Neutrinos through the earth

We can model this as a sector of a circle. Consider a circle, centre $O$ , such that it has a radius of $6,370 km$ and points $x_0$ and $x_1$ on the circumference. Then, by using the cosine rule, we can see that, $\cos(\angle x_0Ox_1) = \frac{2 \cdot 6370000 ^ 2 - 732000^2}{2 \cdot 6370000^2}$ $\cos(\angle x_0Ox_1) = 0.993 ...$ $\angle x_0Ox_1 = 0.114 ...$ Hence, the length of the arc $D$ is given by, $D = 6370000 \cdot 0.114 ...= 732403.3565$ So the difference between the length of $D$ and the neutrino path to $3$ significant figures is, $732403.3565 - 732000 = \fbox{403}$

Let E represent the center of the Earth, C represent the location of CERN, and G represent Gran Sasso mountain. Note that ∆ ECG is an isosceles triangle such that c and g , segments opposite of points C and G , respectively, are legs formed by radii extending from the center of the Earth. Therefore, c = g = 6370 km and e = 732 km, the straight-line distance between points C and G . Define θ = m ∠ E so that by the law of cosines,

$e^{2}=c^{2}+g^{2}-2cg\cos \theta$

$732^{2}=6370^{2}+6370^{2}-2(6370)(6370)\cos \theta$

$535824=81153800-81153800\cos \theta$

$81153800\cos \theta=80617976$

$\theta=\arccos \frac{80617976}{81153800}$

$\theta\approx0.114977$

The arc length from C to G is given by D = 6370 θ ≈ 732.403 km. Therefore, D – 732 ≈ 0.403 km = 403 m.

Firstly,draw out of a full sphere and draw 2 lines from cern and gran sasso to the centre of the circle.Observe now that the area enclosed by the 2 lines now form a sector on the circle. As we known the formula for the arc lenth of a sector is r x thetha where thetha is the angle of the sector. To find the angle all we have to do is some simple trigonometry 0= 2 sin^-1((732x 0.5)/3760) We know this as the triangle can be split into 2 congurent right triangles by circle properties. Lastly,we apply the formula,hence:

D=732.4034km D-732=0.4034km=403.4m