Question: The radius R_h of a black hole is the radius of a mathematical sphere, called the event horizon that is centered on the black hole. Information from events inside the event horizon cannot reach the outside world. According to Einstein’s general theory of relativity,${\mathbf{\text{R}}}_{\mathbf{\text{h}}}\mathbf{=}\mathbf{2}{\mathbf{\text{GM/c}}}^{\mathbf{\text{2}}}$ , where Mis the mass of the black hole and cis the speed of light.

Suppose that you wish to study a black hole near it, at a radial distance of 50R_h. However, you do not want the difference in gravitational acceleration between your feet and your head to exceed 10 m/s² when you are feet down (or head down) toward the black hole. (a) As a multiple of our Sun’s mass M_s , approximately what is the limit to the mass of the black hole you can tolerate at the given radial distance? (You need to estimate your height.) (b) Is the limit an upper limit (you can tolerate smaller masses) or a lower limit (you can tolerate larger masses)?

The radius of the black hole is ${\text{R}}_{\text{h}}=\frac{\text{2GM}}{{\text{c}}^{\text{2}}}$

The distance of the black hole from the event horizon is${\text{50R}}_{\text{h}}\text{.}$

The maximum difference in gravitational acceleration between your feet and the head is

$\left|{\text{da}}_{\text{g}}\right|=10 {\text{m/s}}^{\text{2}}$

We can find the difference between acceleration due to gravity atand. Using this, we can find the mass of the black hole. Also, we can estimate whether our body can tolerate the upper limit or lower limit.

Formula:

$$\begin{gathered} \text{F =}\frac{\text{GMm}}{{\text{r}}^{\text{2}}} \\ {\text{F = ma}}_{\text{g}} \end{gathered}$$

Equating the gravitational force of attraction and the gravitational force by Newton’s second law, we get

$\frac{\text{GMm}}{{\text{r}}^{\text{2}}}={\text{ma}}_{\text{g}}$

This gives

${\text{a}}_{\text{g}}=\frac{\text{GM}}{{\text{r}}^{\text{2}}}$

The maximum difference in gravitational acceleration between your feet and the head is

$$\begin{gathered} \left|{\text{da}}_{\text{g}}\right|=\text{d}\left(\frac{\text{GM}}{{\text{r}}^{\text{2}}}\right) \\ \left|{\text{da}}_{\text{g}}\right|=\frac{\text{2GM}}{{\text{r}}^{\text{3}}}\text{dr} \\ \left|{\text{da}}_{\text{g}}\right|=\frac{\text{2GM}}{{\left({\text{50R}}_{\text{h}}\text{+ h}\right)}^{\text{3}}}\text{dr} \\ =10m/s^2 \end{gathered}$$

Since $\text{h}<<{\text{50R}}_{\text{h}}$we can write $\text{h}~\text{dR}$ and ${\text{50R}}_{\text{h}}+\text{h}~{\text{50R}}_{\text{h}}$\

Hence,

$$\begin{gathered} \left|{\text{da}}_{\text{g}}\right|=\frac{\text{2GM}}{{\text{50}}^{\text{3}}{\text{R}}_{\text{h}}^{\text{3}}}\text{h} \\ =10{\text{m/s}}^{\text{2}} \end{gathered}$$

But,

${\text{R}}_{\text{h}}=\frac{\text{2GM}}{{\text{c}}^{\text{2}}}$

So, $\left|{\text{da}}_{\text{g}}\right|=\frac{\text{2GM}}{{\text{50}}^{\text{3}}{\left(\frac{\text{2GM}}{{\text{c}}^{\text{2}}}\right)}^{\text{3}}}\text{h}$ (1)

$$\begin{gathered} \text{h}=\left|{\text{da}}_{\text{g}}\right|\frac{{\text{50}}^{\text{3}}{\left(\frac{\text{2GM}}{{\text{c}}^{\text{2}}}\right)}^{\text{3}}}{\text{2GM}} \\ =10\times {50}^3{\left(\frac{{\displaystyle \text{2GM}}}{{\displaystyle {\text{c}}^3}}\right)}^2 \end{gathered}$$

The average height of a person is h = 1.5 m . Using this value in the above equation, we have

$1.5m=10\times {50}^3{\left(\frac{2\times 6.67\times {10}^{-11}\times \text{M}}{{\left(3\times {10}^8\right)}^3}\right)}^2$

If we divide it by the mass of the sun, we will get 110.855 solar masses. So, critical black hole mass, which we can tolerate, should be around 111 solar masses.

As seen in equation (1), we have $\left|{\text{da}}_{\text{g}}\right|$ inversely proportional to the mass of the black hole. So, the above-found limit is the lower limit.