Respuesta :
Answer:
Explanation:
Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass 2.3 x 10¹⁹ Kg
Final moment of inertia I₂ = 2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²
For change in period of rotation we shall apply conservation of angular momentum law
I₁ ω₁ = I₂ ω₂ , ω₁ and ω₂ are angular velocities initially and finally .
I₁ / I₂ = ω₂ / ω₁
I₁ / I₂ = T₁ / T₂ , T₁ , T₂ are time period initially and finally .
T₂ / T₁ = I₂ / I₁
(2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²) / 2/5 MR²
1 + 5 / 3 x 2.3 x 10¹⁹ / M
= 1 + 5 / 3 x 2.3 x 10¹⁹ / 5.97 x 10²⁴
= 1 + .0000064
T₂ = 24 (1 + .0000064)
= 24 hours + .55 s
change in length of the day = .55 s .
Conservation of angular momentum allows finding the result for the change in the period of rotation of the earth due to the melting of ice is:
ΔT = 0.55 s
The angular momentum is a quantity that relates the moment of inertia and the angular velocity, for an isolated system it must be conserved.
L = I w
Where the bold letters indicate vectors, L is the angular momentum, I the moment of inertia and w the angular velocity.
They indicate that we approximate the earth to a sphere of radius and that the ice that is at the poles with a mass of m = 2.3 10¹⁹ kg is melted and distributed uniformly throughout the planet.
If we consider the earth an isolated system the angular momentum must be conserved.
L₀ = [tex]L_f[/tex]
I₀ w₀ = [tex]I_f w_f[/tex] (1)
Where L₀ and L_f are the angular moments before and after the ice melts, let's find each term.
The tabulated moment of inertia:
Sphere I = [tex]\frac{2}{5}[/tex] M R²
Spherical shell I = ⅔ M R²
Before the ice melts the whole earth is spherical.
I₀ = 2/5 M [tex]R_e^2[/tex]
After melting the ice, we can suppose the system formed by the sphere of the earth and a shell of the ice
[tex]I_f = \frac{2}{5} M' R^2 + \frac{2}{3} m R^2[/tex]
The mass of the earth M = 5.98 10²⁴ kg and the mass of the ice is m = 2.3 10¹⁹ kg, the relationship is
[tex]\frac{M}{m}[/tex] = 2.6 10⁵
We can see that the mass change due to melting is very small, therefore
M' = M
R = [tex]R_e[/tex]
we substitute
[tex]I_f = \frac{2}{5} M R_e^2 + \frac{2}{3} m R_e^2[/tex]
Angular velocity is the ratio of the angle rotated in time. In this case let's use the rotation time of one day, this time is called the period and the rotated angle is 2π radians.
w = [tex]\frac{\theta}{t}[/tex]
w = [tex]\frac{2\pi }{T}[/tex]
Before the ice melts.
[tex]w_o = \frac{2\pi }{T_o}[/tex]
After the ice melts.
[tex]w_f = \frac{2\pi }{T_f}[/tex]
Let's substitute in equation 1
[tex]\frac{2}{5} M R_e^2 \ \frac{2\pi }{T_o} = (\frac{2}{5} M R_e^2 + \frac{2}{3} m R_e^2) \ \frac{2\pi }{T_f}[/tex]
[tex]\frac{2}{5} \frac{M}{T_o} = \frac{\frac{2}{5} M +\frac{2}{3} m}{T_f}[/tex]
[tex]\frac{T_f}{T_o} = \frac{ \frac{2}{5} M + \frac{2}{3} m }{ \frac{2}{5} M} \\\frac{T_f}{T_o} = 1 + \frac{5}{3} \ \frac{m}{M}[/tex]
Let's calculate.
[tex]\frac{T_f}{T_o} = 1 + \frac{5}{3} \ \frac{2.3 \ 10^{19}}{ 5.98 \ 10^{24}} \\\frac{T_f}{T_o} = 1.00000641[/tex]
The change in period is
ΔT = T_f - T₀
ΔT = T₀ (0.00000641)
The rotation period of the earth is 24 h (3600 s / 1h) = 86 400 s
ΔT = 86400 0.00000641
ΔT = 0.55 s
In conclusion using the conservation of angular momentum we can find the result for the change in the period of rotation of the earth due to the melting of ice is:
ΔT = 0.55 s
Learn more here: brainly.com/question/25303285
