 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2017/11/d/solution

9 0^{\circ}

cot (ϕ /2) = 1

. Thus

b = \frac{G M _{J}}{v _{0}^{2}} = \frac{( 6.6743 \times 1 0 ^{- 11} ) ( 2 \times 1 0 ^{27} )}{( 1 0 ^{4} ) ^{2}} m ≃ 1.33 \times 1 0^{9} m .

The denominator uses the squared speed in

m s^{- 1}

, not its numerical value in

km s^{- 1}

. From b, the corresponding periapsis distance is

r_{m i n} = (2 - 1) b ≃ 5.53 \times 1 0^{8} m

; this differs substantially from the impact parameter because of gravitational focusing.

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