Past exam of the physics 1 paper of the University of Oxford / 2021 resit / 9

Assume the station's mass remains fixed through an instantaneous kick, with subsequent motion governed only by the Earth's Newtonian gravitational field. Put $\kappa =G{M}_E$, and work per unit station mass. The initial circular orbit has specific angular momentum $j$ and specific orbital energy ${\epsilon }_0$ given byLet $\delta$ denote the fractional increase of kinetic energy. The position and potential energy are unchanged during the kick, soUnchanged angular momentum at unchanged position fixes the tangential velocity; the added kinetic energy must be radial. Thus this is a radial kick, not a purely tangential speed increase. The new orbital eccentricity follows fromAt either radial turning point, the Kepler radial energy equation gives $\epsilon =j^2/(2r^2)-\kappa /r$. Its two roots, equivalently the periapsis and apoapsis of the elliptic orbit, areFor $\delta =0.01$, $e=0.1$ andAn inward or outward kick gives the same pair of radial turning points, although it places the station at a different phase of the orbit. These two finite positive radii require $0\le \delta <1$.

Retain the unchanged-angular momentum assumption of the preceding part. The initial orbit has radius $r_o=6400+500=6900\mathrm{km}$, and the desired pericentre distance is $r_p=6400+100=6500\mathrm{km}$. Using $r_p=r_o/(1+e)$ givesThe required fractional increase in initial kinetic energy is thereforeHere $\Delta E=\Delta K>0$ because the instantaneous kick does not change potential energy. Since the initial total orbital energy is $E_0=-K_0$, an alternative convention for the requested fractional energy change givesThe total orbital energy becomes less negative. Explicitly specifying its denominator avoids interpreting a positive energy input as a decrease of energy.

Let $D$ be the separation of the Earth and Moon, and let $x$ be the spacecraft's distance from the Earth's centre towards the Moon. On $0<x<D$, the two gravitational forces oppose each other. Their equality, by Newton's law of universal gravitation, isTaking the positive square root givesThe centre of mass of the two bodies is at $x_G=D{M}_M/(M_E+M_M)$ from the Earth's centre. Thus the gravitational field null between two point masses lies a distancefrom their centre of mass, on the Moon's side. With $M_E/M_M=81$, $x=9D/10$ and $x_G=D/82$, soThe point is $3.84\times 10^7\mathrm{m}$ from the Moon. It is an instantaneous zero of the physical gravitational field, not a co-rotating Lagrange point: a spacecraft remaining fixed relative to the Earth-Moon line would need a nonzero centripetal acceleration. With zero net force it cannot follow that circular co-rotation.