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Speed of a Plane in Various Frames of Reference

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Suppose an airplane flew east along the equator from the Prime Meridian to the International Dateline; that is, exactly half way around the world. What is its final displacement? Is it the distance covered by its route of flight (12,450 miles) or the distance measured straight through the center of the earth (7,920 miles)? Suppose the trip took exactly one day. What would be the displacement as observed from the Sun?

Suppose the airplane flew at a steady 450 miles per hour, with respect to the surface of the Earth. Discuss, in qualitative (non-mathematical) terms, what its velocity would be with respect to other frames of reference: the center of the Earth, the Moon, the Sun, and the center of the Galaxy.

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The displacement as measured with respect to the earth is 7920 miles. The displacement as measured with respect to the sun depends on the time of day at which the flight begins. If the airplane starts at sunrise, it will finish at sundown a day and a half later, since it has crossed 12 time zones. The displacement of the plane in this case is given approximately D = 2 pi R / 365 - 2 r, where R is the distance from the earth to the sun and r is the radius of the earth. Thus we have

D = 2(3.14)(9.3*10^7 mi)/365 - 7920 mi
= 1.6 * 10^6 mi.

Actually the distance the plane travels with respect to the center of the earth makes very little difference, since the distance traveled with respect to the sun is over 100 times greater. Also, the distance to the sun is not constant but varies by about 3% throughout the year, so the ...

Solution Summary

We compute the speed of a plane in the reference frame of the earth, the moon, the sun, and the center of the galaxy.

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