Big Earth

Relevant wiki: Gravitation

The mass of the planet is proportional to $m \propto R^3$ .

The gravitational field at its surface is proportional to $g \propto \frac{m}{R^2} \propto \frac{R^3}{R^2} = R.$

The atmospheric pressure is $P = \rho_{avg} g h$ . Thus $h = \frac{P}{\rho_{avg} g} \propto \frac 1 g \propto \frac 1 R.$

Thus, doubling the radius of the planet must be matched by halving the height of the atmosphere.

More accurately, we could use the barometric formula $p_h \; = \; p_0 e^{-\frac{mgh}{kT}}$ for the atmospheric pressure $p_0$ at height $h$ , where $p_0$ is the sea-level pressure, $m$ is the mass of an atom of air, $g$ is the gravitational acceleration, $k$ is Bolztmann's constant and $T$ is the temperature of the atmosphere.

This model still assumes that gravitational acceleration is constant throughout the atmosphere, (so that the effective height of the atmosphere is small relative to the radius of the planet), but allows for the fact that the atmosphere does not stop dead at some point. We would need to adopt a definition of what level pressure corresponds to still being in the atmosphere (for example, $p_h = \tfrac{1}{1000}p_0$ at the top of the stratosphere). Let us say, for definiteness, then, that the height of the atmosphere is that height $h$ for which $p_h = \tfrac{1}{1000}p_0$ .

Whatever definition we adopt, the height required to obtain a particular pressure ratio (like $\tfrac{1}{1000}$ ) is inversely proportional to $g$ (assuming that $T$ and $m$ are the same for both planets). Since $g$ on BigEarth is twice that on Earth, half the height of atmosphere is required on BigEarth to achieve the same sea-level pressure as on Earth.

I think this one is quite simple, the radius of the earth was doubled and the pressure stayed at 1atm so something must have been halved (like using opposite operations to solve an equation). Since it says that composition, density of the earth and mass of the earth stay the same, that means it has to be the height that halved.

It seems like if one direction doubles, the mass would go up 8 times 4/3(pi)r^3. Changing r from 1 to 2 would make volume go up by 8 times, making gravity 8 times more, not twice as strong. It seems like the atmosphere would be 1/8 as thick to stay at 1 atmospheres. Where am I wrong in my thinking?

Everything is the same except size. Bigger planet means more mass stronger gravity and lower atmospheric height.