Chapter 13, Problem 62AP

(a) Show that the rate of change of the free-fall acceleration with vertical position near the Earth’s surface is

$\frac{dg}{dr}=\frac{2G{M}_E}{R_E^3}$

This rate of change with position is called a gradient.

(b) Assuming h is small in comparison to the radius of the Earth, show that the difference in free-fall acceleration between two points separated by vertical distance h is

$\left|\Delta g\right|=\frac{2G{M}_{E}h}{R_E^3}$

(c) Evaluate this difference for h = 6.00 m, a typical height for a two-story building.

To determine

To show: The rate of change of free fall acceleration with vertical position near the Earth’s surface is $\frac{dg}{dr}=\left(-\frac{2G{M}_{\text{E}}}{R_{\text{E}}^3}\right)$.

Explanation of Solution

Explanation:

The rate of change free fall acceleration with position of any quantity is called gradient. And the free fall acceleration is the acceleration of a body falling freely in a vacuum near the surface of the Earth. it is also called as acceleration due to gravity.

Formula to calculate the acceleration due to gravity at distance $r$ from the Earth surface is,

$g=\frac{G{M}_{\text{E}}}{{\left(R_{\text{E}}+r\right)}^2}$

$G$ is the universal gravitational constant.

$R_{\text{E}}$ is the radius of the Earth.

$M_{\text{E}}$ is the mass of the Earth.

$g$ is the acceleration due to gravity.

$r$ is the distance from the Earth surface.

The differentiate for the above equation with respect to $r$.

$\begin{array}{c}\frac{dg}{dr}=G{M}_{\text{E}}\frac{d}{dr}\left(\frac{1}{{\left(R_{\text{E}}+r\right)}^2}\right)\\ =G{M}_{\text{E}}\left(\frac{-2}{{\left(R_{\text{E}}+r\right)}^3}\right)\\ =\left(\frac{-2G{M}_{\text{E}}}{{\left(R_{\text{E}}+r\right)}^3}\right)\end{array}$

The distance $r$ is very small in compare to the radius of the Earth so the neglect term $r$.

$\frac{dg}{dr}=\left(\frac{-2G{M}_{\text{E}}}{{\left(R_{\text{E}}\right)}^3}\right)$

Conclusion:

Therefore, the rate of change of free fall acceleration with vertical position near the Earth’s surface is $\frac{dg}{dr}=\left(-\frac{2G{M}_{\text{E}}}{R_{\text{E}}^3}\right)$.

To determine

To show: The difference in free fall acceleration with between two points separated by vertical distance $h$ is $\left|\Delta g\right|=\left(\frac{2G{M}_{\text{E}}h}{R_{\text{E}}^3}\right)$.

Explanation of Solution

Explanation:

The force that attracts a body towards the center of the Earth, or towards any other physical body having mass called as gravity.

Formula to calculate the difference in free fall acceleration between two points is,

$\left|\Delta g\right|=g-g^{\prime }$ (I)

$g^{\prime }$ is the acceleration due to gravity at the distance $h$.

Formula to calculate the acceleration due to gravity at the Earth surface is,

$g=\frac{G{M}_{\text{E}}}{{\left(R_{\text{E}}\right)}^2}$

Formula to calculate the acceleration due to gravity at a vertical distance $h$ from the Earth surface is,

$g^{\prime }=\frac{G{M}_{\text{E}}}{{\left(R_{\text{E}}+h\right)}^2}$

$h$ is the distance from the Earth surface.

Substitute $\frac{G{M}_{\text{E}}}{{\left(R_{\text{E}}+h\right)}^2}$ for $g^{\prime }$ and $\frac{G{M}_{\text{E}}}{{\left(R_{\text{E}}\right)}^2}$ for $g$ in equation (I).

$\begin{array}{c}\left|\Delta g\right|=\frac{G{M}_{\text{E}}}{{\left(R_{\text{E}}\right)}^2}-\frac{G{M}_{\text{E}}}{{\left(R_{\text{E}}+h\right)}^2}\\ =G{M}_{\text{E}}\left(\frac{{\left(R_{\text{E}}+h\right)}^2-{\left(R_{\text{E}}\right)}^2}{{\left(R_{\text{E}}+h\right)}^2{\left(R_{\text{E}}\right)}^2}\right)\\ =G{M}_{\text{E}}\left(\frac{\left(2R_{\text{E}}h+h^2\right)}{{\left(1+\frac{h}{R_{\text{E}}}\right)}^2{\left(R_{\text{E}}\right)}^4}\right)\end{array}$

The distance $h$ is very small in compare to the radius of the Earth so the neglect term $h^2$ and the value of $\frac{h}{R_{\text{E}}}<1$ so neglect the term $\frac{h}{R_{\text{E}}}$.

$\begin{array}{c}\left|\Delta g\right|=G{M}_{\text{E}}\left(\frac{\left(2R_{\text{E}}h\right)}{{\left(1\right)}^2{\left(R_{\text{E}}\right)}^4}\right)\\ =\left(\frac{\left(2G{M}_{\text{E}}h\right)}{{\left(R_{\text{E}}\right)}^3}\right)\end{array}$ (II)

Conclusion:

Therefore, the difference in free fall acceleration with between two points separated by vertical distance $h$ is $\left|\Delta g\right|=\left(\frac{2G{M}_{\text{E}}h}{R_{\text{E}}^3}\right)$.

To determine

To determine: The difference in free fall acceleration between two points separated by vertical distance $h=6.0\text{m}$.

Answer to Problem 62AP

Answer: The difference in free fall acceleration between two points separated by vertical distance $h=6.0\text{m}$ is $1.85\times {10}^{-5}\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Explanation:

From equation (II),

$\left|\Delta g\right|=\left(\frac{\left(2G{M}_{\text{E}}h\right)}{{\left(R_{\text{E}}\right)}^3}\right)$

Substitute $5.972\times {10}^{24}\text{kg}$ for $M_{\text{E}}$, $6371000\text{m}$ for $R_{\text{E}}$ $6.0\text{m}$ for $h$ and $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$ to find $\left|\Delta g\right|$.

$\begin{array}{c}\left|\Delta g\right|=\left(\frac{\left(2\times 6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}/{\text{kg}}^{\text{2}}\times 5.972\times {10}^{24}\text{kg}\times 6.0\text{m}\right)}{{\left(6371000\text{m}\right)}^3}\right)\\ =1.85\times {10}^{-5}\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Therefore, the difference in free fall acceleration with between two points separated by vertical distance $h=6.0\text{m}$ is $1.85\times {10}^{-5}\text{m}/{\text{s}}^{\text{2}}$.