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During a short time $dt$, the rocket loses mass $-dm>0$. Conservation of upward momentum, including gravity's impulse and exhaust velocity $v-U$, gives to first orderwhich is the stated rocket equation.

Here $dm/dt=-\alpha$ and $U=U_0{m}_0/m$, soLift-off from rest requires positive initial acceleration,With $v(0)=0$, integration givesThe dimensions are $[m_0]=[m]=M$, $[\alpha ]=M/T$, $[U_0]=[U]=[v]=L/T$, $[g]=L/T^2$, and $[t]=T$. Both displayed terms in $v$ therefore have dimension $L/T$.

Using spherical shells and averaging the squared distance from the axis givesSliding without friction has acceleration $g\sin \alpha$, soFor rolling, $M{a}_{\mathrm{cm}}=Mg\sin \alpha -F$ and $I(a_{\mathrm{cm}}/a)=Fa$, whenceThe rolling acceleration with rotational inertia gives $5g\sin \alpha /7$ for the uniform sphere, and thereforeMechanical energy is conserved in both idealizations: there is no friction in sliding, while static friction does no work at the instantaneous contact point in pure rolling.

For $I=\gamma M{a}^2$, the same calculation gives

The pressure force in the $x$-direction on a rectangle of dimensions $\delta x\times \delta y$ isThe shear stress is ${\tau }_{xy}=\mu u_y$, so the net viscous force isThere is no convective acceleration. Dividing the momentum balance by the area and taking the limit givesThe $y$-momentum balance gives $p_y=0$, so the pressure depends only on $x$.

The tangent planecuts the hyperboloid inso its intersection is the pair of straight linesTheir accelerations vanish, hence are normal to the surface, so both lines are geodesics. Their tangent vectors at their intersection $(1,0,0)$ are $(0,1,1)$ and $(0,-1,1)$, whose inner product is zero. They therefore intersect at a right angle, giving the second case of plane-section geodesics of the unit one-sheet hyperboloid.

There are also geodesics entirely contained in $z>0$. In the coordinatesthe metric isFor a unit-speed geodesic, the Clairaut first integral for a surface of revolution is $c=(1+z^2)\dot{\varphi }$, and the constant-speed equation becomesChoose $c>1$ and initial height $z_0=\sqrt{c^2-1}$. The geodesic initially tangent to the parallel has $\dot{z}=0$, and the displayed identity prevents it from entering $0<z<z_0$. Since the parallel at $z_0>0$ is not itself a geodesic, the curve turns there and otherwise has $z>z_0$. Hence it remains entirely in $z>0$; this is a geodesic trapped in one half of the unit one-sheet hyperboloid.

Finally, the waistis a geodesic and is preserved setwise by every isometry of the hyperboloid. Indeed, its Gaussian curvature isThe value $K=-1$ occurs exactly at $z=0$. Since Gaussian curvature is intrinsic and therefore preserved by isometries, every isometry preserves the waist. Thus the answer to the final question is also yes, with the isometry-invariant waist geodesic of the unit one-sheet hyperboloid as an example.

The planemeets the one-sheet hyperboloid in the two disjoint curvesThey are the two opposite meridians of the surface of revolution. Parametrize the generating profile by arc length. Its acceleration lies in the meridian plane and is orthogonal to the profile tangent; it is also orthogonal to the azimuthal tangent. It is therefore normal to the surface, so both meridians are geodesics. This is the disjoint case in plane-section geodesics of the unit one-sheet hyperboloid.

A regularly parametrized curve $\gamma$ on an embedded surface $\Sigma$ is a geodesic when its covariant acceleration vanishes. Equivalently, in the ambient Euclidean space,for an affine geodesic parameter. A complete geodesic is one whose maximal affine parameter interval is all of $\mathbb{R}$.

Radiation conservation givesConsequentlyEvaluating at $t_0$, where $a=1$, shows thatPutThen $\dot{b}=2\sqrt{A+B{b}^2}$. Choosing the big bang as $t=0$ and integrating givesThereforewhere

At early times, $\sinh (\beta t)\sim \beta t$, so $a\propto t^{1/2}$, as in a radiation-dominated universe. At late times,the expected de Sitter expansion with $H\to \sqrt{\Lambda c^2/3}$.

Acceleration begins when the radiation deceleration and cosmological-constant acceleration balance:Thus $a=\alpha$, which in the exact solution means $\sinh (\beta t_{\Lambda })=1$. HenceThis is the radiation-to-cosmological-constant transition.

The Friedmann equation immediately givesUsing $P=w\rho$ in the acceleration equation and eliminating $\rho a^2$ with Friedmann givesDifferentiate $\Omega -1=k{c}^2{\mathcal{H}}^{-2}$ and substitute this equation:where ${\mathcal{H}}^2+k{c}^2=\Omega {\mathcal{H}}^2$ was used in the last step.

With $\gamma =(1-v^2/c^2{)}^{-1/2}$,Differentiation with respect to proper time givesOrthogonality $U\cdot A=0$ implies that in the instantaneous rest frame $A^{\prime \mu }=(0,a^{\prime })$. For collinear motion a Lorentz boost givesFor constant positive proper acceleration $a^{\prime }$,This hyperbola has future null asymptote $x=ct-c^2/a^{\prime }$. A right-moving signal emitted from $x_0$ at $t=0$ follows $x=ct+x_0$; if $x_0<-c^2/a^{\prime }$, it lies beyond that acceleration horizon and never intersects the worldline.

In local coordinates $u=(u^1,u^2)$ with metric $g_{ij}={\partial }_i\sigma \cdot {\partial }_j\sigma$, the energy isThe Euler–Lagrange equations are equivalentlywhere ${\Gamma }_{ij}^k=\frac{1}{2}{g}^{k\ell }({\partial }_i{g}_{j\ell }+{\partial }_j{g}_{i\ell }-{\partial }_{\ell }{g}_{ij})$.

For the plane-section curve $\eta$, constant speed gives ${\eta }^{\prime \prime }\perp {\eta }^{\prime }$. Both ${\eta }^{\prime \prime }$ and the surface normal lie in $P$, while $P$ is spanned by ${\eta }^{\prime }$ and that normal along the intersection. Hence ${\eta }^{\prime \prime }$ is normal to the surface. Its tangential acceleration vanishes, which is the geodesic equation.

Linearizing the constraint gives $\delta \rho =-2{\rho }_0\delta a/a_0$. Since $\delta p=\delta \rho /3$, the acceleration equation gives $\delta \ddot{a}=2\Lambda \delta a$. Hence perturbations contain $e^{\sqrt{2\Lambda }t}$: the Einstein static universe is unstable.

Define $p(t)=\int x^{\prime }\rho (t,x^{\prime })d^3{x}^{\prime }$. Expanding the retarded potential for source size $d\ll r$ and speed $v\ll c$, with radiation zone $r\gg d,c/\omega$, and using charge conservation gives $B={\mu }_0\ddot{p}(t-r/c)\times \hat{x}/(4\pi rc)$. Then $S={\mu }_0|\ddot{p}\times \hat{x}{|}^2\hat{x}/(16{\pi }^{2}c{r}^2)$ and $dP/d\Omega ={\mu }_0(|\ddot{p}{|}^2-(\hat{x}\cdot \ddot{p}{)}^2)/(16{\pi }^{2}c)$. For circular motion, $P={\mu }_0{q}^2{\omega }^4{R}^2/(6\pi c)$ and $⟨dP/d\Omega ⟩={\mu }_0{q}^2{\omega }^4{R}^2(1+{\cos }^2\theta )/(32{\pi }^{2}c)$. The stated axial and equatorial polarizations follow directly. Relativistically, perpendicular acceleration gives $P={\mu }_0{q}^2{\gamma }^4{\omega }^4{R}^2/(6\pi c)$.

The leading uniform stellar radiation pressure is a differential acceleration relative to a planet, of scale $\beta G{M}_{\star }/a_p^2$. Relative to planetocentric gravity at dust radius $a$, its magnitude isThus lying inside a Hill sphere alone does not guarantee small radiative perturbations. Approximately circular tightly bound dust requires this ratio sufficiently small, as well as weak tides; the uniform radiative force can otherwise excite orbital eccentricity or unbind grains.

A point mass constrained to a fixed-length massless arm under uniform gravitational acceleration is a simple pendulum. With no drag its angular equation is $\ddot{\theta }+(g/\ell )\sin \theta =0$. The small-angle approximation yields simple harmonic motion, whereas the full restoring force is nonlinear. Adding drag gives a damped pendulum.

Disruption of a self-gravitating body when differential gravitational acceleration across it exceeds its binding acceleration. For a body of radius $R$, mass $m$ at distance $r$ from a much more massive body $M$, the competing scales are $GMR/r^3$ and $Gm/R^2$; their equality gives $r\sim R(M/m{)}^{1/3}$, up to geometry and internal-structure factors. The tidal tensor gives the directional differential acceleration. A tidal disruption event near a black hole is a specific astronomical application.

Differential gravitational acceleration, multiplied by the affected mass, across a spatially extended body. In the linear approximation the acceleration difference is $T\Delta \mathbf{r}$, where $T$ is the tidal tensor. Stellar tidal forces limit the domain of a circumplanetary orbit; sufficiently strong ones can produce tidal disruption.