I assume by 107 you mean 10^7, so the satellite is at an altitude of 31000 km.
If we ignore the curvature of the earth, and consider it a plane, then we have an isosceles triangle of height 3.1x10^7m, and a base of 3.1x10^6m, similar to a triangle of base 1, height 10. So, the distance traveled is 3.1x10^6 * sqrt(101) * 2 = 62.3x10^6m.
The speed of light is 3.0x10^8m/s
so, the call takes 6.23x10^7/3.0x10^8 = 2.07x10^-1 = 0.207s.
A communications satellite is in an orbit that is 3.1 multiplied by 107 m directly above the equator. Consider the moment when the satellite is located midway between Quito, Equador, and Belem, Brazil; two cities almost on the equator that are separated by a distance of 3.6 multiplied by 106 m. Find the time it takes for a telephone call to go by the way of satellite between these cities. Ignore the curvature of the earth.
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