Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 316 / 3 / i

Table of contents
- Solution i

Solution

Since the binary separation and mean motion are both one, the Kepler third law gives

μ_{1} + μ_{2} = G (M_{1} + M_{2}) = 1.

The center of mass condition puts the bodies at

r_{M_{1}} = (- μ_{2}, 0), r_{M_{2}} = (μ_{1}, 0)

in the rotating reference frame. Thus the sketch has

M_{1}

,

O

, and

M_{2}

in that order along the

x

axis, with

P = (x, y)

, and

O M_{1} = μ_{2}, O M_{2} = μ_{1},

M_{1} P = r_{1} = (x + μ_{2})^{2} + y^{2},

M_{2} P = r_{2} = (x - μ_{1})^{2} + y^{2} .

In the inertial sketch this entire configuration rotates uniformly about

O

.

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