Shouldn't black holes exert the same gravitational force as an object of similar mass but lower density?

If the earth got shrunk into the size of a peanut, it would turn into a black hole, which would have a higher density but same mass. Since the center of mass of both bodies would be the same, the distance between a far-away object and the centers of mass would be the same. Since both the variables (mass, distance) would be the same, wouldn't the gravitational force exerted by both the earth and the black hole on a far away object be the same?

If this is true, wouldn't light be unable to escape the earth as well, since light can't escape black holes?

• Who says it isn't? Sep 5 '21 at 16:30
• The key feature your forgetting about is the event horizon, anything outside of the event horizon and having a travectory pointing away from the black hole could escape if it has enough velocity. Sep 5 '21 at 18:36
• FWIW, the Schwarzschild radius of the Earth is ~8.870 mm. Sep 5 '21 at 22:18
• Black holes don't have magical super-powered gravity. What they have is a tiny size that lets you get close to the gravity. Sep 6 '21 at 4:33
• The rule that a spherical mass's gravitational field is equivalent to the one created by a point of the same mass does not hold inside the sphere. Sep 6 '21 at 8:44

The parameter you're not considering is the distance.

The Earth is an object with the mass of the Earth $$m_E$$ and the radius of the Earth $$r_E$$ (duh). If you take a black hole with mass $$m_E$$, then its radius will be the radius of a peanut, $$r_p$$.

When shooting a light ray on Earth, the light easily escapes its gravitational pull because it is shot at a distance $$r_E$$ from the center of mass of the object. When shooting a light ray near our black hole, it can't escape its gravitational pull and falls into it, because it is shot from a much shorter distance, $$r_p$$. If you shoot a light ray at a distance of $$r_E$$ from the black hole, it will behave the same way as it does on the Earth. This is the punchline: light can escape black holes from a great enough distance.
Then does this mean that if we go near the center of the Earth and shoot a light ray when we're at a distance of about $$r_p$$ the light will be pulled in the center of the Earth? Of course not, because the mass "inside" a radius $$r_p$$ of the Earth's center is much much smaller than $$m_E$$.

• In fact, if the Earth's mass was evenly distributed, only the mass inside the peanut radius would count at a peanut's distance from the centre of the Earth. Sep 6 '21 at 0:53
• Fun fact, the majority of Earth's mass is in its metallic core. So much in fact that the force of gravity actually increases slightly as you go down towards the core. Only when you hit the actual core does the force start to decrease proportionally to the distance. Sep 6 '21 at 18:22
• @JohnDvorak That is fun! Trying it now... Sep 6 '21 at 23:30

It is the case that for large distances the gravitational field of Earth and its black hole equivalent are the same. The same with any other object. For example, if you replace the sun with a black hole with the mass of the sun, almost every planet in the solar system will continue to move in the same orbit (but of course temperature will drop significantly in each planet).

This situations can already be explained at the level of classical physics. It is proven by Newton's shell theorem that the gravitational field of a point particle of mass $$m$$ and a homogenous sphere of mass $$m$$ have the same gravitational field at distances larger than the radius of the sphere.

As for why light can escape Earth and not the black hole with the same mass, it is related to the force being stronger as the two objects under consideration (light and Earth/BH) are closer.

For a tiny particle trying to escape the surface of Earth, the gravitational field is already reduced by 1/$$R_\oplus$$ where $$R_\oplus$$ is the radius of Earth. Starting inside Earth does not help, as by the same shell theorem we just consider the gravitational field given by the mass below, and as you go deeper, there is less mass and a weaker gravitational field (thus the particle may still escape). For example, there is no gravitational field inside a hollow sphere with the mass of the Earth.

For a tiny particle trying to escape a planet with the same mass as Earth but with the radius of peanut $$r\approx 5\;\mathrm{cm}$$, feels a gravitational field that is $$R_\oplus/r = 10^8$$ stronger and may not be able to escape.

The calculations are slightly more complicated if you consider general relativity (GR), photons and black holes, but the same conclusions apply. For a generalization of the shell theorem in GR see Birkhoff's theorem.

• Please don't mix Newtonian physics and black holes - it's always a bad idea. Sep 5 '21 at 18:21
• @StephenG This question does not need any knowledge of GR to be answered. Sep 5 '21 at 18:22
• The problem is that you are (indirectly) encouraging people to apply Newtonian physics to black holes. We get questions that arise simply because of the confusion this causes. There is, BTW, a similar idea to the shell theorem in GR Sep 5 '21 at 18:28
• @StephenG I have softened the conclusion per your suggestion. Sep 5 '21 at 18:31
• almost every planet”? Sep 7 '21 at 1:55