The radius ${\text{R}}_{\text{h}}$and mass ${\text{M}}_{\text{h}}$of a black hole are related by ${\text{R}}_{\text{h}}=2G{\text{M}}_{\text{h}}/{\text{c}}^2$, wherecis the speed of light. Assume that the gravitational acceleration ${\text{a}}_{\text{g}}$of an object at a distance ${\text{r}}_0=1.001{\text{R}}_{\text{h}}$from the center of a black hole is given by$a_g=\frac{GM}{r^2}$(it is, for large black holes). (a) In terms of ${\text{M}}_{\text{h}}$, find${\text{a}}_{\text{g}}$at.${\text{r}}_0$ (b) Does$a_g$at${\text{r}}_0$ increase or decrease as${\text{M}}_{\text{h}}$increases? (c) What is$a_g$at${\text{r}}_0$for a very large black hole whose mass is$1.55\times {10}^{12}$times the solar mass of$1.99\times {10}^{30}\text{kg}$? (d) If an astronaut of height$1.70\text{m}$is at${\text{r}}_0$with her feet down, what is the difference in gravitational acceleration between her head and feet?(e) Is the tendency to stretch the astronaut severe?

1) The radius of a black hole,$R_h=\frac{2G{M}_h}{c^2},\text{where}c=\text{speedoflight}$

2) Distance of object from black hole,$r_0=1.001R_h$

3) Gravitational acceleration of an object,

$a_g=\frac{GM}{r^2}$

4) Mass of a black hole is $1.55\times {10}^{12}$times the solar mass of$1.99\times {10}^{30}\text{\hspace{0.33em}kg}$

5) The height of the astronaut is$1.70\text{\hspace{0.33em}m}$

The gravitational acceleration of a particle (of mass m) is due solely to the gravitational force acting on it. When the particle is at distance r from the center of a uniform, spherical body of mass M, the magnitude F of the gravitational force on the particle is given by,

$F=G\frac{m_1{m}_2}{r^2}$

From Newton’s second law,

$F=m{a}_g$

Therefore, the gravitational acceleration is,

$a_g=\frac{GM}{r^2}$

Using the equation of gravitational acceleration and the given condition of the radius, we can find the gravitational acceleration and conclude about its proportion with mass. It is easy to find the differential gravitational acceleration.

Formula:

Differential acceleration equation, $d{a}_g=-\frac{2G{M}_E}{r^3}dr$ …(i)

We have been given,

$\begin{array}{l}{R}_h=\frac{2G{M}_h}{c^2}\\ r_0=1.001R_h\\ a_g=\frac{GM}{R^2}\end{array}$

So, substituting values of, $R_h$we get

$\begin{array}{c}{a}_g=\frac{G{M}_h}{{\left(1.001R_h\right)}^2}\\ =\frac{G{M}_h}{{\left(1.001\right)}^2\left(\frac{2G{M}_h}{c^2}\right)}\\ =\frac{c^4}{{\left(2.002\right)}^{2}G{M}_h}\\ =\frac{{\left(3\times {10}^8\text{\hspace{0.33em}m/s}\right)}^4}{4.008\times 6.67\times {10}^{-11}\text{\hspace{0.33em}N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\cdot M_h}\\ =\left(3.02\times {10}^{43}\text{\hspace{0.17em}\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}\right)/M_h\end{array}$

The value of acceleration is $\left(3.02\times {10}^{43}\text{\hspace{0.17em}\hspace{0.17em}kg}\cdot {\text{m/s}}^{\text{2}}\right)/M_h$

As we see from the above result, we conclude that$a_g$ is inversely proportional to.$M_h$ So, $a_g$increases when,$M_h$ decreases.

It is given that mass of black hole is $1.55\times {10}^{12}$times the solar mass of$1.99\times {10}^{30}\text{\hspace{0.33em}kg}$

$\begin{array}{c}{M}_h=\left(1.55\times {10}^{12}\right)\left(1.99\times {10}^{30}\text{\hspace{0.17em}kg}\right)\\ =3.08\times {10}^{42}\text{\hspace{0.17em}\hspace{0.17em}kg}\end{array}$

Substituting the value of$M_h$in equation (ii), we get

$\begin{array}{c}{a}_g=\frac{3.02\times {10}^{43}}{M_h}\\ =\frac{3.02\times {10}^{43}\text{\hspace{0.33em}kg}\cdot {\text{m/s}}^{\text{2}}}{3.08\times {10}^{42}\text{\hspace{0.33em}kg}}\\ =9.8{\text{\hspace{0.33em}m/s}}^{\text{2}}\end{array}$

Hence, the acceleration is $9.8{\text{\hspace{0.33em}m/s}}^{\text{2}}$

Substituting the value in equation (i), we get

$\begin{array}{c}d{a}_g=-2\frac{GM}{{\left(\frac{2.002GM}{c^2}\right)}^3}dr\\ =-\frac{2c^6}{{\left(2.002\right)}^3{\left(GM\right)}^2}dr\end{array}$

Since it is given in problem that$dr=1.70\text{\hspace{0.33em}m}$

$\begin{array}{c}d{a}_g=-\frac{2c^6}{{\left(2.002\right)}^3{\left(GM\right)}^2}\left(1.70\text{\hspace{0.33em}m}\right)\\ =-\frac{2{\left(3.00\times {10}^8\text{\hspace{0.33em}m/s}\right)}^6}{{\left(2.002\right)}^3{\left(6.67\times {10}^{-11}\text{\hspace{0.33em}N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\times \left(1.55\times {10}^{12}\times 1.99\times {10}^{30}\text{\hspace{0.33em}kg}\right)\right)}^2}\left(1.70\text{\hspace{0.33em}m}\right)\\ =-0.732\times {10}^{-14}{\text{\hspace{0.33em}m/s}}^{\text{2}}\end{array}$

Magnitude of $d{a}_g$is$0.732\times {10}^{-14}{\text{\hspace{0.33em}m/s}}^{\text{2}}$

The effect of differential gravitational force is very negligible, as it can be clearly seen from the magnitude of difference acceleration that the value is very small.