Past exam of the physics 1 paper of the University of Oxford / 2017 / 11

Use a reference frame moving with Jupiter, neglect the spacecraft's effect on the planet and neglect the Sun during the short encounter. The Newtonian central force makes the orbit planar and conserves the angular momentum $J=m{r}^2\dot{\vartheta }$. In plane polar coordinates, the kinetic energy is $m({\dot{r}}^2+r^2{\dot{\vartheta }}^2)/2$. Choosing zero gravitational potential energy at infinity givesThe terms are respectively radial kinetic energy, transverse kinetic energy, and attractive gravitational potential energy. The second term is often combined with the third into the radial effective potential. Far from Jupiter,Here $b$ is the perpendicular impact parameter of the incoming asymptote, not the eventual periapsis distance.

Put ${\mu }_J=G{M}_J$. At the periapsis distance, the radial velocity vanishes. Insert $E=m{v}_0^2/2$ and $J=mb{v}_0$ into a:The positive root isRationalizing gives the numerically stable alternative $r_{\min }=b^2{v}_0^2/[\sqrt{{\mu }_J^2+b^2{v}_0^4}+{\mu }_J]$. Attraction makes $r_{\min }<b$. This is a point-planet scattering calculation; a noncolliding physical encounter must also have $r_{\min }$ outside Jupiter's surface. At $b=0$, the limiting formula gives a radial collision, not a flyby.

Let $h=J/m=b{v}_0$ be the specific angular momentum, and measure the orbital angle $\vartheta$ from periapsis. Set $w=1/r$. From $\dot{\vartheta }=h/r^2$, differentiating gives $\dot{r}=-hdw/d\vartheta$. The radial equation of motion $\ddot{r}-r{\dot{\vartheta }}^2=-{\mu }_J/r^2$ becomes the Binet equationIts solution symmetric about periapsis isFor example, the expression for the orbital eccentricity follows by substituting $w$ and $\dot{r}=-h{w}^{\prime }$ into the energy equation: $E/m={\mu }_J^2(e^2-1)/(2h^2)=v_0^2/2$. Since $e>1$, this is a hyperbolic Kepler orbit.

The outgoing asymptote occurs at ${\vartheta }_{\infty }$ with $\cos {\vartheta }_{\infty }=-1/e$, where $\pi /2<{\vartheta }_{\infty }<\pi$. The incoming velocity points inward along the incoming radial asymptote, whereas the outgoing velocity points outward. Therefore the change in velocity direction is $\varphi =2{\vartheta }_{\infty }-\pi$, not $2{\vartheta }_{\infty }$. It follows thatFor $b>0$, this gives $0<\varphi <\pi$: a large impact parameter produces a small deflection, and the ideal point-mass limit $b\to 0^{+}$ approaches a reversal.

A $90^{\circ }$ gravitational scattering angle has $\cot (\varphi /2)=1$. ThusThe denominator uses the squared speed in $\mathrm{m}{\mathrm{s}}^{-1}$, not its numerical value in $\mathrm{km}{\mathrm{s}}^{-1}$. From b, the corresponding periapsis distance is $r_{\min }=(\sqrt{2}-1)b\simeq 5.53\times 10^8\mathrm{m}$; this differs substantially from the impact parameter because of gravitational focusing.

Take Jupiter's heliocentric velocity as ${\mathbf{V}}_J=13\hat{\mathbf{x}}\mathrm{km}{\mathrm{s}}^{-1}$. Conservation of energy in the Jupiter frame makes the incoming and outgoing asymptotic speeds both $10\mathrm{km}{\mathrm{s}}^{-1}$. For the $90^{\circ }$ encounter, an outgoing velocity in Jupiter's direction requires the incoming Jupiter-frame velocity to be perpendicular to it:Use a Galilean transformation back to the Sun frame:The heliocentric speeds are thereforeThe incoming direction makes an angle $\arctan (10/13)\simeq 37.6^{\circ }$ with Jupiter's motion; its two signs describe mirror-image encounter planes. The gravitational assist transfers orbital energy from Jupiter to the spacecraft: the spacecraft's specific kinetic energy increases by $\frac{1}{2}(23^2-16.4^2)\mathrm{k}{\mathrm{m}}^2{\mathrm{s}}^{-2}$, or, using the unrounded squared incoming speed, exactly $130\mathrm{k}{\mathrm{m}}^2{\mathrm{s}}^{-2}=1.30\times 10^8\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$. There is no gain in asymptotic speed in the Jupiter frame; the gain is a consequence of changing frames while the planet is moving.