How do you calculate the mass of the sun, #M_"sun"#, using Kepler's third law (#T^2=(4 pi^2 r^3)/(G M_"sun")#)?

Assume the period of the Earth is #T=3.156xx10^7# seconds and the Earth's distance from the Sun is #1.496xx10^11# meters.

1 Answer
May 13, 2017

Plug the given values into the given equation. Answer: #~~1.98955xx10^30# #"kg"#

Explanation:

Since we are given the equation:
#T^2=(4pi^2r^3)/(GM_(sun))#
where #T=3.156xx10^7# seconds, #r=1.496xx10^11# meters, #pi~~3.14# is the mathematical constant, and #G=6.67xx10^-11# as the gravitational constant, we can first solve for #M_(sun)# with variables then substitute the given values to find #M_(sun)#:

First, we will use variables to solve for #M_(sun)# to avoid the amount of numbers in the equation:
#T^2=(4pi^2r^3)/(GM_(sun))#
#T^2(GM_(sun))=4pi^2r^3#
#M_(sun)=(4pi^2r^3)/(T^2G)#

Now, we can substitute our given values:
#M_(sun)=(4pi^2(1.496xx10^11)^3)/((3.156xx10^7)^2(6.67xx10^-11))#

#~~1.98955xx10^30# #"kg"# rounded to 5 decimals