= Solution
In <geometrized units>, a mass is converted to a length by multiplying its SI value by $G/c^2$, or to a time by multiplying by $G/c^3$. Using the supplied constants, the <solar mass> becomes
$$
\boxed{M_\odot^{(\rm length)}\simeq1.47\times10^3\,\mathrm m,\qquad
M_\odot^{(\rm time)}\simeq4.92\times10^{-6}\,\mathrm s.}
$$
Thus the characteristic scales are kilometres and a few microseconds. With potential zero at infinity, the <Newtonian gravitational potential> at the surface, in units of $c^2$, is
$$
\boxed{\Phi_\odot=-\frac{GM_\odot}{R_\odot c^2}\simeq-2.12\times10^{-6}.}
$$
Its very small magnitude is the relevant weak-field measure. In SI potential units the same number corresponds to about $-1.91\times10^{11}\,\mathrm{m^2\,s^{-2}}$.
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