A Hollow Planet
We all know that the mass of the Earth is what keeps us grounded, through the force of gravity. Anyone on or outside the big blue planet gets pulled down to it. But here’s a wild question you’ve probably never considered: How would gravity behave if you were inside a heavy object? The answer is very surprising, and quite beautiful.
Imagine you took the Earth and somehow scooped out the inside. Or, since we don’t want to damage our planet (for I suppose removing its core would be quite detrimental…), let’s imagine something else.
Say there was a planet out there somewhere, and rather than being a solid sphere, it was actually a spherical shell (essentially a hollowed out planet). To visualize this, I’m going to show only half the planet to stress that the inside is empty. Keep in mind that there is a top half not shown, and that the whole planet is perfectly spherical.
If an astronaut were floating around in space somewhere near this planet, they would be pulled toward it by its gravity. The fact that it is hollowed out won’t change this.
Now here’s where it gets interesting. What if the astronaut weren’t outside the planet, but rather inside the planet? What would gravity do if they found themselves surrounded by the planet’s mass?
The simplest case is one in which the astronaut is at the exact center of the sphere:
A little thought suggests that they wouldn’t feel any gravitational pull and would therefore float. How come?
Given that they’re at the exact center of the sphere, and given that the planet’s mass is equally distributed all around them, the gravitational pull would be equal in all directions. So, whatever mass is pulling them to the left would have a counteracting mass pulling them to the right. All forces cancel out, and the effects of gravity disappear. (Picture a game of tug of war in which equally strong participants are pulling on the astronaut from all directions; the astronaut wouldn’t move.)
Now we’ve gotten to the real meat of the question. Here’s what we really want to know. What happens to the astronaut (in terms of gravity) if they are inside the sphere, but not at the exact center? In which direction would they fall?
Take a moment at think about it.
I’ll put in some filler text…
…before I tell you the answer…
…just in case you want to try and solve this yourself…
You might think that the astronaut gets pulled toward the nearest side. That’s certainly what I thought at first. After all, if they’re positioned toward the left side (as in the picture above), they’re closer to the mass on the left than the right.
Right?
That’s why we’re held to the gravity of the Earth and not that of heavier Jupiter. We’re closer to the Earth.
Except that, in this case, that isn’t true. The astronaut wouldn’t be pulled to the left. They wouldn’t be pulled in any direction. No matter where inside the sphere the astronaut is located, they would be weightless. The effects of gravity would be perfectly canceled out. And they wouldn’t need to be at the exact center for that to happen.
The Shell Theorem
This concept, called the Shell Theorem, was discovered by Isaac Newton nearly four-hundred years ago.
Here’s a succinct summary of the Shell Theorem: The gravitational forces inside the shell of a perfect sphere all cancel out, meaning there is no gravitational pull anywhere inside the sphere.
Now let’s look at why this happens. Without getting into the weeds of the math too much, it is important to note one thing you may remember from high school physics: the force of gravity between two objects is proportional to the inverse square of their distance (1/r2 , where r is the distance between them). Put differently, if you’re one foot away from something, the pull from that something would be four times stronger than if you were two feet away. (Doubling the distance divides the force by 22 ).
Now, back to the astronaut. Recall that in the case where they were perfectly centered, the left and right forces canceled out. My claim (or rather, Newton’s claim) is that the same is true even if they’re not at the center. And here’s a picture that shows why:
That little green circle on the left represents an area of mass pulling to the left. If we took the same angle (shown in red) on the right side, the circle drawn out would be larger. So, even though the right side is farther away from the astronaut, there is more mass contributing to the right-side pull.
How much more?
Some basic math (which I won’t get into here) shows that it’s exactly proportional to the square of the distance. Therefore, even though gravity is 1/r2 weaker on the right, there is exactly r2 more stuff on the right exerting that gravity. Those r2 s cancel each other out all around the sphere, and every pull in every direction is precisely negated by a pull on the opposite side.
A Perfect Universe
Don’t worry if you got lost in the proof. The takeaway is that this unintuitive fact didn’t need to be true. The universe could have made the math much harder than high school trigonometry.
But, our universe is made up of mathematical laws that are much more elegant and understandable than we, frankly, deserve. Einstein once said, “The most incomprehensible thing about the universe is that it is comprehensible.” I couldn’t agree more.
If gravity fell at a rate of 1/r, or 1/r3 , or 1/r2.1 , this wouldn’t work. If we didn’t live in three dimensions of space, this wouldn’t work. But the universe did give us a force of gravity that falls off at 1/r2 , and it did put us in 3D. And the consequence is that anywhere inside of a perfectly spherical mass, there is no net gravitational force at all.
Several months ago, I wrote an article in which I argued that there exist aspects of the universe which are imperfect, like music, and which necessitate a compromise on the part of humans. But seeing proofs of things like the Shell Theorem makes me second guess myself… If things do seem imperfect, it’s much more likely because we humans haven’t yet found a way to understand them.