Question

Tidal Forces near a Black Hole. An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030 kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer.

(a) What is the tension between her ears?

Answer 1

Answer:

$833.4801043*10^6N$ on ear that is closer to Black hole.

$13.83803929*10^6N$ On ear that is farther from Black hole.

Explanation:

This problem can be solved as two masses that are at two different location from a bigger mass whose gravity affects both.

tension is an equal and opposite force that is exerted in response to applied force.

so on ear that is closer to black hole would have tension that is equal in magnitude and opposite in direction to gravitational force that ear experience due to the black hole at that location.

this true for ear that is further away from black hole as well.

(1) Force on ear that is closer to black hole.

                                                 $F =\frac{m_{1}*m_{2} }{r^2} G$

                    $m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

                     $m_{2}= 0.030kg$ is mass of ear that is close to black hole.

                    $G = 6.679*10^-11m^3*kg^-1*s^-2$

                     r = $120km-\frac{6}{100000} km=119.99994km=119999.94m$

Note, we have subtracted because ear is closer to black hole.

plugging all this in formula gives.

                                      $F = 833.4801043*10^6N$

       That is tension of ear.

(2) Force on ear that is further from black hole.

                              $F =\frac{m_{1}*m_{2} }{r^2} G$

                    $m_{1}= 5*1.989*10^30kg$ is mass of Black hole.

                     $m_{2}= 0.030kg$ is mass of ear that is close to black hole.

                    $G = 6.679*10^-11m^3*kg^-1*s^-2$

         this time r is further away from black hole so it would be.

                       $r = 120km+\frac{6}{100000} km = 120000.06m$

Plugging this all in we get

                          $F = 13.83803929N$

and that is tension on ear that is further from black hole.

Notice the tension difference, and order of magnitude of tension,it is enormous .

this astronaut is lethally close to black hole.