4 ms·
Ok lets see if we can get this to match. I get the gamma factor to be gam = 0.9999999985758079 with this, earth sees any time interval on new horizons t' to b
by Trombone12 11y ago
Ok lets see if we can get this to match. I get the gamma factor to be
gam = 0.9999999985758079
with this, earth sees any time interval on new horizons t' to be
\detla t' = \delta t / gam
longer than the time t on earth. Integrating both sides above over time, keeping velocity constant, we get
t' = t/ gam = 9.5 yr / 0.9999999985758079 = 9.500000013529824 yr
this gives an difference between our observed "new horizons time" and out local time of
t' - t = t/gam - t = (1/gam - 1) t = 0.4 s
we disagree by about a factor of 45000.