# With a 10% increase in Earth's mass, would the Karman line move up or down, and by how much?

This is an exercise to better understand the basic physics and math behind scale height and the Karman line. It was inspired by this answer to Why is FAI considering lowering the Karman Line to 80km? (also note they might not be anymore; for more on that see answers to When is/will be the symposium to revisit the Karman line and consider the "McDowell line"?

For the purposes of this question:

1. You can either assume that the total mass of the atmosphere is unchanged, or that the surface pressure is unchanged.
2. You are welcome to use the ideal gas law and a constant temperature for the atmosphere as a function of height and use an atmospheric scale height approximation, or you can get fancier if you like.
3. Keep the Earth's radius the same; assume the iron-nickel core is larger so that the surface gravity has increased to 1.1 times standard gravity $$g_0$$.
4. Stick to Wikipedia's standard definition of the Karman line and do not debate if it is meaningful or not please! Whatever "Karman plane" (i.e. wing loading) yields 100 km for normal Earth, assume it's the same.

Question: With a 10% increase in Earth's mass, would the Karman line move up or down, and by how much?

## It would move Down!

By the definition of the Karman line on wikipedia, the lift force and the "centrifugal force" must be equal to the gravitational force and, therefore, each other. This gives the following equation:

$$\frac{1}{2}\rho v^2C_LS = \frac{v^2m}{R_e+h}$$

Where $$\rho$$ is density, v is velocity, $$C_L$$ is lift coefficient, S is wing area, m is vehicle mass, $$R_e$$ is the Earth's radius, and h is the altitude.

The velocity terms cancel and I assumed that $$C_LS$$ and m canceled because S is entirely a design choice and could be chosen such that $$C_LS$$ was equal to m; regardless, the Karman line should be independent of vehicle (for practical purposes anyway). Following these simplifications, the equation becomes:

$$(R_e+h)\rho=2$$

Applying the Scale Height model you provided gives a means to compute $$\rho$$ as a function of altitude. I tried to solve the resulting equation analytically, but It got rather nasty and so I used a zero finder.

Per the "scale height approximation", the density as a function of altitude is the following:

$$\rho(h) = \frac{M_{air}P_oe^{-h/H}}{RT_{avg}}$$ with $$M_{air}$$ being the mean molecular mass of the atmosphere, $$P_o$$ being the s.l. pressure, R being the gas constant, and $$T_{avg}$$ the mean atmospheric temperature. Pressure ($$P_o$$) as a function of depth in a fluid scales linearly with the gravitational acceleration- I am assuming the height of the atmosphere to remain the same and the surface pressure to scale with the increase in surface gravity.

$$H$$ is the scale height and represents the height in which the pressure decreases by a factor of e. It is defined as the following:

$$H = \frac{RT_{avg}}{M_{air}g}$$

The final equation I used to zero find for the Karman altitude is this:

$$(R_e+h)\frac{M_{air}P_oe^{-h/H}}{RT_{avg}}-2=0$$ With $$P_o$$ and $$H$$ being dependent on surface gravity as defined above.

I first adjusted the "mean atmospheric temperature" until the height was equal to the traditional 100 km mark (with the standard 250K, the value was ~110 km). At 221.55K, I computed a Karman altitude of 100km +/- 5 cm.

Finally I could adjust the Earth's mass. A mass of 1.1 Earth's decreased the Scale Height from 6.47km to 5.88km (effectively increasing the pressure gradient of the atmosphere and providing a lower density at a given altitude) and provided a new Karman Line at 91.5km

EDIT: I would like to say that my algebra in regards to the canceling of $$C_LS$$ with m is a bit iffy. The Karman line is very dependent on vehicle mass and lift characterisics. For the purposes of simplicity, I was assuming the two values were equal, but realized a moment ago that their units do not cancel. The math still holds, but the algebra should include a unit factor containing the missing dimensions.

• This might get way too complicated, but would a heavier earth be able to retain a bigger atmosphere that would change this analysis? – Oscar Smith Jun 9 '20 at 21:36
• @OscarSmith point #1 in my question includes a simplifying assumption to avoid that complication. It's a great question but should probably asked separately, and perhaps in Earth Science SE. – uhoh Jun 10 '20 at 2:18
• Did you include the change in orbital velocity for the new mass? $$v = \sqrt{\frac{GM}{r}}$$ It would be great to include the final equation to which you applied the zero-finder. Thanks! – uhoh Jun 10 '20 at 2:22
• @uhoh yes, I did include that. The equation you’ve listed is derived from $\frac{mv^2}{r}= \frac{mMG}{r^2}$. The RHS of that equation is what the lift force is equated to as well (which is what allowed the first equation I listed to be valid), so as far as the two are related, the earth’s mass only comes into play with the density. I will edit my answer tomorrow to include the final equation. I think it was this: – A McKelvy Jun 10 '20 at 4:29
• continued due to mobile failures I will edit my answer tomorrow to include the final equation. I think it was this: $0 = (R_e + h)\frac{RP_oe^{-h/H}}{M_{air}T} - 2$ with $P_o$ being s.l. Pressure and H being scale height: $H = \frac{RT}{M_{air}g_o}$ with $g_o$ being s.l. gravity (computed with Earth mass as an input). I’ll check these equations and edit my answer tomorrow, like I said. – A McKelvy Jun 10 '20 at 4:38