So, I created a planet that has a civilization that is beyond advanced.
The planet's radius is 10x the radius of Earth, and is has the same density as Earth. What issues would I have with the gravity?
So, I created a planet that has a civilization that is beyond advanced.
The planet's radius is 10x the radius of Earth, and is has the same density as Earth. What issues would I have with the gravity?
The gravity would be 10 times larger than Earth's gravity. Since mass grows by the cube and gravity falls by the square your planet would have 1000 Earth masses and surface gravity of 10g, assuming they have same density. And since your planet is very large it's uncompressed density must be much lower than Earth density.
I hope that you didn't plan to have humans there.
If you want realism take a look at The Trouble with Terrestrials and Mass-Radius Relationships for solid exoplanets:
In general you can't have more than 1.6 Earth radius if you want Earth like planet. Planets larger than 1.6 tend to keep their Hydrogen and become gas giants, sort of Mini-Neptunes. Though Universe is large and there might be exceptions
If you want more handwavium you can use Chthonian Planet that somehow got it's atmosphere stripped and after due to some collision moved back into habitable zone.
Whatever you decide to use 10g is way too much gravity.
Surface gravity $\hat{g}$ is a function of the mass $M$ and radius $r$ of the planet:
$$\hat{g} = \frac{G\cdot M}{r^2},$$
where $G$ is the universal gravitation constant $6.67\times10^{-11}\,\frac{\text{N}\cdot\text{m}^2}{\text{kg}^2}$. If you assume your planet is in hydrostatic equilibrium (a good assumption for any planet with noticeable surface gravity), then mass is in turn a function of radius and density $\rho$:
$$M = \rho\frac{4}{3}\pi r^3.$$
Put these together and you get:
$$\hat{g} = \frac{4}{3}\pi G\rho r.$$
Proof. The radius of earth is 6371 km; the density is 5515 kg/m$^3$.
$$ \hat{g}_{earth} = \frac{4}{3}\pi \left(6.67\times10^{-11}\,\frac{\text{N}\cdot\text{m}^2}{\text{kg}^2}\right) \left(5515 \frac{\text{kg}}{\text{m}^3}\right) \left(6371000 \text{m}\right) = 9.81 \frac{\text{m}}{\text{s}^2}.$$
Surface gravity scales linearly with radius and density. If you double your radius, but want surface gravity to stay the same, you must halve the density of your planet.
Example. If Earth had the density of the moon (3348 kg/m$^3$), then its density would be 0.607 of earth's, so its radius would have to change by 1/0.607 = 1.65 to give the same surface gravity. The new radius of earth would be 10500 km.
Size and Density Reference. From Wikipedia, here is a chart with both size and density for many solar system objects. You can read about what these objects are made of (iron, rocks, ice, hydrogen, etc) to find out what a reasonable density would be. Use a spreadsheet, plug in the equation above, and you can calculate surface gravity for all sorts of fantastical worlds.