

Stretched By a Black Hole?
You Wrote: I am a retired person, non university type, with a keen interest in science, particularly in the physics of the cosmos from quarks to the edge of the universe. I believe that physics is not just for scientists but can be understood and appreciated by interested non academics. My question is to relate a of a person's feet to the a of their head while approaching a black hole to see how it would stretch the person. Thank you for your help. Don Watts
Dear Don, We are happy you had a good experience with our answers. Here I will try to answer your questions.
As a first step, I have to distinguish two different cases. Your first question about the Earth created field falls into the category of weak fields. I will start with the introduction of the gravitational field strength H. It is the gravitational force, that acts in a gravitational field on a unit mass. So, if someone tells you, "Here, at this point I measure a force of 98 Newtons exerted on a 10 kilogram body", you immediately know that H=98/10=9.8 m/s^2. By comparing Newton's law, F=ma, and from what I said above, you figure that H is nothing else, than the requested acceleration of bodies placed in gravitational fields. ( So from now, the gravitational acceleration and the gravitational field strength will be treated as the same things.) In Newtonian physics, If we have a known mass density distribution "o(r)" of the gravitating object, the gravitational field strength H can be calculated by solving the Poisson equation "Laplace operator acting on H is proportional to the mass density distribution o(r) " In general, for an arbitrary distribution of matter, this might be a very difficult task. ( After all, it is a system of partial differential equations). However, in the case, when the matter distribution is spherically symmetric, the solution is simple. It reads H = G*M/r^2, where M is the mass of the object that creates the field, r is the distance from the center of the source, and G is the Newtonian gravitational constant G=6.67*10^11 N*m^2/kg^2. This formula above also gives you the recipe how to calculate the gravitational acceleration ( field strength) at any distance from the Earth. For example: If you plug into this formula for M the mass of the Earth (M=6*10^13 kg), and for the r the radius of the Earth (R= 6380 Km), you will end up with H=9.8 m/s^2. If you plug in the mass of the Earth and the distance you chose ( 0.5 AU), you end up essentially with 0 acceleration. ( In such a far distance, such a light object as our Earth essentially does not create any gravitational field. 2) Your second question about falling to a black hole needs to be answered within the framework of Einstein's general theory of relativity. The reason for that is that in the case of very heavy objects, or in the case of very strong gravitational fields, the Poisson equation for calculation of the gravitational field strength H is NOT VALID anymore. Instead, Einstein's equations must be used. ( I will not show them here, but I assure you, they are also extremely complicated to solve for general mass distributions. ) However, for spherically symmetric sources it is doable. Let us consider a physicist falling into a singularity ( black hole, or collapsing star which will eventually end as black hole). When he is far from the singularity, his head and feet fell almost the same gravitational force, therefore these two parts of his body will accelerate with the same acceleration. However, when he gets closer to the singularity, the gradient of the gravitational field becomes very big, and the feet will be accelerated with a much higher acceleration then his head, and this will have a disastrous impact on him. By using the results from the literature, we examen the acceleration of a little element dm of the physicist's body. Suppose this dm is located in a distance h from the body's center of mass. Then dm's acceleration with respect to the center of mass of the body is given by a=2*M/r^3*h where M is the mass of the gravitational field source, and r is the distance between the source center and the physicists. This above formula gives you the acceleration of the dm with respect to his center of mass. In order to prevent different parts of his body ( for example his head and feet) to drift away from each of others, he has to exert a force F=1/4*m*M*L/r^3 to keep them together. (I assumed his mass is m and the distance between his head and feet is L). If his crosssectional area is A, the above force translates into a pressure p=1/4*m*M*L/(A*r^3) in the longitudinal direction ( updown). By plugging in some typical numbers into this equation, such as m=80 kg, L=1.8 m, A=20cmx20cm=400cm^2, critical pressure that a human body can stand p=100 atmospheres, M = mass of the sun, we end up with the distance r=200 km from the center of the gravitational field source where the physicists body cannot withstand the pressure( tension) anymore. ( Note, in this case, the Schwarchild radius is R_Sch=3km. The R_Sch is the event horizon for the black hole.) For more details please look into some general relativity books. I used Ch. W Misner, K. Thorne, J.A. Wheeler, Gravitation, 1973. I hope my answer helps you to dive even deeper into this beautiful area of physics which studies exotic objects and exotic behaviors. Who knows, maybe what seems exotic today, will be considered perfectly normal tomorrow. Bye Arnold Pompos 
last modified 5/7/1999 physicsquestions@fnal.gov 
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