Chapter 1, Problem 33P

At 45° latitude: the gravitational acceleration as a function of elevation z above sea level is given by $g=a-bz$, where $a=9.807m/s^2$ and $b=3.32\times {10}^{-6}{s}^{-2}$ . Determine the height above sea level where the weight of an object will decrease by 1 percent.

To determine

(a)

The height above the sea level where decrease in the object weight is $1\%$.

Answer to Problem 33P

The height above the sea level where decrease in the object weight is $1\%$ is $29518\text{m}$.

Explanation of Solution

Given Information:

The gravitation constant function is $a-bz$, constant a is $9.807\text{m}/{\text{s}}^{\text{2}}$ and the constant b is $3.32\times {10}^{-6}{\text{s}}^{\text{-2}}$.

Write the expression for the weight of the object above sea level.

  $W_{abovesealevel}=\left(percentagedecrease\right)W_{sealevel}$...... (I)

Since there is a $1\%$ decrease in the weight. Substitute $0.99$ for percentage decrease in Equation (1)

  $W_{abovesealevel}=\left(0.99\right)W_{sealevel}$...... (II)

Write the expression for the gravitational acceleration as a function of elevation.

  $g=a-bz$....... (III)

Here the constant are $a$ and $b$, elevation above sea level is $z$.

Substitute $9.807\text{m}/{\text{s}}^{\text{2}}$ for $a$ and $3.32\times {10}^{-6}{\text{s}}^{\text{-2}}$ for $b$ in Equation (III)

  $g=\left(9.807\text{m}/{\text{s}}^{\text{2}}\right)-\left(3.32\times {10}^{-6}{\text{s}}^{\text{-2}}\right)z$.... (IV)

Since at the sea level the elevation is $0$, substitute $0$ for $z$ in Equation (IV)

  $\begin{array}{l}g=a-b\left(0\right)\\ g=a\end{array}$...... (V)

Write the expression for the weight of the object.

  $W=mg$....... (VI)

Here mass of the object is $m$ and the acceleration due to gravity is $g$.

Substitute $mg$ for $W$ in Equation (II)

  $\begin{array}{l}{\left(mg\right)}_{abovesealevel}=\left(0.99\right){\left(mg\right)}_{sealevel}\\ {\left(g\right)}_{abovesealevel}=\left(0.99\right){\left(g\right)}_{sealevel}\end{array}$...... (VII).

Calculation:

Substitute $9.807\text{m}/{\text{s}}^{\text{2}}$ for $a$ in Equation (V)

  $g=9.807\text{m}/{\text{s}}^{\text{2}}$

Substitute $9.807\text{m}/{\text{s}}^{\text{2}}$ for ${\left(g\right)}_{sealevel}$ and $\left(9.807\text{m}/{\text{s}}^{\text{2}}\right)-\left(3.32\times {10}^{-6}{\text{s}}^{\text{-2}}\right)z$ for ${\left(g\right)}_{abovesealevel}$ in Equation (VII).

  $\begin{array}{l}\left(9.807\text{m}/{\text{s}}^{\text{2}}\right)-\left(3.32\times {10}^{-6}{\text{s}}^{\text{-2}}\right)z=\left(0.99\right)9.807\text{m}/{\text{s}}^{\text{2}}\\ -\left(3.32\times {10}^{-6}{\text{s}}^{\text{-2}}\right)z=\left(0.99\right)9.807\text{m}/{\text{s}}^{\text{2}}-\left(9.807\text{m}/{\text{s}}^{\text{2}}\right)\\ -\left(3.32\times {10}^{-6}{\text{s}}^{\text{-2}}\right)z=-0.098\text{m}/{\text{s}}^{\text{2}}\\ z=29518\text{m}\end{array}$

Conclusion:

The height above the sea level where decrease in the object weight is $1\%$ is $29518\text{m}$.