Chapter 7, Problem 69PQ

(a)

To determine

The proof for the statement that when $d>>l$ , the change in gravitational field can be written as $\Delta \overrightarrow{\text{g}}\left(d\right)=\overrightarrow{\text{g}}\left(d-\frac{l}{2}\right)-\overrightarrow{\text{g}}\left(d+\frac{l}{2}\right)\approx -GM\left(\frac{2l}{d^3}\right)\widehat{\text{r}}$ .

Answer to Problem 69PQ

It is showed that when $d>>l$ , the change in gravitational field can be written as $\Delta \overrightarrow{\text{g}}\left(d\right)=\overrightarrow{\text{g}}\left(d-\frac{l}{2}\right)-\overrightarrow{\text{g}}\left(d+\frac{l}{2}\right)\approx -GM\left(\frac{2l}{d^3}\right)\widehat{\text{r}}$ .

Explanation of Solution

Write the equation for the gravitational field for a mass.

  $\overrightarrow{\text{g}}\left(r\right)=-\frac{GM}{r^2}\widehat{\text{r}}$                                                                                                      (I)

Here, $\overrightarrow{\text{g}}\left(r\right)$ is the gravitational field at the point at a distance $r$ from the mass, $G$ is the universal gravitation constant, $M$ is the mass and $\widehat{\text{r}}$ is the unit vector pointing away from the point.

Write the expression for the difference in gravitational field at a distance $d$ from the source across a length $l$ .

  $\Delta \overrightarrow{\text{g}}\left(d\right)=\overrightarrow{\text{g}}\left(d-\frac{l}{2}\right)-\overrightarrow{\text{g}}\left(d+\frac{l}{2}\right)$                                                                               (II)

Use equation (I) to find the expression for $\overrightarrow{\text{g}}\left(d+\frac{l}{2}\right)$ .

  $\overrightarrow{\text{g}}\left(d+\frac{l}{2}\right)=-\frac{GM}{{\left(d+\frac{l}{2}\right)}^2}\widehat{\text{r}}$

Use equation (I) to find the expression for $\overrightarrow{\text{g}}\left(d-\frac{l}{2}\right)$ .

  $\overrightarrow{\text{g}}\left(d-\frac{l}{2}\right)=-\frac{GM}{{\left(d-\frac{l}{2}\right)}^2}\widehat{\text{r}}$

Put the above two equations in equation (II).

  $\begin{array}{c}\Delta \overrightarrow{\text{g}}\left(d\right)=-\frac{GM}{{\left(d-\frac{l}{2}\right)}^2}\widehat{\text{r}}-\left(-\frac{GM}{{\left(d+\frac{l}{2}\right)}^2}\widehat{\text{r}}\right)\\ =-GM\left(\frac{1}{{\left(d-\frac{l}{2}\right)}^2}-\frac{1}{{\left(d+\frac{l}{2}\right)}^2}\right)\widehat{\text{r}}\\ =-GM\left(\frac{{\left(d+\frac{l}{2}\right)}^2-{\left(d-\frac{l}{2}\right)}^2}{{\left(d-\frac{l}{2}\right)}^2{\left(d+\frac{l}{2}\right)}^2}\right)\widehat{\text{r}}\end{array}$

Expand the numerator of the above equation.

  $\begin{array}{c}\Delta \overrightarrow{\text{g}}\left(d\right)=-GM\left(\frac{d^2+dl+\frac{l^2}{4}-\left(d^2-dl+\frac{l^2}{4}\right)}{{\left(d-\frac{l}{2}\right)}^2{\left(d+\frac{l}{2}\right)}^2}\right)\widehat{\text{r}}\\ \text{=}-GM\left(\frac{2dl}{{\left(d-\frac{l}{2}\right)}^2{\left(d+\frac{l}{2}\right)}^2}\right)\widehat{\text{r}}\end{array}$                                                    (III)

It is given that $d>>l$ . This implies that $l$ in the denominator of the above equation can be neglected compared to $d$ .

Conclusion:

Neglect $l$ in equation (III).

  $\begin{array}{c}\Delta \overrightarrow{\text{g}}\left(d\right)\approx -GM\left(\frac{2dl}{d^2{d}^2}\right)\widehat{\text{r}}\\ \approx -GM\left(\frac{2l}{d^3}\right)\widehat{\text{r}}\end{array}$

Thus, it is showed that when $d>>l$ , the change in gravitational field can be written as $\Delta \overrightarrow{\text{g}}\left(d\right)=\overrightarrow{\text{g}}\left(d-\frac{l}{2}\right)-\overrightarrow{\text{g}}\left(d+\frac{l}{2}\right)\approx -GM\left(\frac{2l}{d^3}\right)\widehat{\text{r}}$ .

(b)

To determine

The difference between the gravitational field of the black hole at the feet and the head if the person falls with feet first into the black hole.

Answer to Problem 69PQ

The difference between the gravitational field of the black hole at the feet and the head if the person falls with feet first into the black hole is $-4.0\times {10}^2\widehat{\text{r}}{\text{ m/s}}^2$ .

Explanation of Solution

The expression for the difference between the gravitational field of the black hole at the feet and the head if the person falls with feet first into the black hole is derived in part (a).

Write the expression for the difference between the gravitational field of the black hole at the feet and the head if the person falls with feet first into the black hole.

  $\Delta \overrightarrow{\text{g}}\left(d\right)\approx -GM\left(\frac{2l}{d^3}\right)\widehat{\text{r}}$                                                                                            (IV)

Conclusion:

Given that the mass of the black hole is one solar mass, the length of the person is $1.5\text{ m}$ and the value of $d$ is $1000\text{ km}$ . The value of the solar mass is $1.99\times {10}^{30}\text{ kg}$ and the value of $G$ is $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ .

Substitute $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2$ for $G$, $1.99\times {10}^{30}\text{ kg}$ for $M$, $1.5\text{ m}$ for $l$ and $1000\text{ km}$ for $d$ in equation (IV) to find $\Delta \overrightarrow{\text{g}}\left(d\right)$ .

  $\begin{array}{c}\Delta \overrightarrow{\text{g}}\left(d\right)\approx -\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(1.99\times {10}^{30}\text{ kg}\right)\left(\frac{2\left(1.5\text{ m}\right)}{{\left(1000\text{ km}\frac{1000\text{ m}}{1\text{ km}}\right)}^3}\right)\widehat{\text{r}}\\ \text{=}-4.0\times {10}^2\widehat{\text{r}}{\text{ m/s}}^2\end{array}$

Therefore, the difference between the gravitational field of the black hole at the feet and the head if the person falls with feet first into the black hole is $-4.0\times {10}^2\widehat{\text{r}}{\text{ m/s}}^2$ .

(c)

To determine

Whether the difference in gravitational field found in part (b) large enough to spaghettify the person.

Answer to Problem 69PQ

The difference in gravitational field found in part (b) is large enough to spaghettify the person.

Explanation of Solution

Spaghettification is the term used by Stephen Hawking to describe what happens to someone who falls feet first into a small but highly massive object. Since the gravitational field at the person’s feet is sufficiently higher than the gravitational field at the head, the person gets stretched out like a spaghetti noodle.

In part (b), it is found that the difference between the gravitational field of the black hole at the feet and the head if the person falls with feet first into the black hole is $-4.0\times {10}^2\widehat{\text{r}}{\text{ m/s}}^2$ . The value of the acceleration due to gravity of Earth is $9.81\text{ m/s}$ so that the calculated gravitational field difference is about $40g$ . A gravitational field difference of $4.0\times {10}^2{\text{ m/s}}^2$ across the body will mean that the feet of the person are accelerating toward the black hole $4.0\times {10}^2{\text{ m/s}}^2$ faster than the person’s head.

Conclusion:

Since the feet of the person are accelerating toward the black hole $4.0\times {10}^2{\text{ m/s}}^2$ faster than the person’s head, the person is thoroughly spaghettified or in other words the difference in gravitational field found in part (b) is large enough to spaghettify the person.