Published by Patrick Mutisya · 14 days ago
Students should be able to state that the current best estimate for the Hubble constant is
\$H_0 = 2.2 \times 10^{-18}\ \text{s}^{-1}\$
The Hubble constant (\$H_0\$) describes the rate at which the Universe is expanding. It relates the recession velocity (\$v\$) of a distant galaxy to its distance (\$d\$) from us:
\$v = H_0 \, d\$
where:
In cosmology the Hubble constant is often quoted in units of km s⁻¹ Mpc⁻¹. The relationship between the two common units is:
\$\$1\ \text{km s}^{-1}\text{Mpc}^{-1} = \frac{1\ \text{km}}{1\ \text{s}\times 1\ \text{Mpc}}
= \frac{10^3\ \text{m}}{1\ \text{s}\times 3.086\times10^{22}\ \text{m}}
= 3.2408\times10^{-20}\ \text{s}^{-1}\$\$
Therefore, the current estimate can also be expressed as:
\$\$H_0 = \frac{2.2 \times 10^{-18}\ \text{s}^{-1}}{3.2408\times10^{-20}\ \text{s}^{-1}\,\text{(km s}^{-1}\text{Mpc}^{-1})}
\approx 68\ \text{km s}^{-1}\text{Mpc}^{-1}\$\$
Different methods give slightly different results, leading to the so‑called “Hubble tension”. The value quoted here, \$2.2 \times 10^{-18}\ \text{s}^{-1}\$ (≈ 68 km s⁻¹ Mpc⁻¹), represents a consensus average of recent measurements.
Suppose a galaxy is observed to have a red‑shift corresponding to a recession speed of \$v = 10\,200\ \text{km s}^{-1}\$. Using \$H_0 = 68\ \text{km s}^{-1}\text{Mpc}^{-1}\$, its distance is:
\$d = \frac{v}{H_0} = \frac{10\,200\ \text{km s}^{-1}}{68\ \text{km s}^{-1}\text{Mpc}^{-1}} \approx 150\ \text{Mpc}\$
| Method | Typical \$H_0\$ Value | Uncertainty (1σ) | Units |
|---|---|---|---|
| Cepheid + Type Ia Supernovae | 73 | ±1.5 | km s⁻¹ Mpc⁻¹ |
| Planck CMB (ΛCDM fit) | 67.4 | ±0.5 | km s⁻¹ Mpc⁻¹ |
| Gravitational‑wave standard sirens | 70 | ±10 | km s⁻¹ Mpc⁻¹ |