Gravitation

Isaac Newton published the inverse-square-law attraction between 2 massive particles:

in his Principia in 1687. Henry Cavendish first measured the value of G in 1798; it is currenly measured to be 6.6726 ± 0.008 x 10-8 cm3/(g x s2) and is the most imprecisely known of the fundamental physical constants. For extended body with density distribution ρ(r), the force exerted on a point mass m is

The gravitational field (acceleration) is g = F/m

Gravitational potential

The gravitational field g can be expressed as the gradient of a scalar potential g = -∇Φ. For a point mass, this gravitational potential is Φ = -GM/r. The constant term in the potential has been chosen so that Φ → 0 as r → ∞. This choice isn't possible for mass distributions that are infinite in extent. The potential from an extended distribution is

The work done to move a particle in a gravitational field is

The gravitational potential energy of particle of mass m is U = mΦ.

Gravitational potential of a spherical shell

Consider the potential at a point a distance r from the center of a spherical shell of radius R

= -GM/r r > R= -GM/R r < R

This tells us that for any spherically symmetric density distribution (which can be considered to be a superposition of shells), the potential outside the distribution is the same as that of a point mass at its center, and the potential inside the distribution is constant (and hence the field g = -∇Φ vanishes).

Gravitational potential of a ring

Consider the potential at a point a distance r from the center of a ring of radius R. Is the equilibrium point at the center of ring stable or unstable?

where λ = M/2πR. Taylor expand this expression for r << R.

We find dU/dr = 0 at r = 0, and d2U/dr2 = -GMm/2R3 < 0. The equilibrium is unstable.

Poisson's equation

The gravitational field g of a point mass varies inversely as the square of the distance, just like the electric field of a point charge, but it is directed inwards, not outwards. We therefore have

Since g = -∇Φ, we can also write

This is Poisson's equation, a partial differential equation that that can be solved for the potential from an arbitrary density distribution. It is known as Laplace's equation when ρ = 0.

Lines of force

Lines of force point along the direction of the vector field. The total number of lines is proportional to the mass of the particle, while their surface density is proportional to the strength of the field. Equipotential surfaces (Φ = const.) are perpendicular to these lines since along an equipotential surface

Ocean tides

Consider a fluid element of mass m at position rm. This fluid element experiences an acceleration

due to the influence of the Earth and Moon, however the Earth experience an acceleration

in the Moon's gravitational field. The relative acceleration is therefore

The 1st term is the Earth's gravitational field, while the 2nd term in the Moon's tidal field. We can also use the gravitational potential to calculate the tidal force as a function of rm.

Let dEM be the distance from the Earth to the Moon and assume that this distance is much greater than the radius of the Earth RE. We then have

implying that

after we have subtracted off the Earth's acceleration. As a day is much shorter than a month, we experience 2 high tides per day as the Earth rotates through the tidal bulges. The Moon's mass is MM = 4/3 π ρM RM

3, implying that the tidal acceleration is proportional to ρM (RM/dEM)3. The angle subtended by the Moon on the sky is (RM/dEM), implying that if 2 objects (like the Moon and Sun) subtend the same solid angle, their relative tidal accelerations will be proportional to their densities. The Sun is about 46% the density of the Moon, implying that its tides are weaker by that same factor. The highest tides each month occur during the new and full Moon (spring tides), when the Sun and Moon act in concert. The weakest tides (neap tides) occur at 1st and 3rd quarter when the solar and lunar tides oppose each other. Due to friction, the tidal bulges lead the Moon, transferring angular momentum from the Earth's rotation to the Moon's orbit. The Moon moves away from the Earth ~1cm/yr, and the Earth will become tidally locked (day = month) in ~60 billion years.

Two-body problem

Let us consider the motion of 2 point masses m1 and m2. Let the positions of these 2 point masses be given by r1 and r2. Newton's 3rd law tells us that if m1 exerts a force F21 on m2, m2 will exert a force F12 = -F21 on m1. This implies that:

This means that we can define a center-of-mass vector:

such that dR/dt = const. We can then boost to a frame in which dR/dt = 0 ⟹ R = const. We can also define the separation vector r = r1 and r2 between our 2 masses, implying that:

A central force depends only on the magnitude of r and can be expressed as F = -∇U(|r|). In this case the energy of the system can be expressed as:

where

is the reduced mass of the system. We have thus reduced the 2-body problem for point masses under the influence of a central force to an effective one-body problem for a single point mass μ in the potential U(r).

Angular momentum

The fact that F(r) is a central force implies:

This in turn implies that the motion is restricted to the plane perpendicular to L. In polar coordinates (r, θ) in this plane:

The constancy of the angular momentum implies that the area per unit time swept out by the particle as it travels along its orbit:

is also constant. This is Kepler's 2nd law of planetary motion. The constancy of the angular momentum reduces the 2-body problem to a 1-dimensional problem in the radial coordinate: