Hubble time and constant [duplicate]

marked as duplicate by John Rennie spacetime StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jul 2 '18 at 11:10
• A quick (inaccurate) calculation using this numerical solution of the Friedmann equation finds the Hubble constant to be $H = \dot{a}/a = (0.6619901-0.6615104)/(6.01-5.99)/0.6612529$ $= 0.03627205264 \text{ Gyrs}^{-1}$ $\approx 35 \text{ km/s/Mpc}^{-1}$. As such $1/H_0 \approx 27.6\text{ Gyrs}$, a wildly different value from the expected $7.4 \text{ Gyrs}$. This is because $1/H_0$ is only a good approximation when $H$ increases similarly to an exponential curve. However, in the early universe this wasn't so, so $1/H_0$ is a bad approximation. – Beta Decay Jul 2 '18 at 11:48