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 Incoming links: Past exam of the physics 1 paper of the University of Oxford / 2017 / 11 / a

Past exam of the physics 1 paper of the University of Oxford / 2017 / 11 / b / Solution 2026-10-25

Put

μ_{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:

\frac{1}{2} v_{0}^{2} = \frac{b ^{2} v _{0}^{2}}{2 r _{m i n}^{2}} - \frac{μ _{J}}{r _{m i n}}, v_{0}^{2} r_{m i n}^{2} + 2 μ_{J} r_{m i n} - b^{2} v_{0}^{2} = 0.

The positive root is

r_{m i n} = \frac{μ _{J}^{2} + b ^{2} v _{0}^{4} - μ _{J}}{v _{0}^{2}} = b^{2} + (\frac{μ _{J}}{v _{0}^{2}})^{2} - \frac{μ _{J}}{v _{0}^{2}} .

Rationalizing gives the numerically stable alternative

r_{m i n} = b^{2} v_{0}^{2} / [μ_{J}^{2} + b^{2} v_{0}^{4} + μ_{J}]

. Attraction makes

r_{m i n} < b

. This is a point-planet scattering calculation; a noncolliding physical encounter must also have

r_{m i n}

outside Jupiter's surface. At

b = 0

, the limiting formula gives a radial collision, not a flyby.

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