4\pi^{2}r^{3}}{GT^{2}}=\frac{4\pi^{2}\left(1.496\times 10^{11}\right)^{3}}{\left(6.674\times 10^{-11}\right)\left(3.156\times 10^{7}\right)^{2}}\approx 1.99\times 10^{30}$ kg. A satellite never reveals its own mass, only that of the body it orbits: an orbit always gives away the mass of the CENTRE.

Each star describes a circle around the centre of mass, with radii $r_{1}$ and $r_{2}$ such that $m_{1}r_{1}=m_{2}r_{2}$ and $r_{1}+r_{2}=d$. For star 1: $\frac{Gm_{1}m_{2}}{d^{2}}=m_{1}\frac{4\pi^{2}r_{1}}{T^{2}}$, so $\frac{Gm_{2}}{d^{2}}=\frac{4\pi^{2}r_{1}}{T^{2}}$. The same equation for star 2 gives $\frac{Gm_{1}}{d^{2}}=\frac{4\pi^{2}r_{2}}{T^{2}}$. Adding: $\frac{G\left(m_{1}+m_{2}\right)}{d^{2}}=\frac{4\pi^{2}\left(r_{1}+r_{2}\right)}{T^{2}}=\frac{4\pi^{2}d}{T^{2}}$, that is $T^{2}=\frac{4\pi^{2}d^{3}}{G\left(m_{1}+m_{2}\right)}$. This is Kepler's third law, in which the SUM of the masses appears. Application, in astronomical units, years and solar masses, where the formula reduces to $m_{1}+m_{2}=\frac{d^{3}}{T^{2}}$: $\frac{27.0}{4.00}=6.75$ solar masses for the pair. Observing the orbit gives only the sum; you have to measure both radii $r_{1}$ and $r_{2}$ to separate the two masses.

The star and the planet revolve around their common centre of mass, so $M_{*}v_{*}=m_{p}v_{p}$: it is this small oscillation of the star that the Doppler shift of its spectrum reveals. First we need the orbital speed of the planet. Radius of its orbit by the third law, with $T=4.00$ days $=3.456\times 10^{5}$ s: $a=\left(\frac{GM_{*}T^{2}}{4\pi^{2}}\right)^{1/3}=\left(\frac{\left(6.674\times 10^{-11}\right)\left(1.989\times 10^{30}\right)\left(3.456\times 10^{5}\right)^{2}}{4\pi^{2}}\right)^{1/3}\approx 7.38\times 10^{9}$ m, about $0.05$ astronomical unit: the planet is scorching hot and skims its star. Speed: $v_{p}=\frac{2\pi a}{T}\approx 1.34\times 10^{5}$ m/s. Mass: $m_{p}=\frac{M_{*}v_{*}}{v_{p}}=\frac{\left(1.989\times 10^{30}\right)(12.5)}{1.341\times 10^{5}}\approx 1.85\times 10^{26}$ kg, that is $\frac{1.85\times 10^{26}}{5.972\times 10^{24}}\approx 31$ Earth masses, or about a tenth of a Jupiter mass. This is a Neptune-type planet in a very tight orbit. Two caveats never to forget: if the orbit is not seen exactly edge-on, the measured radial velocity is only a projection and the method gives only a MINIMUM mass; and a speed of $12.5$ m/s, to be measured on a star dozens of light-years away, is roughly the speed of a cyclist.

$r=6.371\times 10^{6}+8.00\times 10^{5}=7.171\times 10^{6}$ m. The speed is $v=\sqrt{\dfrac{GM}{r}}=7.452\times 10^{3}$ m/s, that is 7.45 km/s. The period is $T=\dfrac{2\pi r}{v}=6046$ s, about 101 minutes.

At the ground: $E_{ground}=-\dfrac{GMm}{R}=-7.507\times 10^{10}$ J. In orbit: $E_{orbit}=-\dfrac{GMm}{2r}=-3.332\times 10^{10}$ J. The energy to supply is $\Delta E=4.175\times 10^{10}$ J, about 41.7 GJ for 1200 kg, that is 34.8 MJ per kilogram placed in orbit.

A point on the equator travels around the circumference of the Earth in one sidereal period: $v_{rot}=\dfrac{2\pi R}{T_{sid}}=\dfrac{2\pi\times 6.371\times 10^{6}}{86{,}164}=464.6$ m/s, about 1670 km/h.

This represents $\dfrac{464.6}{7452}=6.2$ % of the orbital speed. That is little as a percentage, but considerable in practice: every metre per second saved translates into an exponential reduction in the mass of propellant, according to the Tsiolkovsky equation. A launcher heading east from the equator therefore receives 465 m/s for free, which can mean several hundred kilograms of additional payload. This is why Kourou, at 5° latitude, is the European site, and why Cape Canaveral launches out over the Atlantic.

No, the advantage almost entirely DISAPPEARS. A polar orbit passes over the poles, so its plane is perpendicular to the equator: the rotational speed of the Earth, directed east, is then perpendicular to the target orbital speed, and does not add to it. It even becomes a nuisance that has to be cancelled. This is why polar launches are made from HIGH-latitude sites, where $v_{rot}=\dfrac{2\pi R\cos\lambda}{T}$ is smaller: at the latitude of Montreal, $\cos 45.5^{\circ}=0.701$, so only 326 m/s. The choice of launch site therefore depends entirely on the inclination of the target orbit, and not on a universal advantage of the equator.