Laboratory Exercise

Assignment 1: Laboratory Exercise

Answer all questions in the field provided. Scientific notation, if required, can be indicated either
by using E-notation or the power symbol (i.e. 5.2 x 1012 = 5.215 12 = 5.2 x 10″12). You will need to
(a) give a numerical answer (if applicable) and (b) explain briefly in words or mathematical
workings how you obtained your answer for EACH question. Answer in SI units.
Question 1: Let’s consider what the Earth would look like without horizontal energy transport.
Figure 3.9 in the textbook provides curves of isolation (the average flux received throughout the
day on a horizontal surface) at several different latitudes. Using these values and the Stefan-
Boltzmann law, determine the equilibrium temperature at:
(a) The Equator at the start of February
(b) 609 N at the start of February
(c) Using the lnternet, look up the daily average temperatures in February for Whitehorse,
Yukon (60.79N) and Singapore (1.49N) as real-world proxies for the values you
calculated above. NOTE: don ’t forget to cite when? you found the information” If they
are different from the values you calculated in (a) and (b), describe why.
Question 2: One of the factors that will affect atmospheric thickness is temperature.
(3) Using the temperatures you found in Question 1, determine the atmospheric scale
height at Whitehorse and Singapore in February.
(b) Do you think that the height of major atmospheric structures could change from pole to
equator? Why or why not?
Question 3: Spacecraft in orbit around the Earth are subject to heating from both the sun and also
from the Earth itself.
(a) If the flux from the sun is 1368 W in2 and the emission temperature of the Earth is 254
K, estimate the flux the spacecraft receives from the Earth if it is located 200 km above
the emission altitude of the Earth.
(b) ls heating from the planet an important contributor to the heat balance of these

Book Link:
Read from 3.1 Introduction to 4 to get better understanding:

Figure 3.9:

Average daily weather in Whitehorse in February (2017):
Average daily weather in Singapore (2017) in February:

ESSE 1012 3.0 Laboratory Exercise #1.


1. The Shape of the Earth.
Earth is rotating, or spinning, about its rotation axis which passes through the geographic North and
South poles. If the Earth were not spinning, its shape would be very close to a perfect sphere.
However, because of the Earth’s rotation equatorial regions feel a larger centrifugal force than polar
regions. As a result the sphere is flattened at the poles and bulges at the equator (much like a
spinning ball of jelly). The “flattening”, f, of Earth is defined as the difference between the
equatorial radius, re, and the polar radius, rp, expressed as a fraction of the equatorial radius.
1(a) If the equatorial radius of Earth was re = 5, and the flattening of Earth is f = 0.0033528,
compute polar radius, rp . Show your work.
ANSWER: _________________________
1(b) By how many km would the equatorial radius exceed the polar radius?
ANSWER: __________________________Page 2
1(c) The Gas Giants are much more fluid than Earth and consequently have greater flattening due to
spinning about their rotation axes. For example, for Saturn f = 0.100. Given the equatorial radius of
Saturn as re = 50,268 km, compute the polar radius of Saturn.
ANSWER: ____________________________________________
1(d) By how many km does Saturn’s equatorial radius exceed its polar radius?
ANSWER: ____________________________________________

2. Geographic Coordinates.

Locations on Earth’s surface can be specified in terms of latitude and longitude. Parallels (or lines)
of Latitude are parallel to the Earth’s equator and measured in degrees north and south of the
equator. Latitude is 0o at the equator, 90oN at the north pole and 90oS at the south pole. Meridians
(or lines) of Longitude are not parallel to each other. They follow “great circle” paths that run northsouth and intersect as they all pass through the north and south geographic poles. The centre of Earth
is the centre of all great circles on Earth’s surface, including the Meridians of Longitude. The
Meridians of Longitude are perpendicular to the Parallels of Latitude. Longitude is measured in
degrees east and west of the “prime meridian” which passes through a former observatory in
Greenwich, England. Longitude ranges from 0o to 180o E (or 180oW). Each degree is subdivided
into 60 units called minutes, denoted by the symbol ‘, and each minute is further subdivided into 60
smaller units called seconds, denoted “. Thus, the location of the Petrie Science and Engineering
Building can be given in terms of latitude and longitude as: 43o 46′ 24.5″N, 79o 30′ 23.5″W.
Assuming the Earth is a sphere of radius re, it is possible to determine the distance along the surface
between points at different latitude and longitude.Page 3
2(a) Consider two points on the equator at coordinates (lat., long.) of (00 0′ N, 850 30′ W) and (00 0′
N, 870 10′ W). What is the surface distance between these two points, measured in km? Show your
ANSWER: _______________________
2(b) Consider two points with the same longitudes as in (a) but at a latitude of 50o N. What is the
surface distance between these two points along the 500 N parallel of latitude?Page 4
ANSWER: __________________________
2(c) Now consider two points on the same meridian of longitude, but with different latitudes, such as
(5o 0′ 30″ N, 127o W) and (10o 30′ 0″ N, 127o W). Determine the surface distance between these
ANSWER: __________________________
2(d) Would your answer be different at different latitudes if the change in latitude were the same as
previously [e.g. (55o 0′ 30″ N, 127o W) and (60o 30’ 0” N, 127o W)]. If so, by how much?
ANSWER: __________________________
2(e) Would your answer be different at a different longitude, such as 12o E? If so by how much?
ANSWER: ___________________________Page 5
Maps of Earth’s surface are created by making projections of the features of Earth’s spherical surface
onto two-dimensional flat surfaces. The familiar Mercator projection used on road maps and atlases
is an example of a cylindrical projection, wherein Earth’s surface features are projected along rays
from the centre of Earth (imagine a light bulb at the centre of Earth) to the inner surface of a
cylindrical projection surface which has its central axis coincident with the rotation axis of the
planet. Distortion is minimized by having the cylindrical projection surface touch Earth’s surface
along the equator, as illustrated below.

Cylindrical Projection

Projection surface

2(f) Where are distortions of the projected surface features a maximum? (Choose one of: equator,
mid-latitudes, or high latitudes).
ANSWER _____________________

Consider the cylindrical projection surface when it is cut along a vertical line and opened to form a
flat 2D map.
2(g) Do the projected parallels of latitude appear as horizontal lines, circular arcs, or vertical lines?
ANSWER __________________________________________
2(h) Do meridians of longitude project as horizontal lines, circular arcs, or vertical lines?
ANSWER __________________________________________Page 6

3. Size of the Earth

The Earth’s radius was first estimated by Eratosthenes, an early Greek astronomer, from the length of
shadows at different locations. Consider two points at the same longitude on the Earth’s surface but
400 km apart. At noon the Sun is directly overhead at one location, whereas at the other location a
rod 2.5 m high casts a shadow towards the south 15.72 cm long.
Show clearly how the Earth’s radius may be estimated from these observations.
(Hint: Draw a cross section view of the Earth using a compass to draw an arc through the two points
on the Earth’s surface. Assume that rays of sunlight arriving at the two locations travel along parallel
15.72 cm

4. Space Geodesy

The Global Positioning System (GPS) uses a “constellation” of 24 artificial satellites that are
constantly orbiting the Earth at an average altitude of 20,183 km. The satellites emit radio signals
that are captured by special devices called GPS receivers which can process the signals to determine
the location (latitude, longitude) of the receiver on the surface of the Earth to within a fraction of a
second of arc. However, in order to determine a position on the surface of the Earth accurately, the
satellite must be at least 10o above the observer’s horizon. If we set up a very similar constellation
of GPS satellites on Mars, at an altitude of 16,500 km, for what fraction of the orbital period would
the satellites be available for position determination.
(Hint: Assume that the orbit of the satellite passes directly overhead of the observer and that the orbit
of the satellite is circular. Also assume that the radius of the Mars is 3,390 km. The horizon is
defined as a horizontal plane at the point of observation. Use the cross-sectional sketch below
showing the surface of the Earth, the satellite orbit and the horizon.)
GPS Satellite orbit
MarsPage 8

5. The Mass of the Earth.

The gravitational force of attraction, Fg, between two bodies of mass M and m, the centres of which
are a distance r apart is given by the expression
= GMm/r2 (1)
where G is the universal gravitational constant and has a value of 6.668 x 10-11 m3/kg-sec2.
In particular, if the mass of the Earth is M, the gravitational force acting upon an object of mass m on
the Earth’s surface is given by
= mg (2)
where g is the acceleration due to gravity.
5(a) Use the relations above to calculate the mass of the Earth, assuming that g = 9.780 m/sec2
and r = re.
ANSWER: ________________________________
5(b) Use your result for the mass of the Earth to estimate the mean density of the Earth. (Hint:
Approximate the volume of the Earth with that of a sphere with radius re.)
ANSWER: ________________________________
5(c) How does this mean density for the whole Earth compare with those for the minerals found
in the rock samples from the upper mantle (olivine, pyroxene, garnet)?
Densities of Mantle Materials (Mg/m3 = gm/cm3)
Olivine Pyroxene Garnet Typical Mix (.4 Ol., .5 Py., .1 Ga.)
3.31 3.34 4.16 3.39
ANSWER: __________________________________Page 9
6. Planetary Orbits
Useful information: 1 A.U. = 149.6 x 106
Distance from Sun to Mars = 1.52 A.U.
Time for Mars to complete one orbit about the Sun: 687.0 Earth days
1 Earth Year = 3.15 x 107
1 Mile = 1.609 km.
6(a) Assuming a circular orbit for Mars about the Sun, compute the speed at which this planet
moves along its orbit in cm/sec and in km/hr.
ANSWER: ________________ _________________
6(b) Alternatively, for circular orbits around a large mass it is known from celestial mechanics
that the orbital speed is constant with a value given by
Vcircular = √(GM/a) (3)
where a is the radius of the circular orbit. For orbits about the Sun, M = 1.991 x 1030 kg.
Use Equation (3) to compute Vcircular for Mars’ orbit about the Sun.
ANSWER: ______________________________
6(c) The distance between the Sun and Neptune is 30.1 A.U. Compute the orbital speed for
Neptune from Equation (3).
ANSWER: __________________________________Page 10
6(d) Use this value to compute the number of Earth years for Uranus to complete one orbit.
ANSWER: ___________________________________
Geostationary communications satellites orbit Earth above the equator with a period of 24 hours.
Consequently from the point of view of an observer on Earth, the satellite appears to be stationary,
remaining always above the same point on the Equator.
6(e) Using the mass of Earth computed in 5(a) above, and equation (3) for the speed of a satellite in
a circular orbit, compute the height of a geostationary satellite above Earth’s surface at the equator.
ANSWER: ___________________________________Page 11

7. Escape Velocity

The escape velocity for any large planetary mass M is the minimum velocity which any small
object must have if it is to escape the gravitational pull of M. This velocity, Vescape, decreases
with increasing distance, r, from M. The escape velocity for M at a given r is given, in
general, by the expression
escape = √(2GM/r) (4)
where r is the distance of the small object from the centre of M.
7(a) Compute the escape velocity for the Sun at a distance equal to the orbital radius of Earth. What
fraction of the Sun’s escape velocity is the orbital velocity of Earth?
ANSWER: ____________________ _________________________
7(b) Compute the escape velocity for Earth at the Earth’s surface.
ANSWER: _____________________________________
(c) Consider an approximately spherical asteroid with a diameter of 3.00 km, and an average
density equal to that of olivine. How fast would an astronaut have to run in order to jump off
the asteroid and escape its gravitational attraction? Express your answer in (i) km/hr, and (ii)
miles/hr.Page 12
ANSWER: ______________________ ___________________

Laboratory Exercise
The End

Last Updated on June 17, 2018 by EssayPro