Gravitation

Every force in this course so far has been a push or a pull between things in contact, and the one exception, the weight $m g$ that has appeared since the first lesson, has been used without explanation.

What this lesson needs: circular motion, with a centripetal acceleration of $v^{2} / r$ or equivalently $4 π^{2} r / T^{2}$ ; the second law; and the definition of potential energy as minus the integral of a force over distance.

Three laws that were only data

By 1619 Johannes Kepler had extracted three regularities from Tycho Brahe's naked eye observations, the most accurate ever made before telescopes, accurate to about two arcminutes. Each planet moves on an ellipse with the Sun at one focus. The line from the Sun to a planet sweeps equal areas in equal times. And the square of a planet's period is proportional to the cube of the semi-major axis of its orbit, $T^{2} \propto a^{3}$ .

These were descriptions, not explanations. Kepler had no mechanism, and his own proposal, that the Sun sweeps the planets round with a rotating influence, is closer to Aristotle than to Newton. The second law has already been derived in this course, in the lesson on angular momentum, and its derivation used only that the force points at the Sun. The third law is the one that fixes how strong the force is.

Take a circular orbit, which is the special case of an ellipse with $a = r$ . The centripetal acceleration is $4 π^{2} r / T^{2}$ , so the force on a planet of mass $m$ is $F = 4 π^{2} m r / T^{2}$ . If Kepler's third law holds, $T^{2} = C r^{3}$ for some constant $C$ the same for every planet, and substituting,

F = \frac{4 π^{2} m r}{C r^{3}} = \frac{4 π^{2} m}{C} \cdot \frac{1}{r^{2}}

The force must fall off as the inverse square of the distance. This is not a guess and not a fit: given the third law and the second law, no other exponent is possible. The modern data make the third law's constancy plain. Computing $T^{2} / a^{3}$ in SI units gives $2.975 \times 10^{- 19}$ for Mercury, $2.975 \times 10^{- 19}$ for the Earth, $2.970 \times 10^{- 19}$ for Jupiter and $2.978 \times 10^{- 19}$ for Neptune, across a range of orbital radii of nearly eighty to one.

The Moon test

The inverse square law describes the planets. Newton's claim, and the reason the word universal appears in the title of the law, is that the same force holds an apple to the ground. That is a leap, and it can be checked in one calculation, which Newton first did around 1666 and, dissatisfied with the value of the Earth's radius available to him, redid two decades later.

If the same force acts, and it falls off as the inverse square, then the acceleration of the Moon towards the Earth should be smaller than the acceleration of an apple by the square of the ratio of their distances from the Earth's centre.

The Moon's orbital radius is $3.844 \times 10^{8}$ m and its period is 27.32 days, or $2.361 \times 10^{6}$ s. Its centripetal acceleration is

a = \frac{4 π^{2} r}{T^{2}} = \frac{4 π^{2} (3.844 \times 10^{8})}{(2.361 \times 10^{6})^{2}} = 2.723 \times 10^{- 3} {m s}^{- 2}

Compare that with $g = 9.81$ m s⁻² at the Earth's surface: the ratio is 3602. Now the geometric ratio. The Moon is at 60.34 Earth radii, and ${60.34}^{2} = 3640$ . The two numbers agree to within one per cent, and the residual is mostly because the Moon and Earth both orbit their common centre of mass rather than the Earth being fixed.

That agreement is the moment celestial and terrestrial physics became one subject. The same law, with the same exponent and the same constant, governs a falling apple and an orbiting Moon, and there is no separate physics of the heavens.

The universal law

Newton's law of universal gravitation, published in the Principia in 1687, states that every pair of point masses attracts along the line joining them with a force

F = \frac{G m_{1} m_{2}}{r^{2}}

The mass of both bodies appears, by the third law: the force on each is the same, so it cannot depend on one mass alone. And the mass that appears here is the same $m$ that measures inertia in $F = m a$ , which is not obvious at all. It is why all bodies fall at the same rate, since $m g = G M m / r^{2}$ cancels the mass, and it is an experimental fact tested to about one part in $10^{15}$ by the MICROSCOPE satellite in 2022. Einstein took it as the starting point for general relativity rather than as a coincidence.

The constant $G$ is $6.674 \times 10^{- 11}$ N m² kg⁻², and it is the most poorly known of the fundamental constants, uncertain in the fifth digit, because gravity is so weak that the experiment is almost impossible to isolate. Two 1000 kg spheres one metre apart attract with $6.7 \times 10^{- 5}$ N, which is the weight of a grain of sand.

Example. Mars has a mass of $6.417 \times 10^{23}$ kg and a radius of $3.390 \times 10^{6}$ m. What is the surface gravity, and what would a 70 kg astronaut weigh there?

$g = G M / R^{2} = (6.674 \times 10^{- 11}) (6.417 \times 10^{23}) / (3.390 \times 10^{6})^{2} = 3.73$ m s⁻². The astronaut's weight is $70 \times 3.73 = 261$ N, against 687 N on Earth, and their mass is 70 kg in both places.

Now you. The Moon has a mass of $7.342 \times 10^{22}$ kg and a radius of $1.737 \times 10^{6}$ m. What is its surface gravity?

Answer

$g = (6.674 \times 10^{- 11}) (7.342 \times 10^{22}) / (1.737 \times 10^{6})^{2} = 1.62$ m s⁻², which is the value used in the earlier lessons.

The shell theorem

There is a gap in everything above. The law is stated for point masses, and the Earth is not a point: it is a ball 12742 km across, and a person standing on it is 6371 km from the middle and a few metres from some of it. Why should the distance in the formula be measured to the centre?

Newton proved that it should, and the proof delayed the Principia by years. The shell theorem has two parts. A uniform spherical shell attracts an external body exactly as though all its mass were concentrated at its centre. And a uniform spherical shell exerts no net force at all on a body anywhere inside it.

The first part licenses everything: a sphere is a nest of shells, so any spherically symmetric body, however its density varies with depth, pulls external bodies as a point mass at its centre. That is why $g = G M / R^{2}$ works at the surface of a planet, and it is a special property of the inverse square law rather than a general geometric fact.

The second part is more surprising and can be seen without calculus. Stand off centre inside a hollow shell and look in opposite directions. The patch of shell on the near side is close, so its pull is strong, but it is small; the patch on the far side is distant and weak, but proportionally larger. The area of each patch grows as the square of its distance, exactly cancelling the inverse square weakening, so the two pulls are equal and opposite. This holds for every pair of opposite directions, so the total is zero everywhere inside.

Putting the two together gives the field inside a uniform planet. At depth, only the mass within the current radius counts, and that mass is $M (r / R)^{3}$ , so

g (r) = \frac{G M (r / R)^{3}}{r^{2}} = \frac{G M r}{R^{3}}

Gravity falls linearly to zero at the centre, rather than diverging. The real Earth is not uniform, so its interior field actually rises slightly with depth before falling, peaking near the core mantle boundary, which is itself evidence about the density profile.

Weighing the Earth

The law contains $G$ and $M$ only as a product. Timing the Moon gives $G M_{E}$ to great precision and neither one separately, which means that before $G$ was measured nobody knew the mass of the Earth at all.

Henry Cavendish separated them in 1798, using a torsion balance built by John Michell, who died before he could use it. Two small lead balls sit at the ends of a light rod hung from a thin wire; two large lead balls are brought close, and the tiny gravitational attraction twists the wire. The wire's stiffness is calibrated by timing the rod's torsional oscillations, which is the small oscillations machinery of the previous lesson doing real work, and the deflection then gives the force.

Cavendish framed the result as the mean density of the Earth, 5.48 times that of water, against the modern 5.514. That number was the point: it is twice the density of surface rock, which is how it was first known that the interior must be metallic. From it, $M = ρ V = 5.97 \times 10^{24}$ kg, and $G$ follows from $g = G M / R^{2}$ . The apparatus was so sensitive that Cavendish operated it from another room through a telescope, to keep his own body heat from stirring the air, and his value stood essentially unimproved for a century.

Example. Given $g = 9.81$ m s⁻² at the surface, $R_{E} = 6.371 \times 10^{6}$ m and $G = 6.674 \times 10^{- 11}$ , find the mass of the Earth and its mean density.

From $g = G M / R^{2}$ , $M = g R^{2} / G = (9.81) (6.371 \times 10^{6})^{2} / (6.674 \times 10^{- 11}) = 5.97 \times 10^{24}$ kg. The volume is $\frac{4}{3} π R^{3} = 1.083 \times 10^{21}$ m³, so the mean density is $5.51 \times 10^{3}$ kg m⁻³, or 5.51 times water.

Now you. The International Space Station orbits at an altitude of 408 km. What is $g$ there, and what fraction of its surface value is that?

Answer

The orbital radius is $6.371 \times 10^{6} + 4.08 \times 10^{5} = 6.779 \times 10^{6}$ m, so $g = G M / r^{2} = (6.674 \times 10^{- 11}) (5.972 \times 10^{24}) / (6.779 \times 10^{6})^{2} = 8.67$ m s⁻², which is 88 per cent of the surface value. Astronauts on the station are not weightless because gravity is absent; they are weightless because they are in free fall.

Potential energy, and escape

Gravity is a central force and passes the path independence test, so it has a potential energy. Integrating the force from a reference point to a distance $r$ ,

U (r) = - \int_{\infty}^{r} \frac{G M m}{r^{2}} d r^{'} = - \frac{G M m}{r}

taking the zero at infinite separation, which is the only choice that makes the constant natural. The negative sign says that work must be done to separate two bodies, and it means bound systems have negative total energy, a fact the next lesson builds an entire classification on.

Near the surface this must reduce to $m g h$ , and it does. Raising a body from $R$ to $R + h$ changes $U$ by $G M m (1 / R - 1 / (R + h)) = G M m h / (R (R + h))$ , and for $h ≪ R$ that is $G M m h / R^{2} = m g h$ . The familiar formula is the first term of an expansion, valid while the height is small compared with the radius of the planet.

Escape velocity follows in one line. A body just escapes if its total energy is zero, since then it arrives at infinity with nothing left:

\frac{1}{2} m v^{2} - \frac{G M m}{R} = 0 \Rightarrow v_{esc} = \sqrt{\frac{2 G M}{R}}

The mass of the escaping body cancels, so a pebble and a spacecraft need the same speed. For the Earth it is 11.2 km s⁻¹, for the Moon 2.38 km s⁻¹, and for the surface of the Sun 618 km s⁻¹. It is worth being clear about what the number means: it is the speed needed for an unpowered projectile launched from the surface. A rocket under continuous thrust can leave at any speed it likes, and pays for the privilege in propellant.

Escape velocity also explains atmospheres. A gas molecule at temperature $T$ has a typical speed of a few hundred metres per second, well below 11.2 km s⁻¹, but the distribution has a tail, and over billions of years the tail leaks away. Light molecules move faster at a given temperature, which is why the Earth has kept its nitrogen and oxygen but lost nearly all its hydrogen and helium, and why the Moon, with an escape velocity a fifth of the Earth's, has no atmosphere at all.

Example. Jupiter has a mass of $1.898 \times 10^{27}$ kg and an equatorial radius of $6.991 \times 10^{7}$ m. What is its escape velocity, and what does that imply about its composition?

$v_{esc} = \sqrt{2 (6.674 \times 10^{- 11}) (1.898 \times 10^{27}) / (6.991 \times 10^{7})} = 6.02 \times 10^{4}$ m s⁻¹, or 60.2 km s⁻¹, which is 5.4 times the Earth's. Nothing escapes from Jupiter, and it has therefore kept the hydrogen and helium it formed from, in roughly the proportions the Sun has. The rocky planets, which could not hold those gases, are what is left over.

Now you. Ceres, the largest asteroid, has a mass of $9.38 \times 10^{20}$ kg and a radius of $4.696 \times 10^{5}$ m. What is its escape velocity?

Answer

$v_{esc} = \sqrt{2 (6.674 \times 10^{- 11}) (9.38 \times 10^{20}) / (4.696 \times 10^{5})} = 516$ m s⁻¹, slower than a rifle bullet. A visiting spacecraft can leave under gentle thrust, and no atmosphere of any kind could survive there.

What the law has not yet been asked

At this point gravity is fully specified: a central, inverse square, always attractive force between every pair of masses, with a known constant and a potential energy. Everything that has been done with it, though, has assumed a circular orbit or a straight fall.

The general question is what such a force does to a body given any starting position and velocity. It is a genuine differential equation, and it has a complete solution: the possible paths are exactly the conic sections, circle, ellipse, parabola and hyperbola, with the sign of the total energy deciding which. All three of Kepler's laws come out, along with the radius at which a satellite hovers over one spot on the Earth and the cost of a trip to Mars.

The same solution is where the subject finds its edge. Mercury's orbit precesses by 43 arcseconds per century more than Newtonian gravity can account for, and that small discrepancy, measured carefully in the 1850s, was the first hard evidence that this law is an approximation to something else.