×
Back to all chapters

# Gravitational Physics

Here we lay out Newton's law of gravity and crack open the universe of consequences that spring from it.

# Atmospheric thickness

The atmosphere is easy to take for granted, but without it Earth would be uninhabitable due to temperature swings of hundreds of degrees between day and night, a constant bombardment with lethal radiation, and last, but not least, the lack of $$\ce{O2}$$ and $$\ce{CO2}.$$

In a minimal sense, an atmosphere is a thin shell of gas that surrounds a planet, kept from floating off into space by Earth's gravitational field.

In this quiz, we'll develop a simple model for the change in the structure of the atmosphere as a function of the height above the surface. At the end, we'll turn our calculation on its head to predict the height of the atmosphere on a neutron star—with shockingly good accuracy.

We've seen that the gravitational field we feel from a particle shrinks as the inverse square of our distance from it. As we'll prove later in this course, this is true of (spherical) planets as well.

If $$h$$ is the distance from its surface, and $$r_p$$ its radius, then the gravitational field of a planet is $\mathbf{g}_\textrm{planet}(h) = G\frac{M_p}{\left(r_p + h\right)^2}\hat{\mathbf{r}}.$ This works, but is unwieldy. As creatures who exist close to the surface, it would be nice to not always have to deal with an inverse square in order to get things done.

Our first question is: how does the gravitational field change as we move from Earth's surface up to a height $$h?$$

To start to get a handle on this, take a Taylor expansion of $g(h) = G \frac{M_\textrm{Earth}}{(R_\textrm{Earth} + h)^2},$ with respect to $$h$$ and keep terms to first order.

You should get two terms, a constant one that is $$h$$-independent, corresponding to the gravitational field at the surface, and a second that is a height dependent correction: $g(h) = \overbrace{g(\textrm{surface})}^{h-\textrm{independent}} + \overbrace{\Delta g(h)}^{\textrm{correction}} .$ What is the correction term?

The purpose of the Taylor expansion is to have an easy way to calculate the change in the gravitational field with height.

Now, let's use it to look at a scale that's relevant for atmospheric phenomena: the height of the ozone layer, which starts at approximately $$h_\textrm{ozone} = \SI{15}{\kilo\meter}$$ above Earth's surface.

By what fraction does the strength of the gravitational field weaken in moving from the surface into the ozone layer?

Find $$\Delta g(h_\textrm{ozone})/g(\textrm{surface}).$$

As you just found, the strength of gravity stays almost unchanged as we move from the surface up into the ozone layer. Maybe the ozone layer isn't high enough to make a dent in $$g(h)$$ but, surely, gravity is negligible for the astronauts on the International Space Station.

Is it? After all, we see them floating freely inside, doing all sorts of things we Earthlings can't manage.

Find the relative change in gravity at the height of the space station $\Delta g (h_\textrm{ss})/g(\textrm{surface}).$

Details: the International Space Station orbits at $$h_\textrm{ss} = \SI{400}{\kilo\meter}.$$

For all intents and purposes, the force of gravity is constant from the surface of Earth all the way to the top of the atmosphere. This is because the corrective term in the expansion is insignificant next to the surface gravity for all heights near the surface.

As you showed, $$\Delta g(h)/g(\textrm{surface}) = -2h/R_\textrm{Earth},$$ and so deviations are negligible until $$h$$ is on the order of half an Earth radius. In other words, we have $g_\textrm{Earth} \approx g(\textrm{surface}) \equiv G\frac{M_\textrm{Earth}}{R_\textrm{Earth}^2}.$

For near Earth problems, you'll be using this approximation quite a bit, so it is prudent to find its value.

Calculate $$g(\textrm{surface})$$ (in $$\si[per-mode=symbol]{\meter\per\second\squared}$$).

Details:

• $$G \approx \SI[per-mode=symbol]{6.67e-11}{\newton\meter\squared\per\kilo\gram\squared}$$
• $$M_\textrm{Earth} \approx \SI{5.97e24}{\kilo\gram}$$
• $$R_\textrm{Earth} \approx \SI{6.371e6}{\meter}.$$

As we saw in Hydrostatics, as fluid of any kind builds up, points underneath will feel a pressure due to the weight of all the atmosphere that's above them. As Earth's atmosphere is a compressible gas, the shell is densest close to the surface, and thins out as it reaches into outer space.

Over the next few questions, we're going to derive and solve a differential equation that governs the relationship between height and atmospheric pressure, using the simple linear approximation you just developed.

Atmospheric pressure drops as we increase $$h$$ because the local atmosphere no longer needs to support the column of air below $$h.$$ We can use this insight to show how the atmospheric pressure drops after an infinitesimal increase in $$h,$$ $$\Delta h.$$

Let the local pressure at a height $$h$$ above the surface be $$P(h).$$ Find the difference in atmospheric pressure between the layer at $$h$$ and a layer a small distance $$\Delta h$$ above, $\Delta P = P(h+\Delta h) - P(h).$

Recall: the ideal gas law, $$PV = Nk_B T$$, holds here. Let the average mass of an air molecule be $$\overline{m}.$$

You can use your result from the last problem to find a differential equation for $$P$$ in terms of $$h,$$ e.g. a relation of the form $\frac{dP}{dh}= f(h)$ Solve it to find $$P(h),$$ which tells us the precise manner in which atmospheric pressure changes with height.

You've derived a functional form for the atmospheric pressure that we can write in the general form $$P(h) = P_0 f(h/A),$$ where $$f$$ is a monotonically decreasing function that goes to zero for $$h\gg A.$$

Therefore, we can think of $$A$$ as a relevant length scale over which atmospheric pressures decreases.

Calculate the value of $$A$$ on Earth in meters.

Details & Assumptions

• $$k_B = \SI[per-mode=symbol]{1.381\times 10^{-23}}{\joule\per\kelvin}$$
• $$T = \SI{300}{\kelvin}$$
• $$\overline{m} = \SI{12}{amu} = \SI{12}{\gram}/N_A\approx \SI{2e-26}{\kilo\gram}$$

Because $$f$$ is the exponential function, $$A$$ is the scale over which $$P$$ drops by a factor of $$1/e \approx 37\%.$$

For example, once we move five factors of $$A$$ away from the surface, $$P$$ is down to $$e^{-5} \lt 1\%$$ of its terrestrial value.

Let's consider the edge of space the height $$h^*$$ at which atmospheric pressure falls to less than one one hundred thousandth its value at the surface.

Find the value of $$h^*$$ in $$\si{\kilo\meter}.$$

Despite its simplicity, our model makes a quantitative prediction about the scale height of the atmosphere. The constant $$P(0)$$ is a constant of integration that we need to measure at the surface; it is approximately $$\SI[per-mode=symbol]{1e5}{\newton\per\meter\squared}.$$

Here, we treated $$P(0)$$ like it has a constant value. In reality, there are local fluctuations about $$P(0)$$ due to temperature, weather, latitude and other factors that can swing it by as much as 2% in either direction. Therefore, it is better to think of $$P(0)$$ as an average result, so that $$P(0) \rightarrow \langle P(0) \rangle.$$

The validity of our approximation is a pure function of geometry, failing after $$2h/r \approx 0.05,$$ independent of surface temperature or the local strength of gravity.

Therefore, we should expect that our predictions carry over to other celestial objects, e.g. neutron stars whose surface gravity is $$\num{e11}$$ times as strong as on Earth, and has a surface temperature of about $$\SI{1e6}{\kelvin}.$$

Naively applying our result from Earth, find the approximate height of the atmosphere on a neutron star $$h^*.$$

Note: the atmosphere on neutron star is made of carbon atoms, not oxygen so $$m$$ goes from 28 to 12.

Real measurements of the atmosphere on neutron stars find a thickness $$h_\textrm{atm}$$ on the order of several $$\SI{1}{\centi\meter},$$ which suggests that our incredibly naive approach at least captures the right order of magnitude.

In this quiz we analyzed the gravitational field for spherical planets and found a simple approximation that holds as long as we don't move too far away from the surface $$h \ll \frac12 r_\textrm{planet}.$$ As surface phenomena stay well below this threshold, for most applications it is perfectly valid to approximate terrestrial fields by $$g_0.$$

As we saw, $$g_0$$ stays within 2% of $$g(h)$$ even as we breach the edge of space, and is only 5% off the mark at the height of the International Space Station.

As you move on in physics you'll run into more situations like this one where (as long as we're cognizant of the reasons for why) we can discard the full complexity of a theory and use a simpler form of it instead.

×