Past exam of the physics 1 paper of the University of Oxford / 2013 / 11

Let $\mu =G{M}_{\ast }$ and use plane polar coordinates in the orbital plane. The Newtonian gravitational potential energy is $U=-\mu M_p/r$, soA central force keeps the orbit in a plane. The star is treated as stationary, appropriate when $M_p\ll M_{\ast }$; otherwise the two-body reduction uses the reduced mass and the relative separation.

The Lagrangian has no explicit time dependence, so its energy function is constant. With total speed $v^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$, division by $M_p$ givesConsequently, between the initial point $(r_i,v_i)$ and any other point,This energy relation holds for nonradial motion too. Radially outward motion is the stipulated approximation for the subsequent probe calculation; angular momentum is then zero.

Retain the source's idealized solar radial trajectory. Use ${\mu }_{\odot }\simeq 1.32712\times 10^{11}\mathrm{k}{\mathrm{m}}^3{\mathrm{s}}^{-2}$, $r_i=1.5\times 10^9\mathrm{km}$ and $v_i=22\mathrm{km}{\mathrm{s}}^{-1}$. At $r=1.8\times 10^{10}\mathrm{km}$,The speed has decreased because the probe gains gravitational potential energy while moving away from the Sun. This is the supplied simplified model, rather than a reconstruction of Voyager 1's actual three-dimensional trajectory or planetary encounters.

Setting $1/r\to 0$ in b gives the asymptotic speedThe radicand is positive, so the ideal probe has positive total mechanical energy and escapes the Sun. The later dust-cloud model takes this limiting speed as its initial speed.

For constant-area capture of stationary dust, let $x=0$ at cloud entry, $M(0)=M_p$, $v(0)=v_{\infty }$, and define $k=\rho A$ with fixed collecting area $A$. Every swept volume $Adx$ adds stationary mass, soThere is no external longitudinal force, and the incoming dust has zero momentum in the cloud frame. The variable-mass system therefore satisfies $d(Mv)/dt=0$, rather than $M\dot{v}=0$. Using $Mv=M_p{v}_{\infty }$ givesHenceThe kinetic energy decreases during sticking: $dK/dt=-\frac{1}{2}\dot{M}{v}^2$. The lost energy becomes internal energy. Radiation, gravity, sputtering and changes in area are excluded by the ideal complete-collection model. The zero-density limit recovers $M=M_p$, $v=v_{\infty }$ and $x=v_{\infty }t$.

Use a year of $365.25$ days, so $t=9.46728\times 10^8\mathrm{s}$, and $k=\rho A=10^{-11}\mathrm{kg}{\mathrm{m}}^{-1}$. With $M_p=722\mathrm{kg}$ and $v_{\infty }\simeq 1.752\times 10^4\mathrm{m}{\mathrm{s}}^{-1}$, e givesThe corresponding cloud thickness is $x_f=(M_f-M_p)/k\simeq 1.50\times 10^{13}\mathrm{m}$. Checking $M_f{v}_f=M_p{v}_{\infty }$ verifies conservation of momentum; using the initial speed for all thirty years would overestimate the collected mass.