Recall that the gravitational force exerted by the Earth onan object is called the

Chapter 9, Problem 44

(choose chapter or problem)

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 (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 exerts on an object of mass \(m\) is

\(F=\frac{m g R^{2}}{(R+h)^{2}}\)

where \(R\) is the radius of the Earth and \(h\) is the height of the object above the Earth's surface.

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

(b) Show that if \(h=0 \text {, then } F=m g\).

(c) Show that if \(h / R \approx 0, \text { then } F \approx m g-(2 m g h / R)\) .
[Note: The quantity \(2 m g h / R\) can be thought of as a "correction term" for the weight that takes the object's height above the Earth's surface into account.]

(d) If we assume that the Earth is a sphere of radius \(R=4000 m i\) 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 ft)?

Equation  Transcription:

Text Transcription:

F=mgR^2/(R+h)^2

1/(1+x)^2

h/R

h=0

f=mg

h/R approx 0

F approx mg - (2mgh/R)

2mgh/R

R=4000mi