Is there a peak gravitational force between bodies?

Question

Suppose Object A is exerting gravitational force on Object B. Object A increases in mass, and so increases in volume, increasing the gravitational force on Object B. But, since mass occupies space the objects will need to be moved apart in order to keep them from joining. But as Object B is moved away from Object A.

Eventually Object A turns into a black hole, which further complicates the issue. Due to the event horizon, so it's radius is a definite minimum distance between the objects.

Is there some sweet spot of maximum gravity?

Answer 1

Object A increases in mass, and so increases in volume

I'm going to make the assumption that we are adding mass to A by providing more material of the same density $\rho_{A}$, rather than exchanging the current material with denser material or adding varying densities. I don't have to make that assumption, but it seems like it's what you're going for and we'll see where it leads. I'll also let A and B be spherical, to make them planet-like and also stable, so we don't have to worry about their shape changing. (assume a spherical cow, amirite?)

Once A is large enough to touch B, the distance between their centers is $r_{A}+r_{B}$. The mass of A is $\rho_{A}(\frac{4}{3} \pi r_{A}^{3})$. So the gravitational force between them is:

\begin{equation} F=\frac{G \rho_{A} (\frac{4}{3} \pi r_{A}^{3})m_{B}}{(r_{a}+r_{b})^{2}} \end{equation}

It's clear that the top will grow more quickly than the bottom (everything is constant except for the radii). After a certain point, a cube grows faster than a square. Thus, the force will continue to increase, almost linearly. But will A turn into a black hole?

The critical point at which A turns into a black hole is when the following equation is satisfied:

\begin{equation} \frac{m_{A}}{r_{A}}=\frac{c^{2}}{2G} \end{equation}

c is the speed of light, G is the gravitational constant as above.

We also know:

\begin{equation} \frac{m_{A}}{r_{A}}=\rho_{A}\frac{4\pi r^{2}}{3} \end{equation}

(This comes from $V_{A}=\frac{4}{3}\pi r_{A}^{3}$)

Thus, the black hole happens when:

\begin{equation} r_{A}=\sqrt{\frac{3c^{2}}{8\pi \rho G}}=\frac{1.27*10^{13}}{\sqrt{\rho_{A}}} \end{equation}

So the gravity on B increases almost linearly until a black hole is created. The gravity then only keeps increasing (of course, B would be a part of A at this point too). If we relaxed our assumption of constant density of A, the gravitational force would only increase. There is no sweet spot.

Answer 2

Yes there is a maximum gravittational field, although of course the gravitationational force will be unbounded because there is no upper limit to the force you can putinto that hravitational field.

The gravitational field outside a black hole has an upper limit: First notice that the surface gravity of a black hole is actually larger for less massive, smaller, black holes. A classical calculation gives (M is the mass of the black hole):

$F_g=-\frac{GmM}{r^2}$

If we use the Schwarzschild radius $r_s=2GM/c^2$ and replace above we get:

$g=F_g/m=-\frac{c^4}{4GM}$

so the smaller the black hole the larger the surface gravity. A more exact general relativity calculation also gives a dependence with the inverse of M for the surface gravity $\kappa$:

$\kappa=-\frac{1}{4M}$

Thus, maximum gravity field will be given by microblackholes that cannot be smallerthan the plank scale, I did not calculate the numbers, but that shoudl be the upper limit reachable outside a black hole.

Answer 3

There is no maximum gravity.

Assuming constant density, mass grows as $r^3$, while gravity attenuates as $r^{-2}$. Therefore as long as density is constant, the force of gravity between two touching spheres will grow like their radii (or the cube root of their masses), meaning there is no sweet spot. Of course, real matter does compress under enough gravitational force. This would mean that the spheres would get more dense, and gravitational force would start growing faster than $r$.

Answer 4

Object A increases in mass, and so increases in volume.

That's not necessarily true. Some objects such as white dwarfs and neutron stars decrease in volume as mass increases.

Is there some sweet spot of maximum gravity?

The answer depends very much on what kind of object the mass is being added to, on the kind of mass being added to the object, and the nature in which the mass is being added.

Looking at the problem generically, suppose that in the neighborhood of some mass $M_0$ and radius $R_0$, the relationship between size and mass is given by $$R = R_0 \left( \frac M {M_0} \right)^\alpha$$ Assuming the object has not yet entered a state where relativistic calculations are needed, the gravitational acceleration at the surface of the object can be calculated with Newton's law of gravitation, $g=GM/R^2$. Using the above mass-radius relationship, this becomes $$g=g_0 \left( \frac M {M_0} \right)^{1-2\alpha}$$ where $g_0 = GM_0/R_0^2$. A local sweet spot occurs where $dg/dM$ switches from positive to negative. This occurs at $1-2\alpha=0$ or $\alpha = 1/2$. Adding mass will increase surface gravitation if $\alpha < 1/2$, decrease it if $\alpha > 1/2$.

This exponential coefficient $\alpha$ varies considerably depending on the object at hand. The table below shows the coefficient for different types of objects.

$$\begin{matrix} \text{Object} & \alpha & \text{Transition} \\ \text{---------------------------} & \text{------} & \text{-------------------------------------} \\ \text{Large gas giant} & ~0 & \text{Brown dwarf} \\ \text{Brown dwarf} & ~0 & \text{Stellar ignition} \\ \text{Red dwarf} & 0.9 & \text{Non-convective core} \\ \text{Sun-like star} & 1/13 & \text{CNO cycle takes over} \\ \text{Intermediate mass star} & 15/19 & \text{Radiation pressure dominates} \\ \text{Large star} & 1/2 & \text{Eddington limit} \\ \text{---------------------------} & \text{------} & \text{-------------------------------------} \\ \text{Small white dwarf} & -1/3 & \text{Fermi gas becomes relativistic} \\ \text{Large white dwarf} & < -1/3 & \text{Explosion or neutron star} \\ \text{Small neutron star} & -1/3 & \text{Fermi gas becomes relativistic} \\ \text{Large neutron star} & < -1/3 & \text{Explosion, exotic object, or black hole} \\ \text{Black hole} & 1\,\text{(see note)} & \\ \end{matrix} $$ Note: The surface gravity of a black hole is not well-defined. I'm using the Schwarzschild solution for the surface gravity of a black hole whose event horizon is a Killing horizon.

From the first half of the table, there are a couple of "sweet spots" that result from adding hydrogen. The first occurs at the transition from a brown dwarf to a red dwarf, about 1/10 of a solar mass. The second occurs at the transition from the pp chain to the CNO cycle, which occurs at about 2 solar masses. The first sweet spot is much sweeter than the second. The largest mass brown dwarf has a significantly greater surface gravity than does the largest Sun-like star.

There are another couple of sweet spots from the second half of the table, the Chandrasekhar limit of about 1.4 solar masses for a white dwarf, and the corresponding limit of 2 to 3 solar masses for a neutron star. If you can somehow coerce the object to move past those limits without having the object explode, do so. A neutron star just above the Chandrasekhar limit is much smaller in diameter than is a white dwarf just under that limit, and a 2 to 3 solar mass black hole is in turn much smaller in diameter than is a neutron star.

Once you've reached a black hole, you need to stop adding mass. The Schwarzschild radius of a black hole is proportional to mass. Adding mass decreases gravitational acceleration at the Killing horizon.