Solution to Practice Problem Set 2
CIV300/ENV346
Fall 2010
Question# 1: Newton’s law of gravitation states that the gravitational attraction between two bodies is proportional
to the product of their masses and inversely proportional to the square of the distance separating their centers of
mass. The proportionality constant is usually taken as 6.67×10-11 N m2/kg2. Use this law and the fact that an
object on the Earth’s surface experiences an acceleration of 9.81 m/s2 to deduce the mass of the Earth. Use the
acceleration of the Earth in its orbit (from question 3 in last week’s PPS 1) to deduce the mass of the sun.
Compare your answers with generally accepted values of these two astronomical constants.
Solution. Newton's law of universal gravitation is illustrated in Figure 1:
F2
F1
m1
m2
F1 = F2 = G
m1 × m2
r2
G = 6.67 × 10 −11
Nm 2
kg 2
r
Figure 1- The mechanisms and relation of Newton's law of universal gravitation.
i) To deduce mass of the earth:
F =G
m1 × m2
r2
F = m1 g ;
m1 g = G
m1 × m2
r2
⇒
gr 2
m2 =
G
where m2 is the mass of the earth and r is the earth radius which is equal to 6371 km. then:
m2 =
9.81 × (6371 × 10 3 ) 2
6.67 × 10
−11
= 5.97 × 10 24 kg
ii)
V2
m ×m
m1
=G 1 2 2
r
r
9
where m1 is Earth mass, m2 is sun’s mass and r is the radius of earth’s orbit (150 × 10 m)
From question 2 in PPS1,
V2
m
= 6 × 10 − 3 2
r
s
m2 =
(6 × 10 −3 ) × (150 × 10 9 ) 2
6.67 × 10
−11
m2 = 2.02 × 10 30 kg
The generally accepted value for mass of the earth is 5.98*1024 kg and mass of the sun is 1.988*1030 kg;
so we did very well indeed, particularly for the mass of the Earth!
1 Question# 2: A standard dry atmosphere is 78.08% nitrogen, 20.95% oxygen, and most of the balance taken up
as argon. The universal gas constant is usually taken as 8.3145 J/(mol K) (joules per mole kelvin). Using a
suitably averaged molecular weight, derive the “dry air” gas law used in class from the universal gas law relation.
Species
Nitrogen
Oxygen
Argon
Molecular Mass (g/mol)
28
32
39.95
%ppm in dry air
78.08 (fraction 0.7808)
20.95(fraction 0.2095)
~0.97 (fraction 0.0097)
Solution. The average molar mass of dry air will be a weighted average of the molar mass of molecular
oxygen, molecular nitrogen, and argon, weighted by their fraction in dry air.
(0.2095*32 + 0.7808*28 + 0.0097*39.95) = 28.96 g/mol = Mdry air
Mdry air = 0.02896 kg/mol
The “dry air gas law” is the same as the universal gas law, but uses the dry air constant, which is the
universal gas constant (R = 8.314 J/molK) divided by the effective molecular weight of air.
R/Mdry air = 8.314/0.02896 ≈ 287 J/kgK (which is the standard value, and the one used in class).
Question# 3: In PPS 1 (question 5) you derived how much energy the Earth as a whole receives from the sun.
Convert that quantity to (i) an equivalent amount of combustion of gasoline assuming 33 MJ/litre; (ii) to an
equivalent rate of mass extinction using Einstein’s famous equation E = mc2.
17
Solution. i) The energy that the earth receives from the sun= 1 . 74 × 10
J
)
s
W ( or
The energy produced by 1 litter of gasoline is 33 MJ and the equivalent amount of combustion of gasoline
1 . 74 × 10
is:
33 × 10
6
J
)
s
= 5 . 27 × 10
J
(
)
litre
17
(
9
litre
s
(somewhat “cooked” to be about the same as the total mass extinction for the Sun as a whole, as determined in lectures).
ii)
E = mc
2
c = 3× 10
8
m
s
m=
2 1.74 × 1017
(3 × 108 ) 2
= 1.93
kg
s
Question# 4: It is the dead of night and your tall, sleek, spaceship Long Shot is poised for take off on its launching
pad. As you look through your cabin viewport out over the Pacific Ocean, you notice, even with the horizon, a
flashing light at the very top of a communication tower that you know is exactly 100 m tall and 80 km away.
Estimate the height of your viewport above the surface of the Pacific Ocean.
Assume x + y ≈ 80 km
R = 6371 km
R2 + y2 = (R+0.1)2
[note: 0.1 km = 100 m]
y2 = 0.2R + 0.01 = 1274.21
y = 35.7 km; x = 80 – 35.7 = 44.3 km
R2 + x2 = (R + LS)2
[LS = height of Long Shot viewport]
x2 = 2R*LS + LS2; LS2 + 12742LS – (44.32) = 0
The quadratic equation could be employed here, yielding:
LS = (-12742 ± √(127422 + 4 *1*1962.5))/2 = 0.154 km, -12742.2 km
It’s clearly not negative 12742.2 km tall so the Longshot viewport is 154 m tall.
OR, it could be assumed that LS2 is a negligible value relative to the other terms, yielding:
12742LS = 44.32 = 1962.5
Which leads to about the same answer: LS = 0.154 km
3 Question# 5: Your spaceship Long Shot blasts off following a course that moves directly away from both the earth
and the sun (that is, the Earth is directly behind you and the sun directly behind that). As you look behind, you are
still obviously in darkness but the halo of the sun around the earth is growing quickly; in fact, a moment or two later,
you leave the Earth’s shadow behind and are now unmistakably in sunlight. Supplying the required geometrical
constants, determine an expression to estimate how far from the Earth you now have traveled.
Solution. In this problem, it is helpful to visualize the conical shaped shadow formed behind a planet:
Recall Similar Triangles
Dsun/dsun-longshot = Dearth/dearth-longshot
dsun-longshot = dearth-sun + dearth-longshot = 150*109 + dearth-longshot
Dsun/Dearth = (150*109 + dearth-longshot)/dearth-longshot
1.39*109/12.7*106 = 109.4 = (150*109 + dearth-longshot)/ dearth-longshot
108.4 dearth-longshot = 150*109
dearth-longshot = 1.38*109 m
Question# 6: The real purpose of your mission on Long Shot is to begin to erect an orbiting solar collector. This
ambitious project seeks ultimately to construct a huge square solar collector that is 2000 km by 2000 km. The
collector has a control system that continually keeps it oriented perpendicular to the sun’s rays. The solar energy
is efficiently collected, concentrated, and beamed back to Earth with no energy loss. However, despite this
impressive technology, detractors of the project argue that the net effect of the collector will be to raise the surface
temperature of the Earth. (a) Explain how such a large but efficient solar collector would inevitably raise the Earth’s
temperature. (b) Estimate how much energy the solar collector will intercept. Now, making suitable assumptions,
find an expression to estimate how much the Earth’s temperature might rise with this huge solar collector in longterm service.
4 Preliminary Comments. There are a number of important physical realities to be aware of when solving
this problem. These include:
•
The energy balance of the Earth is determined by radiation in (coming from the Sun) and radiation
out.
•
The incoming radiation strikes the disk of the Earth, or the circular projection of the Earth, with an
energy density of 1370 W/m2, where the m2 represent that disk.
•
The outgoing radiation comes from the sphere of the Earth, radiating out in all directions.
•
The total radiation in must equal total radiation out, or an equilibrium temperature can never be
reached (and, of course all planets have their equilibrium average temperature). Even in the case
of our planet, which is warming, on average, the radiation in is still very close to being equal to
the radiation out.
•
The radiation out from the Earth can be estimated at 390 W/m2 (with the m2 representing the
Earth’s sphere) by either dividing 1370 W/m2 by 4 (4 being the ratio of the surface area of a
sphere to the surface area of a circle with the same diameter), or by assuming an average
temperature of 288 K and using the Stephan-Boltzman law to calculate energy radiated per surface
area (j).
•
The Stephan-Boltzman law states that the energy radiated per surface area of an object (j) is
related to temperature in the following way:
o j = εσT4
where ε is the emissivity of the object, σ is the Stephan-Boltzman constant (σ = 5.67*10-8
J/sm2K4), and T is the temperature of the object, in K. To simplify this problem, we will
assume ε =1.
•
All energy will tend towards the greatest entropy, meaning that if we beam energy down to Earth
from a massive space-based solar array, even if we use the energy as electricity, it will eventually
become heat that must be radiated from the Earth back into space.
•
Thus, adding energy from space to the Earth will increase the amount of energy that the Earth
must get rid of (increasing j), and by the Stephan-Boltzman law, will increase the equilibrium
temperature of the Earth.
Solution 1. There are a number of ways to solve Problem 6. One fairly elegant way involved the
proportionality implications of the Stephan-Boltzman law, and an understanding that the energy leaving
the Earth will be in proportion to the area collecting the energy.
The area of the disk of the Earth is:
AEarthcircle = (π/4)Dearth2 = (π/4)(12.7*106)2 = 1.266*1014 m2
The area of the panel is
5 Apanel = (2*106)2 = 4*1012 m2
(AEarthcircle + Apanel)/AEarthcircle = 1.032
Thus, the area collecting energy has increased by 3.2%. The energy, too, both incoming and outgoing,
will increase by this percentage.
Looking to the Stephan-Boltzman law, it can be seen that temperature will increase in proportion to the
4th root of the increase in outgoing energy per area. Thus, temperature will increase in proportion to
4
√1.032 = 1.008
The original temperature, T1 was 288 K (really any estimate near 300 produces the same result).
1.008*T1 = 1.002*288 = 290.3 K.
Thus, the temperature increase would equal 2.3 K.
Solution 2. Alternatively, a more long-winded solution involving energy flux could be attempted:
In this solution, the additional energy coming to Earth from the panel can be calculated using
Apanel*1370 = 4*1012 * 1370 = 5.48*1015 W
Recall that if this much energy is received by the Earth, then this much energy must be radiated back into
space from the Earth. The additional radiation from the Earth, in W/m2, then, is:
5.48*1015 W/AEarthsphere = 5.48*1015 W/ π(12.7*106)2 = 10.8 W/m2
The new rate of radiation from the Earth, then is
390 W/m2 + 10.8 W/m2 = 400.8 W/m2
Using the Stephan-Boltzman Law, and assuming ε = 1:
400.8 = σT4
T = 4√(400.8/(5.67*10-8)) = 289.96 K
In this case, the temperature increase calculated is 1.96 K.
Final Comments. Clearly the two methods give close enough final values; the solar collector, really
regardless of how the energy would be used, would increase the Earth’s temperature by about 2 degrees.
For comparison sake, I chose the size of the solar collector to give a net energy into the Earth about the
same magnitude is the average rate of the transport of energy from the tropics to the poles via the
atmosphere and ocean.
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