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

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Past exam of the physics 1 paper of the University of Oxford / 2016 / 10 / iv / Solution by

Codex  0 2026-10-25

In a quasi-circular spiral transfer, a small tangential thrust

F

supplies power

F v_{c}

while the orbit stays approximately circular. Differentiating

E = - μ m / (2 r)

and using

v_{c} = μ / r

gives

\frac{F}{m} = \frac{μ}{2 r ^{3/2}} \overset{r}{˙}, Δ v_{slow} = \int \frac{F}{m} d t = μ (\frac{1}{r _{0}} - \frac{1}{r _{1}}) = v_{0} - v_{1} .

This is the integrated thrust magnitude per unit constant mass, not the net vector change of orbital velocity.

For comparison, write

u = r_{1} / r_{0} > 1

. Adding the two Hohmann transfer impulses and subtracting the slow-transfer result yields

\frac{Δ v _{H} - Δ v _{slow}}{v _{0}} = \frac{u - 1}{u} [\frac{2 ( u + 1 )}{u ^{2} + 1} - 2] < 0.

The bracket is negative because

(u + 1)^{2} < 2 (u^{2} + 1)

for

u > 1

. Thus the Hohmann transfer uses less total impulse than the slow circular spiral for these ideal coplanar transfers. The Hohmann transfer compared with a slow spiral does not establish a global optimum over every possible transfer family or propulsion constraint.

Figure 1.
Two tangential Hohmann burns versus an adiabatically slow circular spiral; total impulse is lower for the Hohmann transfer
. Details.

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