Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 4 / 10B (Dynamics and Relativity)

Table of contents
- Solution 10B
- a 10B
  - Solution a
- b 10B
  - Solution b

Solution

Binet's equation gives

r = \frac{r _{0}}{1 + e c o s θ}, r_{0} = \frac{l ^{2}}{k} .

Consequently

l = k r_{0}, E = \frac{mk ( e ^{2} - 1 )}{2 r _{0}} .

a

Solution

At infinity the specific energy is

v^{2} /2

and

l = b v

, so

e = 1 + \frac{b ^{2} v ^{4}}{k ^{2}} .

The angle through which the velocity is deflected between the incoming and outgoing asymptotes is

χ = 2 tan^{- 1} (\frac{k}{b v ^{2}}) .

b

Solution

Zero energy gives

e = 1

. Since the periapsis is

d

,

r_{0} = 2 d

and

l = 2 k d

. Put

T = tan (θ /2)

; then

r = d (1 + T^{2}), \frac{d t}{d T} = \frac{2 d ^{3}}{k} (1 + T^{2}) .

Taking

t = 0

at periapsis yields

θ (t) = 2 tan^{- 1} T (t), T^{3} + 3 T = 3 t \frac{k}{2 d ^{3}} .

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