r_{1}

r_{2}

, the given Kepler orbit satisfies

r_{1} = \frac{h ^{2}}{k ( 1 + e )}, r_{2} = \frac{h ^{2}}{k ( 1 - e )} .

Hence

e = \frac{r _{2} - r _{1}}{r _{1} + r _{2}}, h^{2} = \frac{2 k r _{1} r _{2}}{r _{1} + r _{2}} .

Therefore

h = \frac{2 k r _{1} r _{2}}{r _{1} + r _{2}}, C = 2 .

ii

Solution

A circular orbit of radius

r

v_{c} (r) = k / r

. On the transfer ellipse, the velocity is tangential at both apsides and has magnitude

h / r

. The first prograde burst therefore changes the speed by

Δ v_{1} = \frac{k}{r _{1}} (\frac{2 r _{2}}{r _{1} + r _{2}} - 1) .

r_{2}

the satellite must again accelerate in the direction of motion to circularize, by

Δ v_{2} = \frac{k}{r _{2}} (1 - \frac{2 r _{1}}{r _{1} + r _{2}}) .

a = \frac{r _{1} + r _{2}}{2}, b = a 1 - e^{2} = r_{1} r_{2} .

Conservation of angular momentum gives constant areal velocity

h /2

. The satellite sweeps half the ellipse, whose area is

πab /2

, so

t_{2} - t_{1} = \frac{πab /2}{h /2} = \frac{πab}{h} .

Using part (i),

t_{2} - t_{1} = \frac{π ( r _{1} + r _{2} ) ^{3/2}}{2 2 k} = π \frac{a ^{3}}{k} .

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