# Does the interior surface of a Dyson Sphere exert gravity?

The interior of this particular Dyson Sphere is so large that it encapsulates a star at its center. Otherwise it is filled with breathable air which creatures can breath and fly in. I imagine that for the most part it contains a zero gravity environment, except for the areas closest to the star and to the interior surface of the sphere itself. How would I calculate these?

Edit: Thank you for all the answers and insights.

• This has been asked a number of times on this site. I'll find a duplicate in a moment, but the answer is no. As you descend through a sphere, the only gravity you feel is the gravity below you (toward the center). Everything above you is balanced out by the mass "across the way" from you. Therefore, inside a Dyson Sphere there is no gravity due to mass (now... if you can get that sphere spinning...) – JBH Jan 9 '18 at 23:57
• Well... this isn't an exact duplicate but it goes into a ton of detail about the nature of dyson spheres. – JBH Jan 10 '18 at 0:02
• Nope. Gravity within a uniform hollow sphere is always zero. Check out this website for the more mathy version: grc.nasa.gov/www/k-12/Numbers/Math/Mathematical_Thinking/… – Dubukay Jan 10 '18 at 0:07
• The relevant science is The Shell Theorem and inside a symmetric shell, gravity is zero everywhere. More precisely the net force from the shell is zero everywhere inside the shell. – StephenG Jan 10 '18 at 0:19
• I also have doubts about air. I think it would not be physically possible. – Oleg Lobachev Jan 10 '18 at 7:30

A Dyson sphere is a hypothetical megastructure that completely encompasses a star and captures most or all of its power output

I had to look up what a Dyson sphere was, but using this definition from Wikipedia, I'm assuming it's a spherical shell. If you can assume it is a uniform spherical shell, i.e. it has constant density throughout, then the sphere will exert no gravitational force on any object inside it. This can be easily proved using Gauss' law applied to gravity

$$\nabla \cdot g = -4 \pi G\rho$$

For thorough derivations, search 'gravity shell theorem'. This result was derived by Newton and can be easily obtained with only geometry, Newton's law of gravity and patience.

Things get trickier if you have a non-uniform density or do something else to break the symmetry of the problem, but that gets complicated quickly. If you need a gravitational-like force sticking people to the shell, there are much better solutions out there.

• Newton's law of gravity and patience. What goes up... must come down... but only if you wait long enough. – WernerCD Jan 10 '18 at 23:14

Using an equation derived from the Law of Universal Gravitation, you can calculate the force of gravity a distance $x$ away from the center of mass. The equation is here:

$$g = \frac{Gm}{x^2}$$

where $G$ is the gravitational constant, $m$ is the mass of the body, $x$ is the distance from the center of mass (for force at the surface of body, this is just the body's radius), and $g$ is the resultant acceleration due to gravity.

This equation is directly applicable to the question of the star's gravitational effects. Just based on some napkin math, I think you'll find that any point where the radiation from the star is non-lethal will also have negligible gravity from it.

However, this equation isn't very helpful for the Dyson sphere calculations. Sure, you could assume that the sphere is a point mass and do the same calculations, but a) this would be inaccurate as the sphere is hollow, and b) would imply that people on the inside of the sphere would be hanging on 'upside down', which doesn't seem right. So let's take a better look at it.

This research site has a good argument about this. Basically, it works out that for a uniform hollow sphere, all points within the sphere have an effective gravity of zero. This makes sense if you think about it: by being inside the sphere, you are automatically subjected to the net gravitational forces of all the surrounding sphere walls which must, intuitively, sum to zero. The differential equation on the link is the mathematical way of proving that intuition.

In summary

Your inhabitants would experience no gravity due to the sphere itself (while inside it, that is) no matter how massive the sphere's radius, thick the outer wall, or dense the unobtainium it is built out of.

The star, too, would exert little gravity on the inhabitants as (presumably) the inner surface of the sphere is very far away from the star in order to keep radiation at survivable levels. That said, if the radius of the sphere were to be very close to the radius of the star, then yes presumably there would be some gravity due to the star's mass, pulling towards the construction's center of gravity (the center of the star).

Side note...

One thing you could look into is spinning the sphere about a central axis in order to simulate gravity, much like hypothetical space ships/stations could. The acceleration of spinning would have a tendency to "push" people away from the center, resulting in the appearance of gravity on the inner surface.

• There is definitely gravity from the star and people on the sphere would need to be at least in an orbit (i.e. free-fall) assuming they want to stay in their nice habitable zone. In practice, the sphere should be rotating a bit faster than this orbit so that the rotating reference frame gives you the experience of a 'downward force' into the sphere wall. This only works in a narrow ring around the center for a true sphere, with one axis of rotation. It gets more complex if you have multiple rings that rotate independently. – Dan Bryant Jan 10 '18 at 3:12

The previous answers seem to assume a hollow sphere - but this is not the case here:

it encapsulates a star at its center. Otherwise it is filled with breathable air which creatures can breath and fly in.

So what we have here is a gas (and one star) filled balloon with a (presumably metal) outer skin. Leaving aside the fact that a star in an atmosphere would probably behave WILDLY different from one in a vacuum, this means we're looking at a body that has mass everywhere in it (although at different densities), leading to an overall gravity that points to the center... which means, anything inside your dyson sphere would be drawn towards said center, and thus the star.

If you want to avoid all your nice atmosphere and creatures falling to a fiery death (although it might be a long enough fall to starve on the way, not sure there^^) until you DO end up with a hollow shell (except for the star), I'd recommend rethinking where you want to place that atmosphere. If you handwave some force fields that limit the atmosphere to (some number of kilometers/miles/whatnot), you will have a hollow (except for the star, again) shell and can go with any of the other answers here :)

• +1. Not only the creatures etc. inside the sphere would fall into the star, but so would all the air, turning it into a much bigger, much hotter star than it was originally, surrounded by a near vaccum. – Nathaniel Jan 10 '18 at 9:37
• @Nathaniel Well, normal stars spew out huge amounts of material. If you kept them inside a closed container, it might actually give out more atmosphere that it takes in. It might make for some very interesting behaviour overall, even if you just think about the inevitable "reflection" of massive amounts of radiation and matter back into the star... – Luaan Jan 10 '18 at 12:39
• @Luaan that may be true - I'm not sure of the mass balance calculations but you might be right. Also: when it comes down to it, the only reason the Earth doesn't heat up to the temperature of the Sun is that it can radiate heat away into space. The inside of the sphere can't do that, so it will heat up to 6000 celsius or so. (Assuming a Sun-like star.) If that doesn't melt the sphere it will heat up even more because the star itself will not be able to cool down if it can't radiate into space without its radiation being reflected back. – Nathaniel Jan 10 '18 at 13:12
• The star is not in an atmosphere. Imagine the air pressure of a 1 AU column of air! – rek Jan 10 '18 at 17:26
• Incidentally, a 1AU sphere of "breathable air", assuming a uniform air pressure of 0.4 bar, has over 1000 solar masses. The physics of thousand solar mass spheres of mostly nitrogen is not well studied, so whether the resulting ball of gas is hotter or cooler than the star that was there before isn't obvious. – Yurgen Jan 10 '18 at 18:01

Other answers address the question of gravity, so I'll just expand on the atmosphere topic.

### The Problem

The biggest physics issue with your proposed world is the atmosphere.

A star is formed when enough gas is present that the gravity from all of the gas is enough to collapse it down into a dense hot sphere. It would thus collapse any atmosphere around it into itself. First you may think, well maybe the star only collapses the heavy elements and the light ones could still form an atmosphere. However, most of a star is hydrogen, the lightest element. So this is not true.

The density and pressure of a gas grows exponentially towards a the source of gravity, so it is impossible to have an extremely thick atmosphere without reaching pressures where things form plasmas. The exponential growth is because each layer of gas must have a pressure that can support all of the weight above it.

Even if the star initially didn't have enough gravity by itself to suck down a giant atmosphere, even a low density atmosphere would have more mass and thus gravity than a typical star.

### The Solution

If you want to have a giant atmosphere, consider having it on the outside of a shell. So then you'd have a star, a large vacuum, a (possibly transparent (maybe even diamond)) inner shell, an atmosphere, and an optional outer shell.

With this arrangement your gravity in the atmosphere would be:

$$g=\frac{G\,(M_{star}+M_{shell}+ M_{atm})}{r^2}$$

Where $r$ is the distance to the center of the star, $g$ is your gravitational acceleration, $G$ is the universal constant of gravitation, and the $M_{star}$, $M_{shell}$, and $M_{atm}$ are the masses of the star, inner shell, and portion of the atmosphere closer to the star than $r$ respectively.

So now let's take a look at the equations for the atmosphere to see if we can come up with some numbers that will fit your criteria:

First the specific ideal gas law relating temperature $T$, density $\rho$, and pressure $P$:

$$\rho=\frac{P}{RT}$$

Where $R$ is the specific gas constant (for air = $286.9\frac{J}{kg\,K}$)

Let's say the temperature is constant to simplify our analysis, and keep our creatures comfortable.

The change in mass of atmosphere closer than $r$ as $r$ increases will just be the surface area of the sphere of size $r$ times the density at that $r$:

$$\frac{d\,M_{atm}}{dr}=4\,\pi\,r^2\,\rho=\frac{4\,\pi}{RT}\,r^2\,P$$

Then since the pressure of the atmosphere must support the weight of the gas above it, the rate of change is also related to the density:

$$\frac{dP}{dr}=\rho\,g=\frac{P}{RT}\frac{G\,(M_{star}+M_{shell}+ M_{atm})}{r^2}$$

To simplify things a little let's change our variables:

$$M=M_{star}+M_{shell}+ M_{atm}$$

$$\frac{dM}{dr}=\frac{dM_{atm}}{dr}$$

Now we almost have enough information to do a numerical integration; we just need our initial values, and our constants. So let's try:

$$T=25^\circ C$$

$$M_0=M_{sol}= 10^{30} kg$$

$$P_0=1 atm = 10^5 Pa$$

$$r_0 = 1 AU = 1.5\times 10^{11} m$$

Integrating numerically we can get a plot of pressure vs altitude: As you can see, the pressure drops off to less than three quarters of the initial pressure (enough to cause altitude sickness) by about 5000 km. Certainly a thicker breathable atmosphere than the measly 2.4 km that earth has, but let's see if we can do better.

By increasing our starting radius and decreasing the mass of our star and shell we can decrease the rate of pressure drop off, so let's look at a start with near the minimum mass to still be a red dwarf, about a tenth of our sun, and let's start out 100 times as far away: For this system it looks like the breathable atmosphere would extend out to 7000 km: not much of an improvement for the extremes it took to get there. At 100 AU out from the star, you'd probably need an outer shell to insulate and keep your atmosphere warm.

Other concerns are how you'd get an intershell with a curvature of 1 AU to withstand an atmosphere of pressure, even if it was 100 miles thick, it would need to withstand 100 GPa of stress (diamond breaks somewhere in the 70-300 GPa range) Of course 100 mile thick shell of diamond would have a mass of 160 solar masses, but of course with that much mass we'd need to support its own weight in addition to the atmosphere. Turns out we just can't do this, so maybe just hand wave it away? Weightless force field?