Fall 2004
Dr. Mike Fanelli
Solutions to Assigned Problems
CHAPTER 2
PROBLEM 2-1
This problem presents an application of the small angle formula, introduced in
Chapter 1. Given an angular size, and the distance to three solar system
objects, the problem asks: what physical size corresponds to an angular size of 1
arcminute at each of those distances ? In non-math terms, the question is: given
the limitation of the human eye to resolve details, what is the smallest size that
could be “resolved” at the distance of the Moon, the Sun, and Saturn.
ANSWER: Using the formula as given in chapter 1.
True size = distance X
angular diameter (expressed in degrees)
-------------------------------------------------57.3
For the Moon,
(1 arcminute  60 arcminutes per degree)
Size = 384,000 km X ----------------------------------------------------57.3
= 112 kilometers. Implication: we can’t eyeball lunar craters since they have
diameters generally less than 100 kilometers, but we can see the lunar “seas”,
the dark areas which make up the “Man in the Moon”.
For the Sun, the distance = 1 AU = 1.5 x 108 km
???? = 1.5 x
108
(1 arcminute / 60 arcminutes per degree)
km X ---------------------------------------------------57.3
= 43, 600 kilometers. Implication: the Earth’s diameter is about 12,000
kilometers, so we could not even resolve something as big as Earth at the Sun’s
distance.
For Saturn, its average distance from the Sun is 9.5 AU. So its distance from
Earth is (9.5 – 1) AU, when it is at opposition (and closest to Earth). That equals
8.5 AU.
(1 arcminute / 60 arcminutes per degree)
???? = 1.3 x 109 km X ------------------------------------------------------57.3
= 372, 000 kilometers. Note that this value is larger than the planet’s diameter,
which is why we can’t see a disk when we look at Saturn.
PROBLEM 2-8:
How long would it take a radar signal to complete a round trip between Earth and
Mars when the two planets are 0.7 AU apart ?
ANSWER: This problem deals with the speed of a radar signal, which, like light
waves, travels at 300,000 kilometers per second. You are asked to determine a
time interval. The problem is set up in exactly the same manner as if you were
estimating your travel time for an airline trip. What is the distance traveled ?
How fast are you going ?
Distance = 0.7 AU (Earth-Mars distance) X 150,000,000 kilometers per
AU X 2 (the signal travels to Mars and back, a roundtrip)
= 2.1 x 108 kilometers
Time = distance divided by the rate of travel
= 2.1 x 108 kilometers  300,000 km per sec.
= 700 seconds = 11.7 minutes
PROBLEM 2-14:
Use Newton’s Law of gravity to calculate the force of gravity between you and
Earth, converting your answer to pounds. What do you normally call this force ?
ANSWER: The force of gravity is expressed:
F(grav) = G  (m  M)  r2
The task here is to assign the correct value to each variable in this expression,
then evaluate the expression.
G is the “constant of gravitation”; its value is given in the text:
G = 6.67 x 10-11 newtons meter2 / kg2
“Big” M is the mass of Earth = 6 x 1024 kg.
“r” is the separation between you and the center of the Earth, otherwise known
as the radius of the Earth. r = 6400 kilometers.
To evaluate the expression, you need to assign a value for little “m”, which is
YOUR mass. I’ve used 70 kilograms for an “average” student mass. You must
also be sure that similar units are expressed the same way. Here you need to
convert the Earth’s radius, which is given in kilometers to meters, because both
“G” and newtons are expressed in meters.
Inserting these values in the equation,
F(grav) = 684 newtons.
If 1 lb = 4.45 newtons then 0.225 lbs = 1 newton (dividing each side by 4.45).
Therefore, 684 newtons x 0.225 lbs per newton = 153 lbs.
153 lbs = the force of gravity on this person, which is normally called a person’s
weight.