Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 316 / 3 / ii

Immediately after release, the grain has the comet's position and velocity, but its effective stellar gravitational parameter is

μ_{d} = (1 - β) μ

\frac{v ^{2}}{2} - \frac{μ}{r} = - \frac{μ}{2 a},

the grain energy is

\frac{v ^{2}}{2} - \frac{( 1 - β ) μ}{r} = - \frac{μ}{2 a} + \frac{β μ}{r} = - \frac{( 1 - β ) μ}{2 a _{d}} .

Therefore

a_{d} = \frac{a ( 1 - β )}{1 - 2 β a / r} .

h^{2} = μ a (1 - e^{2})

. Applying

h^{2} = μ_{d} a_{d} (1 - e_{d}^{2})

then gives

e_{d}^{2} = \frac{e ^{2} - 2 β + β ^{2} + 2 β a ( 1 - e ^{2} ) / r}{( 1 - β ) ^{2}} .

The paper instead prints

e^{2} - 2 e β + β^{2}

in the first three terms. That expression is incompatible with both its printed

a_{d}

and conservation of

h

, except at

e = 1

. The next part is the result obtained from the printed eccentricity, so both consequences are recorded below.

 Ancestors (10)

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