Chapter 13, Problem 58A

If you stood atop a ladder that was so tall that you doubled your distance from Earth’s center, how would your weight compare with its present value?

To determine

To compare:Actual weight with the present value.

Answer to Problem 58A

Weight of the body is decreased by its one-fourth as compared with weight of the body on the surface of the earth

Explanation of Solution

Given:

A body stood at a height equal to the radius 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,

  $g=\frac{GM}{R^2}$

Acceleration due to gravity at height h from the surface of the earth is,

  $g_h=\frac{GM}{{(R+h)}^2}$

Where,

  $g_h$ = acceleration due to gravity at height h .

In this case, the height above the surface of the earth is equal to R so $R+h=R+R=2R$

The ratio will be

  $\begin{array}{l}\frac{g_h}{g}=\frac{R^2}{{(R+R)}^2}\\ \frac{g_h}{g}=\frac{R^2}{{(2R)}^2}\\ \frac{g_h}{g}=\frac{R^2}{4R^2}\\ \frac{g_h}{g}=\frac{1}{4}\end{array}$

  $g_h=\frac{1}{4}g$

This shows that acceleration due to gravity above the surface of the Earth is 1/4 th to the acceleration due to gravity on the surface of the earth (distance R).

Multiply above equation by mass of the body so as to get the weight of the body.

  $\begin{array}{l}m{g}_h=\frac{1}{4}mg\\ W_h=\frac{1}{4}{W}_S\end{array}$

Conclusion:

Therefore, weight of the object is one fourth of the value when compared with the weight of the object on the surface of the earth.