Respuesta :
Answer:
change is imperceptible
w_f = 7.272 10⁻⁵ rad / s
Explanation:
For this exercise we can use the conservation of angular momentum.
Initial. Before the crash
L₀ = I w₀
final. After the crash
L_f = I w_f + p r
where the moment is
p = mv
As the system is formed by the two bodies, the forces during the impact are internal, therefore the angular momentum is conserved
L₀ = L_f
I w₀ = I w_f + m v r
w_f = w₀ - [tex]m \frac{ v \ r }{I}[/tex]
We can approximate the Earth to a sphere, so its angular momentum is
I = 2/5 M r²
we substitute
w_f = w₀ - [tex]\frac{5 \ m \ v}{2 \ M \ r}[/tex]
We can find the angular velocity of the Earth with the duration of a spin which is the period of one day
w₀ = 2π / T
T = 24 h (3600 s / 1h) = 86 400 s
w₀ = 2π / 86400
w₀ = 7.272 10⁻⁵ rad / s
let's calculate
w_f = 7.27 10⁻⁵ - [tex]\frac{5 \ 250 \ 10^{13} \ 33.0 \ 10^{3} }{ 2 \ 5.98 \ 10^{24} \ 6.37 10^{6} }[/tex]
w_f = 7.272 10⁻⁵ - 1.0829 10⁻¹³
w_f = 7.27199999 10⁻⁵
this change is imperceptible
w_f = 7.272 10⁻⁵ rad / s
