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?
Best Answer
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 $m_E$.