# Human between two bodies of large mass

What if a person is put between two identical dense spherical bodies (well, not as dense as black holes, but definitely 100 times more than earth) with mass $M_1=M_2=10^{30}kg$, exactly in the middle between them ( vertical axis parallel with "human height vector" and the line connecting two spheres' centres passes exactly trough person's centre of mass). Let's say $M_1$ and $M_2$ are mounted onto some very strong poles and they are not approaching each other, but are stationary at $1$ meter.

I am aware that the forces would cancel out at exactly the middle, but human body is not symmetrical and it's mass is not equally distributed across it's volume. Would the human in question be torn apart or would it just deform? What would happen if the human body was completely symmetrical and with equal mass distribution, and could withstand such forces without being deformed? Would the person maybe spin?

I am not sure if this will affect the overall outcome, but let's say the mass of the person in question is $m=80~kg$.

• the short answer is "tidal forces would rip you apart". it's that simple. note that (say) merely the difference in a human's height would rip the object apart. Nov 8, 2014 at 15:00

If you are 1m away from something as massive as $10^{30}$ kg, then you are dealing with a black hole. The event horizon of such an object would be 1.5km. You certainly couldn't get two to within a few metres of each other without them having merged.

So your question is just too hypothetical to be worth answering. It is well known however (look up spaghettification, or spelling variants) that a human would be torn apart by tidal forces as they approached such a black hole. The presence of another body pulling in the opposite direction would only enhance the effect.

• Okay, what if we alter the data and say distance is 10 kilometers?
– user16688
Nov 8, 2014 at 9:49
• And maybe it's not a non- deformable, symmetric human? I can't understand the purpose of your question. If everything is perfectly symmetric then the "human" will be in an (unstable) equilibrium. Nov 8, 2014 at 9:56
• Well, the first part of the question relates to a regular human, while the second part relates to a symmetrical superhuman. The question just popped up in my mind and I wanted to know. Maybe I'm not the only one and maybe bunch of people wants to know.. And my apologies again. By completely symmetrical I meant only "vertical symmetry i.e we would get two identical shapes only if we cut the human with a plane passing between their eyes and kidneys.( internal organ symmetry is required)
– user16688
Nov 8, 2014 at 10:11

As the other answer states, this is not a feasible situation.

Considering Newtonian gravity and fields, (i.e. masses like the sun, and distances like the planetary system) the boundary condition "center" is not enough. One has to define how the person and the gravitational sources entered, or, alternatively state that at the center of mass of the system the person has zero momentum.

This is because this is the case of unstable equilibrium , similar to unstable equilibrium copied from the link

The slightest perturbation, will send the symmetric human either on one or the or the other massive bodies, and also induce a spin if not applied to the center of mass of the symmetric human and perpendicular to the symmetry axis of the human.