As thermonuclear fusion proceeds in its core, the Sun loses mass at a rate of 3.
ID: 2025816 • Letter: A
Question
As thermonuclear fusion proceeds in its core, the Sun loses mass at a rate of 3.64 109 kg/s. During the 5,000-yr period of recorded history, by how much has the length of the year changed due to the loss of mass from the Sun? Suggestions: Assume the Earth's orbit is circular. No external torque acts on the Earth–Sun system, so the angular momentum of the Earth is constant.Other than figuring out the mass of the sun and how much mass has been ultimately burned during the 5000 years, I have no idea where to start!
Explanation / Answer
F_g=GmM/R^2 where M is the mass of the sun and m is the mass of the earth.
F_g = F_c
F_c=mv^2/R
mw^2R=GMm/R^2
w=2*pi/t
(2pi/T)^2=GM/R^3
T^2=(4pi^2/G)(R^3)/M
Take log on both sides
2log(T)=log((4pi^2/G)(R^3))-log(M)
Teke derrivative on both sides
2*dT/T=-dm/M
dm=-3.64109 kg/s * 5000yrs * 365 day/yr * 24 h/day * 3600 s/h
=5.73*10^20
T=365*24*3600=31536000
M=1.989*10^30
So change in time is
dt=(5.73*10^20)(31536000)/(2*5.1*10^28)
dt=0.0179