Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 316 / 2 / d

Table of contents
- Solution d

Solution

For

e = 0

, regard

T_{p} (x) = \frac{1}{x} + 2 cos I x .

Its minimum occurs at

x = (cos I)^{- 2/3}

and equals

3 (cos I)^{2/3}

. A circular orbit is therefore possible for

T_{p} = 5/2

only if

3 (cos I)^{2/3} \leq \frac{5}{2} .

The smallest required orbital inclination is consequently

I_{m i n} = arccos (\frac{5}{6})^{3/2} ≃ 40. 5^{\circ} .

Solved by gpt-5.6-sol high.

 Ancestors (10)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version