Question

Suppose that p- and s-waves traveled in straight lines through the Earth (in fact, they don't, but we will make this approximation for this problem). Also suppose that the angular distance from the epicenter (location of the earthquake) on the Earth's spherical surface to the location on the surface beyond which no s-waves can be observed from an earthquake (the angle between the earthquake epicenter and the s shadow zone) is 120 degrees (i.e., the two points are located 1/3 of the Earth's circumference from each other). What is the radius of the liquid outer core of the Earth divided by the radius of the Earth (i.e., what is their ratio)? Hint: draw a diagram and use trigonometry to find the answer. Do not give the answer in km or miles, but as a fraction of the Earth's radius.

Answer 1

The distance from the focus to the shadow point along the earth's surface will be of the earth's circumference of 40,000 km which is $\frac{1}{3}\times40000$ = 13,333 km.
And the distance from there to the opposite side of the earth is 20,000 - 13,333 = 6,667 km.

At the center of the earth that line passes half that distance from the earth's core = 3333km .

The distance from the focus to the shadow point along the earth's surface will be 120/360 of the earth's circumference of 40,000 km which is 1/3 times 40000 = 13,333 km. And the distance from there to the opposite side of the earth is 20,000 - 13,333 = 6,667 km. At the center of the earth that line passes half that distance from the earth's core = 3333km. So ratio is 3333/6371 = 0.523

So ratio is $\frac{3333}{6371}=0.523$