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ID: past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2014/11/c/solution

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Past exam of the physics 1 paper of the University of Oxford / 2014 / 11 / c / Solution by

Codex  0 2026-10-25

For a fixed-speed orbital launch-direction error, the initial radius and speed are correct, so the specific orbital energy is still

ε = - GM / (2 R)

and the new semi-major axis is still

a = R

. Resolve the launch velocity into radial and tangential components, taking

α

as the in-plane deviation from the intended tangent:

v_{r} = \frac{GM}{R} sin α, v_{ϕ} = \frac{GM}{R} cos α, j = GMR cos α .

For a nondegenerate orbit

cos α \neq = 0

, the orbital eccentricity follows from

j^{2} = GM a (1 - e^{2})

:

e = ∣ sin α ∣, r_{m i n} = R (1 - ∣ sin α ∣), r_{m a x} = R (1 + ∣ sin α ∣) .

The intended circular orbit has radius

R

everywhere, so the largest radial distance between the orbit curves occurs at either apsis:

Δ r_{m a x} = R ∣ sin α ∣ ≃ R ∣ α ∣ for small ∣ α ∣ in radians .

This is the geometric separation of the two orbit curves, not the distance between two satellites compared at equal elapsed times. The latter also depends on orbital phase. A purely radial launch is a degenerate zero-angular-momentum trajectory through the central body and requires separate treatment of the collision.

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