Past exam of the physics B course of the University of Cambridge / 2011 / Paper 1 / Section B / B8

A galilean transformation relates inertial frames in uniform relative translation. For $\mathbf{V}$ the velocity of the primed origin,Absolute time and invariant acceleration make Newton's second law retain its form for the usual nonrelativistic interactions. For a particle of mass $m$, ${\mathbf{p}}^{\prime }=\mathbf{p}-m\mathbf{V}$ and $K^{\prime }=K-\mathbf{V}\cdot \mathbf{p}+m{V}^2/2$. Momentum and kinetic energy are frame dependent, although an isolated system's conservation laws hold in either inertial frame. Relative velocities and collision predictions are consistent between frames. A rotating or accelerating frame instead requires inertial-force terms. Maxwell equations with a fixed light speed are not Galilean invariant; Lorentz transformations replace these relations when relativistic effects matter.

Kepler's laws of planetary motion state that, in the relative two-body description, bound trajectories are ellipses with the attracting body at a focus, the radius vector sweeps equal areas in equal times, and the square of the period is proportional to the cube of the semimajor axis for a fixed total mass. Their Newtonian derivation is especially informative. With relative coordinate $\mathbf{r}$ and ${\mu }_g=G(M+m)$, the inverse-square acceleration is $-{\mu }_g\mathbf{r}/r^3$. Its zero torque conserves $\mathbf{h}=\mathbf{r}\times \dot{\mathbf{r}}$, fixing a plane and the areal velocity $h/2$.

For $u=1/r$, the orbit equation is $u^{\prime \prime }+u={\mu }_g/h^2$ as a function of polar angle. ThereforeBound negative-energy motion has $0\le e<1$ and semilatus rectum $h^2/{\mu }_g=a(1-e^2)$. Dividing ellipse area by areal velocity givesFor a planet much lighter than its star, replace $M+m$ by $M$. Zero-energy and positive-energy encounters are parabolic and hyperbolic rather than periodic ellipses. Perturbations from other bodies and departures from point-mass Newtonian gravity can precess or otherwise modify the orbit; Kepler's laws are a two-body approximation, not an exact rule for every observed planetary path.

A tidal force measures the difference of gravitational acceleration across an extended object. A freely falling reference frame removes a spatially uniform acceleration, but not its gradient. For a distant point mass $M$ at distance $R$ and direction $\hat{\mathbf{n}}$ from the reference centre, a small internal displacement $\mathbf{x}$ givesThe tidal tensor has eigenvalues $2GM/R^3$ radially and $-GM/R^3$ in each transverse direction: radial stretching and transverse squeezing. Its trace vanishes in the source-free region, consistent with the gravitational Gauss law. An associated relative potential is ${\Phi }_{\mathrm{tide}}=-GM[3(\hat{\mathbf{n}}\cdot \mathbf{x}{)}^2-x^2]/(2R^3)$.

The scale is $GMl/R^3$ across size $l$, not the centre's absolute gravity $GM/R^2$. This distinction explains how a more distant massive object can exert a large force but a modest tide. Opposite-side stretching supports the elementary two-bulge picture; real ocean tides also involve rotation, basin geometry and fluid dynamics, so the static gradient alone does not predict coastal heights or timing. The same gradient stresses satellites and can disrupt a self-gravitating body when it competes with the body's own gravitational restoring acceleration.

A rapidly spinning gyroscope approximately has angular momentum $\mathbf{L}=I_3{\omega }_3\hat{\mathbf{e}}$ along its axle. With no torque in an inertial frame, this vector retains its direction. A transverse torque changes its direction according to $d\mathbf{L}/dt=\mathbf{\tau }$; it does not initially push the axle in the direction of the applied force.

For a fixed pivot, centre of mass $l\hat{\mathbf{e}}$ from it and gravity $-Mg\hat{\mathbf{z}}$, the torque is $Mgl\hat{\mathbf{z}}\times \hat{\mathbf{e}}$. In slow steady precession about the vertical, $d\mathbf{L}/dt=\Omega \hat{\mathbf{z}}\times \mathbf{L}$, givingThe factor $\sin \theta$ appears both in torque and in the radius traced by the angular-momentum tip, so it cancels for a nonvertical axle. Reversing spin reverses this precession. The full symmetric-top steady condition is $I_1\cos \theta {\Omega }^2-I_3{\omega }_3\Omega +Mgl=0$; the approximation neglects its quadratic term. Sudden release can also produce nutation, so large spin alone does not guarantee a steady sweep.

A wheel-on-pivot experiment can compare precession periods at different measured spin speeds and reverse the spin. Keep the pivot and centre of mass fixed; bearing friction, spin decay, an initial push and support motion can mimic departures from the predicted inverse-spin relation. A freely gimballed direction reference and a gravity-loaded precessing wheel test different torque conditions.^[ref]