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The potential factors as $V=x^2(3-2x)$, so its zeros are $0$ and $3/2$, with a double zero at the origin. Its derivatives are $V^{\prime }=6x(1-x)$ and $V^{\prime \prime }=6-12x$. Thus there is a local minimum $(0,0)$, a local maximum $(1,1)$ and an inflection at $(1/2,1/2)$. The potential tends to $+\infty$ on the far left and $-\infty$ on the far right. These features determine the left-hand sketch.

Put $v=\dot{x}$. Newton's second law gives the phase plane system $\dot{x}=v$, $\dot{v}=-V^{\prime }=6x(x-1)$. Its conserved mechanical energy and trajectories areThe linearization at $(0,0)$ has eigenvalues $\pm i\sqrt{6}$, and the surrounding energy curves are closed, so it is a center equilibrium. At $(1,0)$ the eigenvalues are $\pm \sqrt{6}$, giving a saddle equilibrium. The figure shows the cubic potential barrier phase portrait; arrows point right when $v>0$ and left when $v<0$.

For $0<E<1$, there are three turning-point roots of $V(x)=E$. The two leftmost enclose a closed orbit around the origin: the particle oscillates between them. There is also a separate open component to the right of the third root, where an incoming particle turns and escapes to $+\infty$. The closed orbits run clockwise in the $(x,v)$ plane.

For $E=1$, the factorization $1-V(x)=(x-1{)}^2(2x+1)$ gives the separatrixThe branch with $-1/2\le x<1$ is a homoclinic orbit: it leaves the saddle asymptotically, turns at $x=-1/2$ and returns asymptotically. On $x>1$, one branch approaches the saddle and the other departs towards infinity. Approaching the saddle takes infinite time. The saddle itself is also an equilibrium trajectory.

For $E>1$, the particle can cross the barrier. There is one left turning point below $-1/2$; an incoming particle passes through the well, turns on the left and escapes to the right. For $E<0$, only a right-hand open component exists, with its turning point beyond $3/2$. At $E=0$, the stationary centre coexists with a right-hand open trajectory turning at $3/2$. Thus negative energy does not imply trapping here: the cubic potential is unbounded below on the right.

Take upward velocity as positive, and consider the closed material system consisting of the rocket and the fuel it ejects during a short interval $dt$. Initially its mass and momentum are $m$ and $mv$. Let $\delta m>0$ be the ejected mass, so the rocket's mass change is $dm=-\delta m$. At the end its velocity is $v+dv$, while the ejected gas has inertial velocity $v-u$ to first order; its downward relative velocity need not be downward in the inertial frame.

The final total momentum, to first order, isNeglect air resistance and take the gravitational acceleration $g$ constant. The external gravitational impulse on this material system is $-mgdt$ to first order. Newton's second law therefore gives $mdv-u\delta m=-mgdt$, orThe positive thrust is $-u\dot{m}$ because fuel loss makes $\dot{m}<0$. Applying $d(mv)/dt$ to the rocket alone without accounting for the escaping momentum would give the wrong equation.

For constant $u$ and positive mass, divide by $m$ and integrate from release at $t=0$:Thus the gravity-corrected rocket equation is equivalent to

Define the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ as the number of $k$-element subsets of an $n$-element set. Equivalently it is $n!/[k!(n-k)!]$: there are $n(n-1)\cdots (n-k+1)$ ordered lists of $k$ distinct elements, and each subset gives $k!$ such lists. The empty subset gives $\left(\genfrac{}{}{0ex}{}{n}{0}\right)=1$.

Distinguish one element of the set. A $k$-element subset either omits it, giving $\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$ possibilities, or contains it and chooses its other $k-1$ elements from the remaining set, giving $\left(\genfrac{}{}{0ex}{}{n-1}{k-1}\right)$ possibilities. These cases are disjoint and exhaustive, proving Pascal's identity:For the requested sum, fix $n\ge 0$ and use mathematical induction on $k$. At $k=0$, both sides equal one. If the assertion holds at $k$, adding the next summand giveswhere the last equality is Pascal's identity. Thus the induction proves the hockey-stick identity for all nonnegative $n,k$:

For a fixed central gravitational source, the radial and transverse components of Newton's second law in polar coordinates areThe second equation gives $r^2\dot{\theta }=h$, the constant specific angular momentum. Take $h>0$ by choosing the orientation of $\theta$, and put $u=1/r$. Writing $u^{\prime }$ and $u^{\prime \prime }$ for derivatives with respect to $\theta$,Substituting into the radial equation and dividing by $-h^2{u}^2$ proves the Binet equationAt perihelion the velocity is tangential. The escape velocity is $v_0=\sqrt{2GM/r_0}$, so $h=r_0{v}_0$ and $h^2=2GM{r}_0$. Consequently $k=1/(2r_0)$. The initial conditions are $u(0)=1/r_0$ and $u^{\prime }(0)=0$. Solving the linear ordinary differential equation gives the parabolic Kepler orbitUsing angular momentum conservation again,On the outgoing branch $r=2r_0$ is first reached at $\theta =\pi /2$. Integrate from release at perihelion, using $w=\tan (\theta /2)$:More generally the same integral gives $t=(2r_0^2/h)(w+w^3/3)$, the Barker equation for this parabolic orbit.

Work on the Riemann sphere, including zero and infinity. For $j(z)=1/z$, two iterations give the identity and one does not, so $\mathrm{ord}(j)=2$. For $k(z)=1/(1-z)$, direct composition givesNeither $k$ nor $k^2$ is the identity; in particular $0\mapsto 1\mapsto \infty \mapsto 0$. Thus

To prove the Möbius conjugacy normal forms, write a Möbius transformation as $f(z)=(az+b)/(cz+d)$ with $ad-bc\ne 0$. A finite fixed point satisfies $c{z}^2+(d-a)z-b=0$. If $c\ne 0$, this quadratic has two roots counting multiplicity, and infinity is not fixed. If $c=0$, the transformation is affine, infinity is fixed, and there is one further finite fixed point unless its slope is one. A nonidentity slope-one affine map is a nonzero translation with only infinity fixed. Thus any nonidentity Möbius transformation has one or two distinct fixed points.

If there are two, say $p,q$, choose a Möbius transformation $h$ taking them to zero and infinity. For finite $p,q$ one can use $h(z)=(z-p)/(z-q)$; the cases involving infinity use an affine map or reciprocal. The conjugate $hf{h}^{-1}$ fixes zero and infinity. Fixing infinity sets its denominator's linear coefficient to zero, and fixing zero sets its numerator's constant term to zero; hence it is $z\mapsto \mu z$ with $\mu \ne 0$.

If there is only one fixed point, send it to infinity. The conjugate has the affine form $az+b$. If $a\ne 1$, $b/(1-a)$ would be another fixed point, which is impossible; therefore $a=1,b\ne 0$. Conjugating this translation by $z\mapsto z/b$ yields $z\mapsto z+1$. The identity is already the scaling $\mu =1$. This proves the classification, and invertibility excludes $\mu =0$.

Conjugation carries fixed-point sets bijectively. The translation $z+1$ has exactly one fixed point, infinity. A nonidentity scaling has the two fixed points zero and infinity, while the identity fixes every point. Therefore $z+1$ is not conjugate to any scaling.

Finally let $f$ have finite positive order $n$. A nonzero translation has infinite order, since its $k$th iterate is $z+k$ after normalization. Thus $f$ is conjugate to $z\mapsto \mu z$, where $\mu$ has multiplicative order exactly $n$. For $n>1$, zero and infinity each form a singleton orbit. At a finite nonzero $z$, an iterate returns precisely when ${\mu }^{k}z=z$, equivalently ${\mu }^k=1$; the least positive return time is $n$. Conjugacy preserves orbit sizes. Hence the finite-order Möbius orbit sizes areFor $n>1$ exactly two points have size-one orbits; every other point has orbit size $n$.