 Solution

ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2015/4/i/solution

To escape the Solar system in the single-Sun approximation, the heliocentric specific orbital energy must reach zero. At Earth's orbital radius, the required heliocentric escape speed is

2 v_{E}

. The smallest change from Earth's initial velocity is prograde, parallel to its motion:

w_{escape} = (2 - 1) v_{E} .

Combining the terrestrial and solar energy requirements gives

u_{escape, min} = v_{e, E}^{2} + (2 - 1)^{2} v_{E}^{2} ≃ 16.65 km s^{- 1} .

The asymptotic heliocentric trajectory is a parabolic Kepler orbit; larger launch speeds can give positive solar orbital energy.

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