 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2016-resit/10/solution

j = b v

v^{2} /2

. At closest approach the radial velocity vanishes, so the effective potential gives

\frac{v ^{2}}{2} = \frac{b ^{2} v ^{2}}{2 r _{0}^{2}} - \frac{GM}{r _{0}} .

r_{0} = \frac{G ^{2} M ^{2} + b ^{2} v ^{4} - GM}{v ^{2}} = \frac{b ^{2} v ^{2}}{GM + G ^{2} M ^{2} + b ^{2} v ^{4}}, v_{m a x} = \frac{b v}{r _{0}} = v^{2} + \frac{2 GM}{r _{0}} .

The second form for

r_{0}

avoids subtracting two nearly equal numbers for a small impact parameter. These formulas describe a nonradial orbit;

b = 0

gives a collision rather than a finite nonzero pericentre in the point-planet model.

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