Use Kepler's third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and the Sun is 1.496 * 1011 m. Earth's orbital period around the Sun is 365.26 days. 6.34 * 1029 kg 1.99 * 1030 kg 6.28 * 1037 kg 1.49 * 1040 kg

its 1.99 x10^30 just answered it

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%)