Field of discrete masses. Newton's law of universal gravitation and superposition give the gravitational field away from the source points:It is the force per unit test mass, with units , and points toward each positive source mass.
Infinite circular cylindrical rod. Let be the radius of the circular cross-section, so , and write for the linear mass density. Choose a coaxial cylindrical Gaussian surface with radius and length . Translation and rotation symmetry make . The end caps have zero field flux because their normals are axial; on the curved surface the outward-normal component is the constant negative value . Gauss's law for gravity therefore givesThe gravitational field of an infinite uniform cylinder inside the material instead has enclosed mass , giving . The exterior result assumes a circular cylindrical rod; area by itself would not specify the exact near field of a noncircular cross-section.
Gaussian-cylinder cross-section for an infinite rod and the circular-speed curve of a finite line rod
. Details. In the left panel the dashed circle is a transverse section of the coaxial Gaussian surface; the field points inward and the surface normals outward. Its side area is .
For an exterior circular orbit, Newton's second law equates the force required for centripetal acceleration to gravity:The orbital speed is independent of radius and of the orbiting object's mass, for orbits outside the infinite rod.
Finite rod. Taking the supplied field as the gravitational field of a finite uniform line segment, the corresponding balance at radius isSince , its two limits areThe first reproduces the infinite-rod result: a nearby observer sees a very long source. The second is the familiar circular orbit speed about a point mass, since a distant observer cannot resolve the rod's length. The dimensionless curve isIt starts at a constant plateau and decreases smoothly, approaching an dependence far away. For , , consistent with the sketch.
Derivation of the finite-rod field. Put the source along on the axis. In its central perpendicular plane, an element is at distance . Its field magnitude is , and its inward radial component has the extra factor . The axial components cancel between and . ThusReplacing by givesThe integral treats the source as a line mass. For a finite cylinder of nonzero cross-section, the displayed finite-rod formula is a thin-rod approximation, not an exact field for all . Its near-rod limit should therefore be read as , keeping the orbit outside the material; the apparent line-source singularity at does not describe the interior of a real cylinder.
For a light test particle of mass outside a spherical source of mass , Newton's law of universal gravitation gives . It is a central force, so its torque about the source vanishes and angular momentum is conserved. The motion therefore lies in a fixed plane when . Its gravitational potential energy is , and conservation of mechanical energy givesHere is specific orbital energy and is specific angular momentum. A Kepler orbit has polar equationThe angle is measured from periapsis. The conic section is an ellipse for and , a parabola for and , and a hyperbola for and . The circle is the elliptical case. The special case is radial motion and must be handled as a degenerate limit. Bound ellipses have the source at one focus, which is Kepler's first law.
For a bound orbit of semi-major axis , . Combining this with the energy equation gives the vis-viva equation:The particle therefore moves faster near periapsis and slower near apoapsis. Conservation of angular momentum also gives constant areal velocity, , expressing Kepler's second law. Using the ellipse's area and gives the orbital periodwhich is Kepler's third law.
A circular orbit follows from centripetal acceleration , giving and . The minimum escape velocity from radius follows from the marginally unbound condition :These ideal orbits neglect drag force, other gravitating bodies and changes in the central source. For two comparable masses, the relative two-body problem uses instead of . An orbiting object is continuously in free fall; the inward gravitational force changes its direction even when its speed is constant. These inverse-square results assume the physical variation of gravity with radius, rather than extending a near-surface constant- approximation to an entire orbit.
Newton's law of universal gravitation gives the force on a test mass a distance from a point source. Gauss's law for gravity statesThe inverse-square point field has the stated flux through a surrounding sphere. Superposition extends it to distributed sources; the divergence theorem relates the integral and local forms. Symmetry is still needed to turn a flux into a pointwise field value.
For the gravitational field of an infinite uniform cylinder, a coaxial cylindrical Gaussian surface has no end-cap flux and constant inward field on its curved surface. The enclosed mass per unit length is inside and outside. HenceIts magnitude rises linearly from zero, reaches at the surface, then decreases as . The field is continuous at , while its radial derivative changes because the density has a sharp edge.
Use signed distance along the diameter hole. A rod displaced by occupies , so its force is . At equilibrium symmetry makes . Differentiating the moving endpoints givesThus the leading restoring force and oscillation frequency areFor an explicit positive-displacement check, put and integrate the inside and outside pieces for :Its first-order term is , agreeing with the endpoint argument. The result is the diameter-rod oscillation in cylindrical gravity; it assumes a sufficiently long source cylinder and a narrow hole whose removed mass can be neglected.
Past exam of the quantum condensed matter course of the University of Cambridge 2004 Section A 5 Solution Created 2026-10-18 Updated 2026-10-20
For a low circular orbit, Newton's law of universal gravitation gives . Using yieldsThus the Moon and Mars have roughly comparable mean densities: low circular-orbit period measures mean density. Equal periods would mean equal mean densities in this approximation. This does not require their masses or radii to be similar. The orbit altitudes must be small compared with the body radii, and the bodies are being treated as approximately spherical.
Three-body orbit Created 2026-10-25 Updated 2026-11-05
A three-body orbit is a trajectory governed by the gravitational three-body problem. For three point masses at positions , Newton's law of universal gravitation givesBoth attracting bodies act simultaneously; switching between separate Kepler orbits is instead a patched conic approximation. In the restricted three-body problem, one mass is negligible, so its trajectory responds to the two primary bodies without appreciably changing their motion. Taking their mutual orbit to be circular gives the circular restricted three-body problem.[ref]
Introduction to Celestial Mechanics by Richard Fitzpatrick. Online Circular restricted three-body problem, equations 9.5–9.9 and the negligible-third-mass assumption
For example, an asteroid influenced by the Sun and a planet can deviate from any single fixed Kepler orbit. Around a synchronously rotating Moon, a maintained trajectory involving the planet–moon system is likewise distinct from a simple moon-centred stationary Kepler orbit. The Hill sphere estimates where the moon's local gravitational dominance can support that simpler approximation; it is not a sharp boundary beyond which all three-body trajectories are impossible.

