Chapter 5, Problem 62P

(a)

To determine

The difference in the value of $g$ between his head and his feet.

Answer to Problem 62P

The difference in the value of $g$ between his head and his feet is $\underset{\_}{5.6\times {10}^{-6}\text{m}/{\text{s}}^2}$.

Explanation of Solution

Write the expression for the $g$ value of his feet.

    $\begin{array}{c}{g}_{\text{feet}}=\frac{G{M}_{\text{Earth}}}{r_{\text{Earth}}^2}\\ =g\end{array}$        (I)

Here, $g_{\text{feet}}$ is the force of gravity at his feet, $G$ is the gravitational constant, $M_{\text{Earth}}$ is the mass of the Earth, $r_{\text{Earth}}$ is the radius of Earth.

Write the expression for the $g$ value of his head.

    $g_{\text{head}}=\frac{G{M}_{\text{Earth}}}{{\left(r_{\text{Earth}}+h\right)}^2}$        (II)

Here, $g_{\text{head}}$ is the force of gravity at his head, $h$ is the height of the person.

Use equation (II) and (I) to solve for the ratio of value of $g$ between his head and his feet.

    $\begin{array}{c}\frac{g_{\text{head}}}{g_{\text{feet}}}=\frac{\frac{G{M}_{\text{Earth}}}{{\left(r_{\text{Earth}}+h\right)}^2}}{\frac{G{M}_{\text{Earth}}}{r_{\text{Earth}}^2}}\\ =\frac{r_{\text{Earth}}^2}{{\left(r_{\text{Earth}}+h\right)}^2}\end{array}$        (III)

Use equation (I) in (III) to solve for $g_{\text{head}}$.

    $\begin{array}{c}{g}_{\text{head}}=\left(\frac{r_{\text{Earth}}^2}{{\left(r_{\text{Earth}}+h\right)}^2}\right)g_{\text{feet}}\\ =\left(\frac{r_{\text{Earth}}^2}{{\left(r_{\text{Earth}}+h\right)}^2}\right)g\end{array}$        (IV)

Write the expression for the difference in the value of $g$ between his head and his feet.

    $\begin{array}{c}{g}_{\text{feet}}-g_{\text{head}}=g-\left(\frac{r_{\text{Earth}}^2}{{\left(r_{\text{Earth}}+h\right)}^2}\right)g\\ =g\left(1-\frac{r_{\text{Earth}}^2}{{\left(r_{\text{Earth}}+h\right)}^2}\right)\end{array}$        (V)

Conclusion:

Substitute $6.37\times {10}^6\text{m}$ for $r_{\text{Earth}}$, $6.0\text{ft}$ for $h$, $9.8\text{m}/{\text{s}}^2$ for $g$ in equation (V) to find the difference in the value of $g$ between his head and his feet.

    $\begin{array}{c}{g}_{\text{feet}}-g_{\text{head}}=\left(9.8\text{m}/{\text{s}}^2\right)\left(1-\frac{{\left(6.37\times {10}^6\text{m}\right)}^2}{{\left(6.37\times {10}^6\text{m}+6.0\text{ft}\times \frac{0.3048\text{m}}{1\text{ft}}\right)}^2}\right)\\ =5.6\times {10}^{-6}\text{m}/{\text{s}}^2\end{array}$

Therefore, the difference in the value of $g$ between his head and his feet is $\underset{\_}{5.6\times {10}^{-6}\text{m}/{\text{s}}^2}$.

(b)

To determine

The difference in the gravitational acceleration between his head and his feet.

Answer to Problem 62P

The difference in the gravitational acceleration between his head and his feet is $\underset{\_}{37.6\text{m}/{\text{s}}^2}$.

Explanation of Solution

Write the expression for the magnitude of acceleration near the black hole at the person’s feet.

    $a_{\text{bh,}\text{feet}}=\frac{G{M}_{\text{bh}}}{r_{\text{E}}^2}$        (VI)

Here, $a_{\text{bh,}\text{feet}}$ is the acceleration near the black hole at the person’s feet, $M_{\text{bh}}$ is the mass of the black hole.

Write the expression for the magnitude of acceleration near the black hole at the person’s head.

    $a_{\text{bh,}\text{head}}=\frac{G{M}_{\text{bh}}}{{\left(r_{\text{E}}+h\right)}^2}$        (VII)

Here, $a_{\text{bh,}\text{head}}$ is the acceleration near the black hole at the person’s head.

Use equation (VI) and (VII) to solve for the ratio of value of acceleration between his head and his feet.

    $\begin{array}{c}\frac{a_{\text{bh,}}{}_{\text{head}}}{a_{\text{bh,}}{}_{\text{feet}}}=\frac{\frac{G{M}_{\text{bh}}}{{\left(r_{\text{Earth}}+h\right)}^2}}{\frac{G{M}_{\text{bh}}}{r_{\text{Earth}}^2}}\\ =\frac{r_{\text{Earth}}^2}{{\left(r_{\text{Earth}}+h\right)}^2}\end{array}$        (VIII)

Use equation (VIII) and (VI) to solve for $a_{\text{bh,}}{}_{\text{head}}$.

    $\begin{array}{c}{a}_{\text{bh,}}{}_{\text{head}}=\left(\frac{r_{\text{Earth}}^2}{{\left(r_{\text{Earth}}+h\right)}^2}\right)a_{\text{bh,}}{}_{\text{feet}}\\ =\left(\frac{r_{\text{Earth}}^2}{{\left(r_{\text{Earth}}+h\right)}^2}\right)\left(\frac{G{M}_{\text{bh}}}{r_{\text{E}}^2}\right)\end{array}$        (IX)

Write the expression for the difference in the value of acceleration between his head and his feet.

    $\begin{array}{c}{a}_{\text{bh,}}{}_{\text{feet}}-a_{\text{bh,}}{}_{\text{head}}=\left(\frac{G{M}_{\text{bh}}}{r_{\text{E}}^2}\right)-\left(\frac{r_{\text{Earth}}^2}{{\left(r_{\text{Earth}}+h\right)}^2}\right)\left(\frac{G{M}_{\text{bh}}}{r_{\text{E}}^2}\right)\\ =\left(\frac{G{M}_{\text{bh}}}{r_{\text{E}}^2}\right)\left[1-\left(\frac{r_{\text{Earth}}^2}{{\left(r_{\text{Earth}}+h\right)}^2}\right)\right]\end{array}$        (X)

Conclusion:

Substitute $6.37\times {10}^6\text{m}$ for $r_{\text{Earth}}$, $6.0\text{ft}$ for $h$, $9.8\text{m}/{\text{s}}^2$ for $g$, $20\times 1.99\times {10}^{30}\text{kg}$ for $M_{\text{bh}}$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^2$ for $G$ in equation (X) to find the difference in the value of acceleration between his head and his feet.

    $\begin{array}{c}{a}_{\text{bh,}}{}_{\text{feet}}-a_{\text{bh,}}{}_{\text{head}}=\left[\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^2\right)\left(20\times 1.99\times {10}^{30}\text{kg}\right)}{{\left(6.37\times {10}^6\text{m}\right)}^2}\right]\left[1-\frac{{\left(6.37\times {10}^6\text{m}\right)}^2}{{\left(6.37\times {10}^6\text{m}+6.0\text{ft}\times \frac{0.3048\text{m}}{1\text{ft}}\right)}^2}\right]\\ =37.6\text{m}/{\text{s}}^2\end{array}$

Therefore, the difference in the gravitational acceleration between his head and his feet is $\underset{\_}{37.6\text{m}/{\text{s}}^2}$.