Gravitational fields of two bodies of identical mass but different densities

Question

At a distance, is there any difference between the gravitational fields of two bodies of identical mass but different densities?

As an extreme example, take a planet orbiting at a distance of say 20 AU from a star of say 50 solar masses. Such a star undergoing core collapse can become a black hole without any supernova explosion. What would happen to the planet in orbit around this star? There would certainly be an impact – “day” would become permanent “night”, as the light and warmth of the star would disappear. But the mass that the planet was gravitationally bound to is still there in the same location, the only difference is that it’s now compressed into a single point. Does this change in density make any difference to the gravitational field?

NB: I wanted to use the extreme example of direct core collapse to a black hole, but I realise "density" isn't the right word when referring to a singularity, so if that affects the potential answer, also consider the comparison between two solar systems each with a planet of identical mass orbiting at identical distance around a star of identical mass, but one star is main sequence and the other is a neutron star.

Answer 1

If you assume a spherically symmetric star, then the gravitational field outside it is the same, regardless of the distribution/density of the mass inside it. It doesn't even need to be a constant density.

It is easier to see this with the gravitational analog of Gauss's law for electrostatics:

$$ \int_S \vec g \cdot d\vec S=4\pi Gm $$

Here, $\vec g$ is the gravitational field strength (or gravitational acceleration) at a point $\vec r$, and $d\vec S$ is a surface element of the closed surface $S$ where we are integrating. $m$ is the total mass in the volume enclosed by $S$.

Assuming $S$ is a sphere of radius $r$ that shares its center with the star, we can see that $\vec g$ must be constant over it and point radially inwards (where else would it point, if everything is radially symmetric? And it must point inwards because gravity is always attractive). The element $d\vec S$ points radially outwards. Therefore,

$$ \int_S \vec g \cdot d\vec S = \int _S -g~dS= -g \int_S dS = -g~4\pi r^2$$

Now if $S$ completely encloses the star, we equate this with $4\pi GM$, $M$ being the mass of the star. This yields $$g=-GM/r^2.$$ As you see, this is independent of the distribution of mass inside the star, as long as it is spherically symmetric.

Answer 2

@sagardipak is absolutely right if they are spherical, the radius does not matter as long as you are outside.

What can make a difference is that if one is rotating and the other not, or one is spherical and the other has an Equatorial bulge, and thus a quadrupole moment. Their total masses can be the same, but their densities vary in second case, and most likely the first case. The gravitational fields will be different, usually slightly. It is that way outside the Earth since the equator has a larger radius than the polar distance.

But also note, a dipole makes no difference in the static potential (if done wrt center of mass) and cannot cause any gravitational wave (because of conservation of momentum). For that we'd need a changing quadrupole moment. Rotating Kerr black holes which are axisymmetric have no changing quadrupole moment. The quadrupole moments in black hole or neutron star binaries is just the 'dumbbell' type rotation of one star around the other (and viceversa). The Kerr metric in a rotating black hole is definitely a different gravitational field that when there is no rotation, for objects of the same mass.