Examples: Dimensional Analysis
Astro 111 - Order of Magnitude: Fall 2015
1) Getting in the Ballpark. What is the pressure at the center of a self-gravitating
sphere of mass M and radius R?
Let’s attack this problem as sloppily as we can in order to show the power of just
knowing the physics at hand.
We know that one way to think of pressure is as a force per unit area, P = F/A.
What is the relevant force at hand? Gravity of course, since we said the sphere is
self-gravitating. Then we only need Newton’s Law of Gravitation to get the relevant
force in terms of our variables,
GM 2
.
R2
Now we need an area. The only relevant length scale in the problem is R, so let’s
divide through by R2 .
Fg
GM 2
Pc ∼ 2 =
R
R4
Actually integrating the equation for hydrostatic equilibrium (a concept we’ll get
to later) will get us the exact answer, which turns out to be smaller by a factor
of 3/8π. For fun, try re-doing this argument by thinking of pressure of an energy
density! Getting comfortable with thinking of pressure in this way will be useful
later in the course.
F = Fg =
2) Getting onto the Field. a) Derive Kepler’s Third Law using dimensional arguments alone.
Here we’ll take a more systematic approach to utilizing dimensional analysis.
Consider a test mass orbiting a mass M at a distance a. Kepler’s Third Law relates
the orbital period of our test mass Porb to the distance a. We want to combine
variables in some clever way in order to solve for the orbital period, which has units
of seconds. As we’ll see, this reduces down to a simple algebraic problem, but the
key to these types of problems is knowing the relevant variables and fundamental constants (for a more mathematical treatment of what this means, look up the
Buckingham Pi Theorem).
Here the relevant variables and fundamental constants are M, a, and G. Let’s assume that we can solve for Porb in terms of these three, i.e.
Porb ∼ M α aβ Gγ .
We can solve for the exponents then by knowing that we want to end up with units
of seconds, i.e.
[s] = [g]α [cm]β [cm3 g−1 s−2 ]γ
we know that we do not want cm in our final answer, so let’s use the fact that
exponents add to make sure they add to 0, i.e. that the cm disappear:
cm : β + 3γ = 0
Examples: Dimensional Analysis
Astro 111 - Order of Magnitude: Fall 2015
Doing the same thing for grams and seconds (which we do want!), we end up with
three equations and three unknowns:
cm : β + 3γ = 0
g:α − γ = 0
s:
− 2γ = 1
Solving for our variables we find α = γ = −1/2 and β = 3/2. Plugging into
our original guess yields
3 1/2
a
Porb ∼
,
GM
which is correct to within a factor of 4π 2 (which is technically more than an order
of magnitude, but at least we now understand how it scales with the distance and
mass).
b) Derive the “radius” of a black hole from dimensional analysis. What would it
be for an Earth mass black hole?
Let’s use our newfound tool of dimensional analysis to tackle this one. Here the
relevant variables will include G and M , since we’re dealing with the gravitational
force (RBH is what we’re trying to solve for, so it’s not in this list). However, a new
physical constant, the speed of light c, enters into our constants since a black hole
can be defined as an object whose escape velocity vesc = c. With this, we have
RBH ∼ Gα M β cγ .
Just like last time, we can solve for the exponents by making sure the units work
out:
[cm] = [cm3 g−1 s−2 ]α [g]β [cms−1 ]γ
This yields the system of equations
cm : 3α
+ γ = 1
g : −α + β
= 0
s : −2α
− γ = 0
which is solved with α = β = 1 and γ = −2. Plugging back in, we find
GM
RBH ∼ 2
c
which is only off by a factor of 2, so I’m calling it a success.
To solve for what the radius of an Earth mass black hole is, we must first calculate the mass of the Earth. I don’t know this off-hand, but I do know that the
circumference of the earth is about 25,000 miles, and a factor of 1.6 between miles
and kilometers yields a circumference of 40,000 km. Thus
4 × 109
cm ≈ 109 cm.
2π
If you’re worried about 4/2π = 1, remember that this is order of magnitude. To
quote my former order of magnitude professor, “lower your standards.” Given a
R⊕ ≈
Examples: Dimensional Analysis
Astro 111 - Order of Magnitude: Fall 2015
radius (and hence a volume), all we need is a density to get the mass of the Earth.
The density of water in cgs is conveniently 1 g cm−3 , and the Earth is essentially a
giant rock. I know that rocks sink in water, so a density of 1 g cm−3 would yield
a lower bound on the mass of the Earth. The density of rock depends on the type
of rock of course but mostly hovers around a few times that of water, so let’s say
hρ⊕ i ≈ 3 g cm−3 . Plugging in we find
3 hρ i
G4πR⊕
⊕
≈ 0.1cm,
3c2
which is only a fraction of the size of a marble.
RBH,⊕ ≈