Binary and Many-Body Gravitational Motion
For two isolated masses M1 and M2 separated by a, the relative orbit obeys T^2=4 pi^2 a^3/[G(M1+M2)] and M1 r1=M2 r2; symmetric many-body circular motion requires the vector sum of mutual forces to equal each body's centripetal force.
Why this shows up in the exam
Binary stars · Masses at polygon vertices · Escape from multiple gravitating sources
Learn the idea
Mutually orbiting bodies move about their common centre of mass under the same internal force. In a binary or symmetric cluster, no body is automatically fixed; symmetry locates the shared centre and combines the inward pulls.
🧠 Memory hook: Use separation for gravity and centre-of-mass radius for motion.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- M1 r1 = M2 r2 — centre-of-mass distance relation
- T² = 4 pi² a³/[G(M1+M2)] — binary period using separation a
- sum F_radial = m v²/r_orbit — symmetric many-body circular balance
How to approach it
- 1Locate the common centre of mass
- 2Resolve all mutual forces radially
- 3Apply centripetal balance or total-potential energy
Common slip-ups that cost marks
- •Fixing one comparable mass
- •Using one body's orbital radius as separation
- •Adding scalar forces in a polygon
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Original chapter practice
Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.
A satellite moves in a circular orbit where GM = 4 x 10^14 m^3/s^2 and orbital radius is 2 x 10^6 m. Find its orbital speed.
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