HOMEWORK # 7 - SOLUTIONS
1. First, we need to find the volume of ice that makes up the icecap of Antarctica. We can find this by
multip lying the surface area of the icecap by its depth (units of area times units of length gives units of
volume): vo lume of ice = (0. 5% of Earth’s surface area) x (depth of ice) = 0.005 × A × (3 km) , where A
is the Earth’s s urface area. If we assume that the density of ice is equal to the density of water (which is a
good approximation), then the water from the melted ice will have an equal volume. The water will run off
Antarctica and spread over the oceans. T herefore, instead of being limited to 0.5% of the Earth’s surface, the
water will be spread across 70% of the surface. The vo lume of the water can be expressed as (70% of the
Earth’s surface area) × (depth of water) = 0.7 × A × d, where d is the depth of the additional water (what
we’re trying to find). Yet the volume must remain the same as it started out as , therefore we can make an
equality: 0 .005 × A × (3 k m) = 0.7 × A × d . Not ice that A cancels out so t hat we do not need to
calculate the surface area of the Earth. Doing the algebra we get d = 0 .021 km = 21 m. Therefore, the sea
level will rise by about 21 meters! In the case of the Arctic icecap, which floats on the ocean, the ice is
already displacing a volume of water nearly equal to the vo lume of ice. Therefore melting the ice of the
Arctic will no t change the sea level (in other words, the ice is already contributing to the sea level).
2. We can use the half-life equation from the book, (fraction of remaining parent ma terial) = f = ( 0.5)t /T
where t is the age and T is the half-life of the parent nuclei. T o find t we need to do some algebra. Take the
logarithm of both sides, and using basic l ogarithm rules and algebra we get:
log f =log (0 .5)t/T = ( t/T) log(0.5)
t = T (log f) / ( log 0 .5).
Now if 75% of the U-238 parent material (T = 4.5 billion years) has decayed then that leaves 25% left or a
fraction of 0.25. Therefore the age t = ( 4.5 bill ion years) (log 0 .25) / (log 0.5) = 9 billion years . W ith
90% of the U-238 decayed that leaves a fraction of 0.10 left. The age is t = (4 .5 billion years) (log 0.10) /
(log 0.5) = 15 billions years. So we’d have to wait 6 bil lion more years.
3. All three planets began with s imilar atm ospheres which included the gases left over from so lar system
formation, including hydrogen and helium. These gases are too li ght to stick around so the planets quickly
lost these first atmospheres. The second atmo spheres were created by out-gassing from volcanoes and since
the three planets have similar in terior composition, their second atmospheres were also similar. The
atmospheres were largely nitrogen and carbon dioxide. On Mars, much of the atmosphere left because of its
lower gravity and its further distance from the Sun froze the water that remained leaving i t with a dry, thin
atmosphere composed mo stly of nitrogen and carbon dioxide. On Venus , it was too hot for liquid water so
that all the water is in vapor form. Venus is mas sive enough to ho ld onto its atmosphere and is left with a
very hot, thick atm osphere. On Earth, the temperature was ju st righ t for liquid water, which in the form of
oceans absorbed some of the carbon dioxide. Liquid water allowed life to evolve on Earth. Photosynthetic
life converted some of the remaining carbon dioxide into oxygen and eventually left us with our final
atmosphere composed mo stly of nitrogen and oxygen gas.
4. First we need to find the molecular speed of nitrogen on Mercury. This depends on the molecular weight
of nitrogen (mmol = 28) and the temperature on Mercury (T = 700 K , if we take the highest temperature
since that is when the atmosphere will leak the most and we can expect an atmosphere will distribute the
temperature more evenly). The molecular speed is vrms = 0.157 ( T / mmol) = 0.157 (700 /28) = 0.785 km/s
= 785 m/s. Now we need to set t his speed to one-sixth the escape speed. The escape speed is
vesc =11 .2(M/R) where M and R are in units of Earth’s mass and radius.
If the mass increases due to density and the radius stays constant, then we put in the value for R and solve
for M:
0.785 = (1/6)(11.2)(M / 0.38)
M = 0.15 Earth’s mases. This would however mean that Mercury would have to have an improbably large
density, so a better way to s olve this is to keep the density constant and increase the radius.Then the mas s
of the planet is
M=volume of the sphere*avg. density = 4/3 (pi) ρ R^ 3
So we have
785 m/s = (1/6) (2 GM/R)^(1/2)
785 m/s = (1/6)[(2G)(4/3) (pi)ρ R^3/R]^(1/2) = (1/6) [2G (pi)ρ R^2]^1/2
Given the average density, ρ=5430 kg/m^3 , we can solve for R:
