= Solution
For an <electron-positron collider> of fixed bending radius $R$, <synchrotron radiation> rapidly becomes the principal obstacle. Multiplying the supplied radiated power by the revolution period $2\pi R/c$ gives
$$
U_0=\frac{e^2\gamma^4}{3\epsilon_0R},\qquad U_0[\mathrm{GeV}]\simeq8.85\times10^{-5}\frac{E[\mathrm{GeV}]^4}{R[\mathrm m]}.
$$
A bending radius of $3\ \mathrm{km}$ gives losses of about $2.95\ \mathrm{GeV}$ per turn at $100\ \mathrm{GeV}$ per beam, but $115\ \mathrm{GeV}$ per turn at $250\ \mathrm{GeV}$. At $100\ \mathrm{GeV}$, a $5\ \mathrm{mA}$ beam already radiates about $15\ \mathrm{MW}$. The radiofrequency system must replace this <energy>, and the vacuum chamber must absorb the radiation. The centre-of-mass <energy> is twice the beam <energy> for equal opposite beams.
For a <collider> with two <proton> beams, the same <energy> gives a much smaller <Lorentz factor>. At fixed <energy> and radius the radiated power scales as $m^{-4}$, so the suppression relative to <electrons> is $(m_e/m_p)^4\simeq8.8\times10^{-14}$. The main practical limit is the dipole <magnetic field> through magnetic rigidity:
$$
p[\mathrm{GeV}/c]\simeq0.2998\,B[\mathrm T]R[\mathrm m].
$$
With $R\simeq2.8\ \mathrm{km}$ and $B\simeq8.3\ \mathrm T$, the beam <energy> is about $7\ \mathrm{TeV}$, or $14\ \mathrm{TeV}$ in the centre of mass. Fields near $16\ \mathrm T$ would give approximately $13.4\ \mathrm{TeV}$ per beam at that radius. These are realistic superconducting-magnet scales; a $27\ \mathrm{km}$ circumference includes straight sections, so its bending radius is less than $C/(2\pi)$. The <proton> radiation loss near $7\ \mathrm{TeV}$ is only a few keV per turn, though high beam currents still create substantial cryogenic heat loads. Magnet technology, aperture, protection and stored beam <energy> therefore dominate before the electron-like radiation limit.
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