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At step $n$, the parent and daughter rest masses areThe decay is allowed when $M\ge m$, which holds because $M=\lambda m$ with $\lambda >1$. Hence all $N$ stages occur:The initial rest mass is $NM+m_0$ and the final rest mass is $Nm+m_0$. Thusof rest mass is converted into kinetic energy, corresponding to energy $Nm(\lambda -1)c^2$.

The four-momentum of a massive particle iswhile for a massless particleFor $P_1=P_2+P_3$, invariance givesIn the rest frame of particle 1, $P_1=(m_{1}c,0)$, sowith $E_1=m_1{c}^2$. Equal and opposite final momenta are real exactly whenthe threshold condition for a relativistic two-body decay.

For $u=0$, the equation becomesso $\omega /m$ is constant. With $m(t)=m_0-\mu t$,It reaches zero when the water is exhausted, at

With total mass $M={\sum }_i{m}_i$, defineThe angular momentum about the centre of mass isExpanding, using ${\sum }_i{m}_i(x_i-R)=0$ and $P=M\dot{R}$, gives

To first order in $r_0$,The expanded orbit equation is thereforeIts required family of solutions iswith any $0<A<1$. Successive periapses occur when $\omega \theta$ changes by $2\pi$, so their angular separation isThe periapsis advance per revolution is consequently $2\pi K{r}_0+O(r_0^2)$.

ForBinet's equation becomesA circular orbit has constant $u=u_c$ and therefore satisfiesThe right side has maximum $4/(27r_0)$ at $u_c=1/(3r_0)$. Thus there are two circular orbits when $0<K<4/(27r_0)$, one marginal orbit at equality, and none above it.

Writing $u=u_c+\eta$ and linearizing givesHence the outer orbit $u_c<1/(3r_0)$, equivalently $r_c>3r_0$, is stable; the inner orbit is unstable. These are the circular-orbit branches of the shifted inverse-radius potential.

For $V=-km/r$, one has $dV/dr=km{u}^2$, soAfter choosing the angular origin, the solution isFor $0\le e<1$, using $r=\sqrt{x^2+y^2}$ and $r+ex=p$ givesThis is an ellipse, with the force centre at one focus, as described by a Kepler orbit.

A central force conserves angular momentumPut $u=1/r$. Since $\dot{\theta }=\ell u^2/m$,where primes now denote differentiation with respect to $\theta$. Substitution into the radial equation gives the Binet equation

For any rotation $R$ about the centre, ${\varphi }_R(x)=\varphi (Rx)$ satisfies the same equation and the same Dirichlet data because $V,f,g$ are rotationally invariant. Uniqueness from part (a) gives ${\varphi }_R=\varphi$ for every $R$. Thus $\varphi$ is constant on spheres and is a function of $r$ alone.

Suppose ${\varphi }_1,{\varphi }_2$ solve the Dirichlet problem and put $u={\varphi }_1-{\varphi }_2$. Then ${\nabla }^{2}u=0$ in $V$ and $u=0$ on $S$. The divergence theorem applied to $u\nabla u$ gives Green's identityThus $\nabla u=0$, so $u$ is constant; its zero boundary value makes it zero. The Dirichlet solution is therefore unique.

For homogeneous Neumann data the same calculation again shows that $u$ is constant, but the boundary condition does not determine that constant. Hence, whenever a Neumann solution exists, adding any constant produces another solution. This is the standard uniqueness of Poisson equation distinction.