Chapter 9.10, Problem 46ES

Recall that the gravitational force exerted by the Earth on an object is called the object’s weight (or more precisely, its Earth Weight). If an object of mass m is on the surface of the Earth weight). If an object of mass m is on the surface of the Earth (mean sea level), then the magnitude of its weight is mg, where g is the acceleration due to gravity at the Earth’s surface. A more general formula for the magnitude of the gravitational force that the Earth on an object of mass m is $F=\frac{mg{R}^2}{{(R+h)}^2}$ where $R$ is the radius of Earth and $h$ is the height of the object above the Earth’s surface.

(a) Use the binomial series for $1/{\left(1+x\right)}^2$ obtained in Example 4 of Section 9.9 to express $F$ as a Maclaurin series in power of $h/R.$

(b) Show that if $h=0,$ then $F=mg.$

(c) Show that if $h/R\approx 0,\text{then}F\approx mg-(2mgh/R).$ [Note: The quantity $(2mgh/R$ Can be thought of as a “correction term� the Earth’s surface into account.]

(d) If we assume that the Earth is a sphere of radius $R=4000$ mi at mean sea level, by approximately what percentage does a person’s weight change in going from mean sea level to the top of Mt. Everest $29,028\text{ft})?$