Answer:
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/T...................................2
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
