G = Gravitational constant M = Mass of central object m = Mass of satellite R = Distance of satellite from the central object V = Velocity of satellite F = Gravitational force = G M m / R^2 E = Gravitational potential energy = Integral[Force dR] = -G M m / RThe velocity of a circular orbit is obtained by setting gravitational force equal to centripetal force.

G M m / R = m V^2 / R Circular orbit velocity = SquareRoot(G M / R)The escape velocity is obtained by setting gravitational energy equal to kinetic energy.

G M m / R = 1/2 m V^2 Escape Velocity = SquareRoot(2 G M / R) Escape velocity = SquareRoot(2) * Circular orbit velocity Escape Circular (km/s) orbit (km/s) Earth 11.2 7.9 Mars 5.0 3.6 Moon 2.4 1.7

For an object on a circular orbit,

Gravitational energy = -2 * Kinetic energyThe relationship between the kinetic and gravitational energy doesn't depend on R. If a satellite inspirals toward a central object, the gain in kinetic energy is always half the loss in gravitational energy.

The total energy is negative.

Total energy = Gravitational energy + Kinetic energy = .5 * Gravitational energy = - .5 G M m / R Angular momentum = m V R = m SquareRoot(G M R)As R decreases, both energy and angular momentum decrease. In order for a satellite to inspiral it has to give energy and angular momentum to another object.

A typical globular cluster consists of millions of stars. If you measure the total gravitational and kinetic energy of the stars, you will find that

Total gravitational energy = -2 * Total kinetic energyjust like for a single satellite on a circular orbit.

Suppose a system consists of any number of objects interacting by gravity (could be from 2 to infinity). If the system has reached a long-term equilibrium, then the above statement about energies is true, no matter how chaotic the orbits of the objects. This is the "Virial theorem". It also applies if additional forces are involved. For example, the protons in the sun interact by both gravity and collisions and the virial theorem holds.

Gravitational energy of the sun = -2 * Kinetic energy of protons in the sun.

D = Density R = Radius M = Mass = Density * Volume = 4/3 Pi D R^3 A = Acceleration at the surface = G M / R^2 = (4/3) Pi G D RAcceleration is proportional to R

Density Radius Gravity g/cm^3 (Earth=1) m/s^2 Earth 5.52 1.00 9.8 Venus 5.20 .95 8.87 Uranus 1.27 3.97 8.69 Mars 3.95 .53 3.71 Mercury 5.60 .38 3.7 Moon 3.35 .27 1.62 Titan 1.88 .40 1.35 Ceres 2.08 .074 .27

M = Mass of Earth m = Mass of an object near the Earth's surface R = Radius of Earth G = Gravitational constant F = Force at the Earth's surface = G M m / R^2 = g m g = Acceleration at Earth's surface = G M / R^2 h = Height above Earth's surface E = Gravitational energy of an object at distance R+h from the Earth = - G M m / (R+h) ~ - (G M m / R) (1 - h/R) ~ - G M m / R + G M m h / R^2 ~ - G M m / R + m g hEarth the Earth's surface, we may approximate the gravitational energy as

E = m g h if h << RThis can also be obtained from the gravitational force.

E = Force * h = m g h

Suppose you want to estimate the gravitational energy of a uniform-density sphere. The variables that the energy can depend on are G, M, and R, and the only combination of these variables that has units of energy is

Energy ~ G M^2 / RThe correct formula from calculus is

Energy = .6 G M^2 / ROrder-of-magnitude estimations tend to give the correct exponents on the variables, but not the dimensionless number accompanying them. The dimensionless number (in this case .6) usually has order-of-magnitude 1.

Another example is the formula for the drag force on a sphere moving through a fluid. Such a formula can depend on

D = Density of the fluid A = Cross-sectional area of the sphere V = Velocity of the sphereThe combination of D, A, and V that gives units of force is

Force ~ D A V^2The correct formula from fluid dynamics is

Force = .5 D A V^2

P = Pressure (units of Energy/volume) E = Kinetic Energy per volume Pressure = 2/3 E

Suppose we assume the sun is a uniform-density sphere of protons.

m = Mass of proton = 1.67e-27 kg M = Mass of sun = 1.99e30 kg V = Mean velocity of protons (thermal speed) E = Mean kinetic energy of protons = .5 m V^2 Ek = Total kinetic energy of protons Eg = Gravitational potential energy of solar protons k = Boltzmann constant = 1.38e-23 Joules/Kelvin T = Temperature of protons R = Radius of sun = 6.96e8 metersFor a gas the thermal speed V is defined such that

Mean kinetic energy = E = .5 m V^2The gravitational potential energy of a uniform-density sphere is

Eg = .6 G M^2 / RFrom the Virial theorem,

Eg = -2 EkFor a gas in thermal equilibrium, every degree of freedom has mean energy .5 k T. A proton moving in 3 dimensions has 3 degrees of freedom, hence

.5 m V^2 = 1.5 k T www.jaymaron.com/oom/gas.html

Gravity causes collapse and gas pressure resists collapse. The larger an object, the more effective gravity is compared to gas pressure, and so if an object becomes sufficiently large it will always collapse. The minimum size for collapse is the "Jeans length".

G = Gravitational constant m = Mass of proton = 1.67e-27 kg M = Mass of a sphere of protons D = Density of sphere R = Radius of sphere Rj = Jeans length Ve = Escape velosity Vt = Thermal velocity Vs = Sound speed k = Boltzmann constant = 1.38e-23 Joules/KelvinIn this analysis we will neglect dimensionless constants and focus on units. For example, the escape velocity

Ve^2 ~ G M / R ~ G D R^2The thermal speed is such that

m Vt^2 ~ k T Expanded discussion at www.jaymaron.com/oom/gas.htmlThe sound speed has the same order of magnitude as the thermal speed

Vs ~ Vt For air, Vs = .63 VtThe Jeans length Rj is the size of a sphere such that gravity and gas pressure are in balance.

Ve ~ Vt G D Rj^2 ~ k T / m Rj^2 ~ k T / (m G D)Expressed in terms of the sound speed,

Rj^2 ~ Vs^2 / (G D)

G = Gravitational constant M0 = Mass of star M = Mass of planet m = Mass of moon R = Orbital radius of a planet around a star r = Orbital radius of a moon around a planet H = Planet Hill radiusWithout loss of generality we can set G = M0 = 1. We also assume that the star is vastly heavier than the planet and that the planet is vastly heavier than the moon.

m << M << 1If the planet and moon have the same orbital period,

r = R M^(1/3)This gives the magnitude of a planet's range of gravitational influence.

If a satellite is at the L1 or L2 Lagrange points, an are the balance points

This value of r is called the "Hill radius", and is a measure of the gravitational influence of a planet. Moons within 1/3 of the Hill radius are stable and moons outside this distance are vulnerable to being stolen by the star.

For the Earth's moon,

r = .257 * Hill radius.As the moon spirals outward, it will eventually be stolen by the sun.

M = Mass of central object m = Mass of satellite R = Distance of satellite from central object F = Force of gravity = G M m / R^2 E = Gravitational energy = Integral[Force dR] = -G M m / RSuppose a satellite is on a circular orbit of radius R.

Eg = Gravitational potential energy Ek = Kinetic energy Et = Total energyWhat is the relationship between Eg, Ek, and Et?

Vc = Velocity of a satellite on a circular orbit of radius R Ve = Escape velocity of a satellite at a distance R from the central objectWhat is the relationship between Vc and Ve?

Distance from Earth from the sun = 1 astronomical unit Distance of Mars from the sun = 1.52 astronomical units Time for Earth to orbit sun = 1 yearAssume that the Earth and Mars have circular orbits. How long does it take for Mars to orbit the sun?

Suppose a planet has a density equal to that of the Earth.

D = Density = 5.52 g/cm^3 for the Earth R = Radius of the planet r = Radius of the Earth A = Gravitational acceleration on a planet of radius RWhat is A as a function of (R/r)?

Uranus is the lightest gas giant with a mass of 14.5 Earth masses. If Uranus had the same density as the Earth, what would be the gravity at the surface? This is an upper limit to the gravity that could conceivably exist on a rocky world.

Using data from the web, what is the mass of the Milky Way and Andromeda galaxies, and the distance between them? What is the acceleration of Andromeda toward the Milky Way?

Suppose Andromeda starts at rest and accelerates uniformly toward the Milky Way. If the acceleration is constant, how long does it take to travel a distance equal to the distance to the Milky Way?

Based on data for the Orion Nebula, what would you estimate is the Jeans length for the nebula?

What is the Jeans length and mass for the interstellar medium nearby the sun?