Question

Suppose a meteor of mass 2.50 x 1013 kg is moving at 33.0 km/s relative to the center of the Earth and strikes the Earth. Suppose the meteor creates the maximum possible decrease in the angular speed of the Earth by moving toward the west and striking a point on the equator tangentially. What is the change in the angular speed of the Earth due to this collision

Answer 1

change is imperceptible

w_f = 7.272 10⁻⁵ rad / s

Step-by-step 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 -
`m ( v \ r )/(I)`

We can approximate the Earth to a sphere, so its angular momentum is

I = 2/5 M r²

we substitute

w_f = w₀ -
`(5 \ m \ v)/(2 \ M \ r)`

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 -
`(5 \ 250 \ 10^(13) \ 33.0 \ 10^(3) )/( 2 \ 5.98 \ 10^(24) \ 6.37 10^(6) )`

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