Chapter 13, Problem 59A

Suppose you stood atop a ladder that was so tall that you were three Earth radii from Earth’s center. Show that your weight would be one ninth its present value.

To determine

To show: Change in the weight of body is one-ninth of its present value.

Explanation of Solution

Given:

Given body is at a distance thrice the radius of earth from the center of the earth.

Formula used:

Weight of a body is nothing but the product of mass of the body and acceleration due to gravity ( g ).

  $W=mg$

Weight of an object is equal to the gravitational force acting on the object of mass m $mg=\frac{GMm}{R^2}$

(1)

Where,

Gis the Gravitational constant= $\text{6}{\text{.67×10}}^{\text{-11}}\frac{{\text{Nm}}^{\text{2}}}{{\text{Kg}}^{\text{2}}}$

M is Earth’s mass

m is object’s mass

R is Earth’s radius

Calculation:

From equation (1), acceleration due to gravity on the surface of the earth is,

  $g=\frac{GM}{R^2}$

So acceleration due to gravity at height H from the surface of the earth is,

  $g_H=\frac{GM}{{(R+H)}^2}$

Where,

$g_H$ is the acceleration due to gravity at height H

In this case the height above the surface of the Earth is equal to 2R so $R+2R=3R$

The ratio will be

  $\begin{array}{l}\frac{g_H}{g}=\frac{R^2}{{(3R)}^2}\\ \frac{g_H}{g}=\frac{R^2}{9R^2}\\ \frac{g_H}{g}=\frac{1}{9}\\ g_H=\frac{1}{9}g\left(2\right)\end{array}$

This proves that acceleration due to gravity at a distance three times the radius of the Earth is one-ninth the acceleration due to gravity at a distance R from the center of the earth.

Multiply equation (2) by mass of the body so as to obtain the weight of the body at height H.

  $\begin{array}{l}m{g}_H=\frac{1}{9}mg\\ W_H=\frac{1}{9}{W}_S\end{array}$

Conclusion:

Therefore, weight of the body is decreased by its one-ninth when compared with the weight of the body on the surface of the earth.