Return to Answer

This formula assumes a constant gravitational acceleration over the whole height of the gas column - a reasonable assumption for Earth, as the atmosphere is thin compared to to the size of the planet.

For a simple argument, if you assumed composition is the same, it's clear that the altitude change needed for pressure to change by a certain factor is less on Jupiter than on Earth. Gravity is stronger, and the radius of Jupiter is larger. Both of these factors result in a smaller characteristic height for the atmosphere. That gives some indication that the above quote may be less correct than the reverse proposition. But there are many more complications.

Obviously, Jupiter has a lot more light gas in its atmosphere than Earth. This is partly because Earth's gravity isn't perfect at holding in Hydrogen against the radiation in space. So this pushes in the other direction - making Jupiter's atmosphere have a longer characteristic length.

But the characteristic length isn't the full story. Obviously Earth has a nice (agreed upon) ground level. Jupiter might have a phase change that vaguely resembles this, but even if there is an abrupt density change, it won't be anything like Earth. So this raises the question of how we should define the region of Jupiter's atmosphere in the first place. Could we accept internal metallic Hydrogen in the center as "atmosphere"? Probably not, but that's just the most extreme case.

Differential equations correctly describing the system can be found in other answers. For completeness, they are:

$$ \frac{d}{dr} P(r) = g(r) \rho(r) $$

But let's go into the special case solutions, because I think this is what the OP is asking for. In the case of constant gravity, ideal uniform gas, and constant temperature, the atmospheric pressure follows:

$$ P(h) = P(0)\, e^{ - \frac{ h }{ H } } $$

Where $H$ here is the characteristic length. For Earth, this is about 7.4 km. It depends on the temperature, so it's always in flux.

Now, if we add in the tidal component of gravity, we can obtain a new expression. This is assuming a $\frac{1}{r^2}$ form of gravity, instead of a constant. In this case, the solution to the differential equation yields the following pressure profile:

$$ P(r) = P(r_0)\, e^{ \frac{r_0}{H} \left( \frac{r_0}{r} - 1 \right) } $$

where $H$ is the same characteristic length. $r_0$ is the radius of the planet itself, and $r$ (radius) replaces $h$ (altitude). This equation is accurate for the specific case of:

1. constant temperature
2. constant composition (homogenous)
3. contribution to gravitational field by air itself can be neglected

1 and 2 are terrible assumptions, obviously. For all planets, there tends to be a local minimum of temperature in the atmosphere, and deeper into it, the temperature will be many many times higher, which gives a magnitude of just how bad the constant temperature assumption is.

The homogeneous assumption is also very bad, because the gases stratify subtly. It's not like oil-water separation, but the higher you go, the higher the concentration of the lighter elements will be. In fact, this is a major justification for why the gas giant atmospheres are such pure Hydrogen and Helium to begin with.

Nonetheless, this equation for pressure is still technically better than the constant gravity assumption. But that's not the most relevant on Jupiter. In fact, it's probably more relevant of a correction on Earth.

This formula assumes a constant gravitational acceleration over the whole height of the gas column - a reasonable assumption for Earth, as the atmosphere is thin compared to to the size of the planet.

For a simple argument, if you assumed composition is the same, it's clear that the altitude change needed for pressure to change by a certain factor is less on Jupiter than on Earth. Gravity is stronger, and the radius of Jupiter is larger. Both of these factors result in a smaller characteristic height for the atmosphere. That gives some indication that the above quote may be less correct than the reverse proposition. But there are many more complications.

Obviously, Jupiter has a lot more light gas in its atmosphere than Earth. This is partly because Earth's gravity isn't perfect at holding in Hydrogen against the radiation in space. So this pushes in the other direction - making Jupiter's atmosphere have a longer characteristic length.

But the characteristic length isn't the full story. Obviously Earth has a nice (agreed upon) ground level. Jupiter might have a phase change that vaguely resembles this, but even if there is an abrupt density change, it won't be anything like Earth. So this raises the question of how we should define the region of Jupiter's atmosphere in the first place. Could we accept internal metallic Hydrogen in the center as "atmosphere"? Probably not, but that's just the most extreme case.

Differential equations correctly describing the system can be found in other answers. For completeness, they are:

$$ \frac{d}{dr} P(r) = g(r) \rho(r) $$

But let's go into the special case solutions, because I think this is what the OP is asking for. In the case of constant gravity, ideal uniform gas, and constant temperature, the atmospheric pressure follows:

$$ P(h) = P(0)\, e^{ - \frac{ h }{ H } } $$

Where $H$ here is the characteristic length. For Earth, this is about 7.4 km. It depends on the temperature, so it's always in flux.

Now, if we add in the tidal component of gravity, we can obtain a new expression. This is assuming a $\frac{1}{r^2}$ form of gravity, instead of a constant. In this case, the solution to the differential equation yields the following pressure profile:

$$ P(r) = P(r_0)\, e^{ \frac{r_0}{H} \left( \frac{r_0}{r} - 1 \right) } $$

where $H$ is the same characteristic length. $r_0$ is the radius of the planet itself, and $r$ (radius) replaces $h$ (altitude). This equation is accurate for the specific case of:

1. constant temperature
2. constant composition (homogenous)
3. contribution to gravitational field by air itself can be neglected

1 and 2 are terrible assumptions, obviously. For all planets, there tends to be a local minimum of temperature in the atmosphere, and deeper into it, the temperature will be many many times higher, which gives a magnitude of just how bad the constant temperature assumption is.

The homogeneous assumption is also very bad, because the gases stratify subtly. It's not like oil-water separation, but the higher you go, the higher the concentration of the lighter elements will be. In fact, this is a major justification for why the gas giant atmospheres are such pure Hydrogen and Helium to begin with.

Nonetheless, this equation for pressure is still technically better than the constant gravity assumption. But that's not the most relevant on Jupiter. In fact, it's probably more relevant of a correction on Earth.

This formula assumes a constant gravitational acceleration over the whole height of the gas column - a reasonable assumption for Earth, as the atmosphere is thin compared to to the size of the planet.

For a simple argument, if you assumed composition is the same, it's clear that the altitude change needed for pressure to change by a certain factor is less on Jupiter than on Earth. Gravity is stronger, and the radius of Jupiter is larger. Both of these factors result in a smaller characteristic height for the atmosphere. That gives some indication that the above quote may be less correct than the reverse proposition. But there are many more complications.

Obviously, Jupiter has a lot more light gas in its atmosphere than Earth. This is partly because Earth's gravity isn't perfect at holding in Hydrogen against the radiation in space. So this pushes in the other direction - making Jupiter's atmosphere have a longer characteristic length.

But the characteristic length isn't the full story. Obviously Earth has a nice (agreed upon) ground level. Jupiter might have a phase change that vaguely resembles this, but even if there is an abrupt density change, it won't be anything like Earth. So this raises the question of how we should define the region of Jupiter's atmosphere in the first place. Could we accept internal metallic Hydrogen in the center as "atmosphere"? Probably not, but that's just the most extreme case.

Differential equations correctly describing the system can be found in other answers. For completeness, they are:

$$ \frac{d}{dr} P(r) = g(r) \rho(r) $$

But let's go into the special case solutions, because I think this is what the OP is asking for. In the case of constant gravity, ideal uniform gas, and constant temperature, the atmospheric pressure follows:

$$ P(h) = P(0)\, e^{ - \frac{ h }{ H } } $$

Where $H$ here is the characteristic length. For Earth, this is about 7.4 km. It depends on the temperature, so it's always in flux.

Now, if we add in the tidal component of gravity, we can obtain a new expression. This is assuming a $\frac{1}{r^2}$ form of gravity, instead of a constant. In this case, the solution to the differential equation yields the following pressure profile:

$$ P(r) = P(r_0)\, e^{ \frac{r_0}{H} \left( \frac{r_0}{r} - 1 \right) } $$

where $H$ is the same characteristic length. $r_0$ is the radius of the planet itself, and $r$ (radius) replaces $h$ (altitude). This equation is accurate for the specific case of:

1. constant temperature
2. constant composition (homogenous)

These are terrible assumptions, obviously. For all planets, there tends to be a local minimum of temperature in the atmosphere, and deeper into it, the temperature will be many many times higher, which gives a magnitude of just how bad the constant temperature assumption is.

The homogeneous assumption is also very bad, because the gases stratify subtly. It's not like oil-water separation, but the higher you go, the higher the concentration of the lighter elements will be. In fact, this is a major justification for why the gas giant atmospheres are such pure Hydrogen and Helium to begin with.

Nonetheless, this equation for pressure is still technically better than the constant gravity assumption. But that's not the most relevant on Jupiter. In fact, it's probably more relevant of a correction on Earth.

This formula assumes a constant gravitational acceleration over the whole height of the gas column - a reasonable assumption for Earth, as the atmosphere is thin compared to to the size of the planet.

For a simple argument, if you assumed composition is the same, it's clear that the altitude change needed for pressure to change by a certain factor is less on Jupiter than on Earth. Gravity is stronger, and the radius of Jupiter is larger. Both of these factors result in a smaller characteristic height for the atmosphere. That gives some indication that the above quote may be less correct than the reverse proposition. But there are many more complications.

Obviously, Jupiter has a lot more light gas in its atmosphere than Earth. This is partly because Earth's gravity isn't perfect at holding in Hydrogen against the radiation in space. So this pushes in the other direction - making Jupiter's atmosphere have a longer characteristic length.

But the characteristic length isn't the full story. Obviously Earth has a nice (agreed upon) ground level. Jupiter might have a phase change that vaguely resembles this, but even if there is an abrupt density change, it won't be anything like Earth. So this raises the question of how we should define the region of Jupiter's atmosphere in the first place. Could we accept internal metallic Hydrogen in the center as "atmosphere"? Probably not, but that's just the most extreme case.

Differential equations correctly describing the system can be found in other answers. For completeness, they are:

$$ \frac{d}{dr} P(r) = g(r) \rho(r) $$

But let's go into the special case solutions, because I think this is what the OP is asking for. In the case of constant gravity, ideal uniform gas, and constant temperature, the atmospheric pressure follows:

$$ P(h) = P(0)\, e^{ - \frac{ h }{ H } } $$

Where $H$ here is the characteristic length. For Earth, this is about 7.4 km. It depends on the temperature, so it's always in flux.

Now, if we add in the tidal component of gravity, we can obtain a new expression. This is assuming a $\frac{1}{r^2}$ form of gravity, instead of a constant. In this case, the solution to the differential equation yields the following pressure profile:

$$ P(r) = P(r_0)\, e^{ \frac{r_0}{H} \left( \frac{r_0}{r} - 1 \right) } $$

where $H$ is the same characteristic length. $r_0$ is the radius of the planet itself, and $r$ (radius) replaces $h$ (altitude). This equation is accurate for the specific case of:

1. constant temperature
2. constant composition (homogenous)
3. contribution to gravitational field by air itself can be neglected

1 and 2 are terrible assumptions, obviously. For all planets, there tends to be a local minimum of temperature in the atmosphere, and deeper into it, the temperature will be many many times higher, which gives a magnitude of just how bad the constant temperature assumption is.

The homogeneous assumption is also very bad, because the gases stratify subtly. It's not like oil-water separation, but the higher you go, the higher the concentration of the lighter elements will be. In fact, this is a major justification for why the gas giant atmospheres are such pure Hydrogen and Helium to begin with.

Nonetheless, this equation for pressure is still technically better than the constant gravity assumption. But that's not the most relevant on Jupiter. In fact, it's probably more relevant of a correction on Earth.

This formula assumes a constant gravitational acceleration over the whole height of the gas column - a reasonable assumption for Earth, as the atmosphere is thin compared to to the size of the planet.

For a simple argument, if you assumed composition is the same, it's clear that the altitude change needed for pressure to change by a certain factor is less on Jupiter than on Earth. Gravity is stronger, and the radius of Jupiter is larger. Both of these factors result in a smaller characteristic height for the atmosphere. That gives some indication that the above quote may be less correct than the reverse proposition. But there are many more complications.

Obviously, Jupiter has a lot more light gas in its atmosphere than Earth. This is partly because Earth's gravity isn't perfect at holding in Hydrogen against the radiation in space. So this pushes in the other direction - making Jupiter's atmosphere have a longer characteristic length.

But the characteristic length isn't the full story. Obviously Earth has a nice (agreed upon) ground level. Jupiter might have a phase change that vaguely resembles this, but even if there is an abrupt density change, it won't be anything like Earth. So this raises the question of how we should define the region of Jupiter's atmosphere in the first place. Could we accept internal metallic Hydrogen in the center as "atmosphere"? Probably not, but that's just the most extreme case.

Differential equations correctly describing the system can be found in other answers. For completeness, they are:

$$ \frac{d}{dr} P(r) = g(r) \rho(r) $$

But let's go into the special case solutions, because I think this is what the OP is asking for. In the case of constant gravity, ideal uniform gas, and constant temperature, the atmospheric pressure follows:

$$ P(h) = P(0)\, e^{ - \frac{ h }{ H } } $$

Where $H$ here is the characteristic length. For Earth, this is about 7.4 km. It depends on the temperature, so it's always in flux.

Now, if we add in the tidal component of gravity, we can obtain a new expression. This is assuming a $\frac{1}{r^2}$ form of gravity, instead of a constant. In this case, the solution to the differential equation yields the following pressure profile:

$$ P(r) = P(r_0)\, e^{ \frac{r_0}{H'} \left( \frac{r_0}{r} - 1 \right) } $$

where $H'$ is a different characteristic length. $r_0$ is the radius of the planet itself, and $r$ (radius) replaces $h$ (altitude). This equation is accurate for the specific case of:

1. constant temperature
2. constant composition (homogenous)

These are terrible assumptions, obviously. For all planets, there tends to be a local minimum of temperature in the atmosphere, and deeper into it, the temperature will be many many times higher, which gives a magnitude of just how bad the constant temperature assumption is.

The homogeneous assumption is also very bad, because the gases stratify subtly. It's not like oil-water separation, but the higher you go, the higher the concentration of the lighter elements will be. In fact, this is a major justification for why the gas giant atmospheres are such pure Hydrogen and Helium to begin with.

Nonetheless, this equation for pressure is still technically better than the constant gravity assumption. But that's not the most relevant on Jupiter. In fact, it's probably more relevant of a correction on Earth.

This formula assumes a constant gravitational acceleration over the whole height of the gas column - a reasonable assumption for Earth, as the atmosphere is thin compared to to the size of the planet.

For a simple argument, if you assumed composition is the same, it's clear that the altitude change needed for pressure to change by a certain factor is less on Jupiter than on Earth. Gravity is stronger, and the radius of Jupiter is larger. Both of these factors result in a smaller characteristic height for the atmosphere. That gives some indication that the above quote may be less correct than the reverse proposition. But there are many more complications.

Obviously, Jupiter has a lot more light gas in its atmosphere than Earth. This is partly because Earth's gravity isn't perfect at holding in Hydrogen against the radiation in space. So this pushes in the other direction - making Jupiter's atmosphere have a longer characteristic length.

But the characteristic length isn't the full story. Obviously Earth has a nice (agreed upon) ground level. Jupiter might have a phase change that vaguely resembles this, but even if there is an abrupt density change, it won't be anything like Earth. So this raises the question of how we should define the region of Jupiter's atmosphere in the first place. Could we accept internal metallic Hydrogen in the center as "atmosphere"? Probably not, but that's just the most extreme case.

Differential equations correctly describing the system can be found in other answers. For completeness, they are:

$$ \frac{d}{dr} P(r) = g(r) \rho(r) $$

But let's go into the special case solutions, because I think this is what the OP is asking for. In the case of constant gravity, ideal uniform gas, and constant temperature, the atmospheric pressure follows:

$$ P(h) = P(0)\, e^{ - \frac{ h }{ H } } $$

Where $H$ here is the characteristic length. For Earth, this is about 7.4 km. It depends on the temperature, so it's always in flux.

Now, if we add in the tidal component of gravity, we can obtain a new expression. This is assuming a $\frac{1}{r^2}$ form of gravity, instead of a constant. In this case, the solution to the differential equation yields the following pressure profile:

$$ P(r) = P(r_0)\, e^{ \frac{r_0}{H} \left( \frac{r_0}{r} - 1 \right) } $$

where $H$ is the same characteristic length. $r_0$ is the radius of the planet itself, and $r$ (radius) replaces $h$ (altitude). This equation is accurate for the specific case of:

1. constant temperature
2. constant composition (homogenous)

These are terrible assumptions, obviously. For all planets, there tends to be a local minimum of temperature in the atmosphere, and deeper into it, the temperature will be many many times higher, which gives a magnitude of just how bad the constant temperature assumption is.

The homogeneous assumption is also very bad, because the gases stratify subtly. It's not like oil-water separation, but the higher you go, the higher the concentration of the lighter elements will be. In fact, this is a major justification for why the gas giant atmospheres are such pure Hydrogen and Helium to begin with.

Nonetheless, this equation for pressure is still technically better than the constant gravity assumption. But that's not the most relevant on Jupiter. In fact, it's probably more relevant of a correction on Earth.