= Solution
For <pressureless matter> and zero <cosmological constant>, the <Friedmann acceleration equation> gives $\ddot a/a=-4\pi G\rho/3$. The <deceleration parameter> therefore satisfies the <dust relation between deceleration and density>, $q=4\pi G\rho/(3H^2)$. Evaluating today,
$$
\boxed{\rho_0=\frac{3H_0^2q_0}{4\pi G}.}
$$
Substitute this into the present <Friedmann equation> to obtain $H_0^2+kc^2/a_0^2=2q_0H_0^2$, or
$$
\boxed{\frac{kc^2}{a_0^2}=(2q_0-1)H_0^2.}
$$
In particular, $q_0>1/2$, $q_0=1/2$, and $q_0<1/2$ correspond to positive, zero, and negative spatial curvature.
Back to article page
