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Notes for Week 5.
Chapter 4 (Venus) and Chapter 5 (Earth) involve less math than was lurking
behind the scenes in Chapter 3. The topics I am going to amplify are:
1. The density structure of a planet's atmosphere.
2. Heating by accretion and impact.
3. Heating by radioactive decay.
The atmosphere:
In hydrostatic equilibrium, gravity compresses the atmosphere and the pressure
gradient opposes gravity.
PdA
g = GMm
r2
Pressure forces from
the sides cancel
(P+dP)dA =
(P + dP dr) dA
dr
m = ρ dAdr
The difference between the upward force on the bottom of the volume element
and the downward force on the top is what keeps gravity from pulling the
element downward. Thus:
(P+dP/dr dr) dA - dA = (dP/dr) dr dA = GMρ dA dr / r2
cancelling dr dA and dividing by r:
(1/ρ) dP/dr = - GM/r2 = - g
Now if the gas is all at the same temperature T we can use the ideal gas
equation,
P =
kT / mH
where m H is the mass of a hydrogen atom and µ is the mean molecular mass in
units of mH
so dP/dr = d( ρ kT / µmH )/dr =(kT / µmH ) dρ /dr
Plugging in and rearranging:
©L. A. Willson
printed 14:04, September 25, 1997
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(1/ρ) dρ/dr = - (GM/r2 )(µmH /kT) = µmH g/kT
As long as the thickness of the atmosphere is small compared to r, the right
hand side may be considered constant, and we get
ρ = ρ o e-∆r/H where ∆r is the height in the atmosphere and H is the density scale
height.
A little juggling shows that:
H = (kT / mH g)
and
(-1/H) = (1/ ) d /dr
In fact this second equation for H is in a form that can easily be generalized: the
scale of variation of some quantity Q is |(1/Q) dQ/dr|. In the case of density we
have the minus sign because the density decreases with increasing r.
2. Heating by accretion and impact.
Wnen a solid body hits Earth, the energy released is
1/2 m vimpact2
This energy goes to heating, melting, and/or vaporizing the mass m; heating,
melting, perhaps vaporizing, and producing waves in material originating on
Earth; and possibly into ejecting some Earth material into space.
A good rule of thumb: For the atmosphere to be able to slow or stop an
incoming object, the intercepted mass of atmosphere must be about as big as
the mass of object.)
So what is vimpact?
Conservation of energy:
Total energy = 1/2 mv 2 -GMm/r
Evaluate at ∞: 1/2 m v∞2
Total energy is constant, so
1/2 mv2 -GMm/r =1/2 m v∞2
and
vimpact 2 = 2GM/R + v ∞2
Note that this does not depend on m, as long as we ignore the effect of the
atmosphere.
©L. A. Willson
printed 14:04, September 25, 1997
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The atmosphere will slow, stop, or vaporize small incoming bodies; large ones
will not be affected.
The above allows us to compute the energy released when a small body hits
Earth - anything small enough not to change the total mass and radius of the
planet. But of course when the planet was growing, these did change. To
compute that, we simply find the total energy released in accumulating the
planet. This must be
E tot > (3/5) GM 2 /R
Why the >? Well - what do we know about the relative velocities (v∞) of the
particles that accumulated? They need not all have been zero!
In Chapter 5 this is used to compute the temperature that the planet would have
reached had it accumulated in an instant. The resulting values, 375,000 for
Earth and 325,000 for Venus, are clearly ridiculous! So is instantaneous
accumulation.
To see what would happen if it accumulated slowly:
Recall that 4πr2 σ T4 = rate of loss of energy to radiation
Suppose it accretes at a steady rate; its temperature will approach the
equilibrium value
Accretion Power = total accretion energy/duration of accretion = 4πr2 σ T4
3. Heating by differentiation:
In HW4 you estimated the energy difference between an undifferentiated and a
differentiated body:
∆GPE =∆[ ∫ (GMr(r)/r) 4πr2 dr]
which gave a fairly messy expression.
On page 99 they give an simpler, approximate, expression for the temperature
change when a core is formed:
∆T = (1-f)f∆r (GM/3RρC) where f = fraction of mass in the core,
which means
∆GPE ~ (1-f)f (GM2 /3R)(∆ρ/ρ)
©L. A. Willson
printed 14:04, September 25, 1997
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4. Heating by radioactive decay.
The heat produced by the radioactive decay of a radioactive element may be
expressed as:
(Energy/decay) dN/dt = Energy/sec
but
dN/dt = -λ N
and N = N o e-λt
=>
Energy/sec = (Energy/decay)(-λ No e-λt )
=> exponentially decaying heat source
Note that some things were big heat sources when the solar system was young,
and others are nearly as important today as they were then. What distinguishes
these two cases? (Table5.3)
©L. A. Willson
printed 14:04, September 25, 1997