Kepler’s
third law formula: T^2=4pi^2*r^3/(GM)
We’re trying to find M, so:
M=4pi^2*r^3/(G*T^2)
M=4pi^2*(1.496
× 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)
M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))
Let’s work
with the units:
(m^3)/((N*m^2/kg^2)*days^2))=
=(m^3*kg^2)/(N*m^2*days^2)
=(m*kg^2)/(N*days^2)
=(m*kg^2)/((kg*m/s^2)*days^2)
=(kg)/(days^2/s^2)
=(kg*s^2)/(days^2)
So:
M=1.48× 10^40(kg*s^2)/(days^2)
Now we need to convert days to seconds in order to cancel
them:
1 day=24 hours=24*60minutes=24*60*60s=86400s
M=1.48× 10^40(kg*s^2)/((86400s)^2)
M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)
M=1.48× 10^40kg/86400^2
M=1.98x10^30kg
The
closest answer is 1.99
× 10^30
(it may vary
a little with rounding – the difference is less than 1%)
