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.

r^2 = LT/2mπ

Explanation:

centripetal force

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.

F = m(v^2/r)

v=ωr

ω= 2πf, 2π/T2

substituting back into v

v = 2πr/T

3

taking te value of the linear velocity v into  equation 1

F = m((2πr/T)2/r)

4

F = 4m π2r/T2

gravitational force

 F = GMm/r^2  5

comparing 4 and 5

GMm/r^2 =4m π2r/T2

GMT^2  = 4 π^2 r^3

GMT^1/2  = 4π^2(L/2mπ)^3/2

6

derivative of  equation

6

GM(½T^-1/2dT/dt) + GT^1/2 dM/dt = 0

make dT/dt subject of the formula

dT/dt = -dM/dt (2T/M)

7

∆T/∆t ≈ -dM/dt (2T/M)

∆T ≈ -  ∆t    (dM/dt)   (2T/M)

M=mass of the sun

∆T ≈ -5000(3.64´109)(2(1)/1.991x1030)

∆T ≈ 1.83 x 10-17 yr^2/sec

       3.16x10^7 sec in one year

∆T ≈ 0.0183 sec

L = r x p

L = r x mv

    r ≈ ┴ p

L = mvr

L = m(2πr/T)r

L = 2m π r2/T

r2 = LT/2mπ

gets less

a displacement distance is size of displ