For the Love of Physics (44 page)

In summary, equation 5 tells us that if we know the distance from any planet to the Sun (
r
), the orbital period of the planet (
T
), and
G
, we can calculate the mass of the Sun (
m
₁
).

Orbital periods were known to a high degree of accuracy long before the seventeenth century. The distances between the Sun and the planets were also known to a high degree of accuracy long before the seventeenth century but only on a
relative
scale. In other words, astronomers knew that Venus’s mean distance to the Sun was 72.4 percent of Earth’s and that Jupiter’s mean distance was 5.200 times larger than Earth’s. However, the absolute values of these distances were an entirely different story. In the sixteenth century, in the day of the great Danish astronomer Tycho Brahe, astronomers believed that the distance from the Earth to the Sun was twenty times smaller than what it actually is (close to 150 million kilometers, about 93 million miles). In the early seventeenth century Kepler came up with a more accurate distance to the Sun, but still seven times smaller than what it is.

Since equation 5 indicates that the mass of the Sun is proportional to the distance (to a planet) cubed, if the distance
r
is too low by a factor of seven, then the mass of the Sun will be too low by a factor of 7
³
, which is 343—not very useful at all.

A breakthrough came in 1672 when the Italian scientist Giovanni Cassini measured the distance from the Earth to the Sun to an accuracy of about 7 percent (impressive for those days), which meant that the uncertainty in
r
³
was only about 22 percent. The uncertainty in
G
was probably at least 30 percent. So my guess is that by the end of the seventeenth century the mass of the Sun may have been known to an accuracy no better than 50 percent.

Since the relative distances from the Sun to the planets were known to a high degree of accuracy, knowing the absolute distance from the Sun to the Earth to 7 percent accuracy meant that the absolute distances to the Sun of the other five known planets could also be calculated to that same 7 percent accuracy by the end of the seventeenth century.

The above method to calculate the mass of the Sun can also be used to measure the mass of Jupiter, Saturn, and the Earth. All three planets had known moons in orbit; in 1610 Galileo Galilei discovered four moons of Jupiter, now known as the Galilean moons. If
m
₁
is the mass of Jupiter, and
m
₂
the mass of one of its moons, then we can calculate the mass of Jupiter, using equation 5, in the same way that we can calculate the mass of the Sun, except that now
r
is the distance between Jupiter and its moon, and
T
is the orbital period of that moon around Jupiter. The four Galilean moons (Jupiter has sixty-three moons!) have orbital periods of 1.77 days, 3.55 days, 7.15 days, and 16.69 days.

Accuracies in distances and in
G
have greatly improved over time. By the nineteenth century
G
was known to about 1 percent accuracy. It is now known to an accuracy of about 0.01 percent.

Let me show you a numerical example. Using equation 5, let’s calculate together the mass of the Earth (
m
₁
) by using the orbit of our Moon (with mass
m
₂
). To use equation 5 properly, the distance,
r
, should be in meters, and
T
should be in seconds. If we then use 6.673 × 10
^–11
for
G
, we get the mass in kilograms.

The mean distance to the Moon (
r
) is 3.8440 × 10
⁸
meters (about 239,000 miles); its orbital period (
T
) is 2.3606 × 10
⁶
seconds (27.32 days). If we plug these numbers into equation 5, we find that the mass of the Earth is 6.030 × 10
²⁴
kilograms. The best current value of Earth’s mass is close to 5.974 × 10
²⁴
kilograms, which is only 1 percent lower than what I calculated! Why the difference? One reason is that the equation we used assumed that the Moon’s orbit is circular, when in fact it is elongated, what we call elliptical. As a result, the smallest distance to the Moon is about 224,000 miles; the largest is about 252,000 miles. Of course,
Newton’s laws can also easily deal with elliptical orbits, but the math may blow your mind. Perhaps it already has!

There is another reason why our result for the mass of the Earth is a little off. We assumed that the Moon circles around the Earth and that the center of that circle is the center of the Earth. Thus in equations 1 and 3, we assumed that
r
is the distance between the Earth and the Moon. That is correct in equation 1; however, as I discuss in more detail in
chapter 13
, the Moon and the Earth actually each orbit the center of mass of the Moon-Earth system, and that is about a thousand miles below the Earth’s surface. Thus
r
, in equation 3, is a little less than
r
in equation 1.

Since we live on Earth, there are other ways of calculating the mass of our home planet. One is by measuring the gravitational acceleration near the surface. When dropped, any object of mass
m
(
m
can have any value) will be accelerated with an acceleration,
g
, close to 9.82 meters per second per second.
*
Earth’s average radius is close to 6.371 × 10
⁶
meters (about 3,960 miles).

Now let’s revisit Newton’s equation 1. Since
F
=
ma
(Newton’s second law), then

Here,
r
is the radius of the Earth. With
G
= 6.673 × 10
^–11
,
g
= 9.82 meters per second per second, and
r
= 6.371 × 10
⁶
meters, we can calculate
m
earth in kilograms (you try it!). If we simplify equation 6 somewhat, we get

I find that
m
earth is 5.973 × 10
²⁴
kilograms (impressive, right?).

Notice that the mass,
m
, of the object we dropped does not show up
in equation 7! That should not surprise you, as the mass of the Earth could not possibly depend on the mass of the object that you drop.

You might also be interested in knowing that Newton believed that the average density of the Earth was between 5,000 and 6,000 kilograms per cubic meter. This was not based on any astronomical information; it was completely independent of any of his laws. It was his best “educated” guess. The average density of the Earth is, in fact, 5,540 kilograms per cubic meter. If you allow me to write Newton’s guess as 5,500 ± 500 kilograms per cubic meter, his uncertainty was only 10 percent (amazing!).

I do not know if Newton’s guess was ever taken seriously in his day, but suppose it was. Since Earth’s radius was well known in the seventeenth century, its mass could have been calculated to an accuracy of 10 percent (mass is volume times density). Equation 7 could then be used to calculate
G
also to an accuracy of 10 percent. I am telling you this because it intrigues me that, accepting Newton’s guess for the mean density of the Earth, at the end of the seventeenth century the gravitational constant,
G
, could already have been known to an accuracy of 10 percent!

INDEX

absorption lines,
237
–39

Academic Earth,
x

Academy of Sciences, Bavaria,
155

Academy of Sciences, St. Petersburg,
146

Academy of Sciences, Soviet,
253

accretion disk,
231
,
245

adenosine triphosphate (ATP),
169

“Aerodynamic Lift, Bernoulli Effect, Reaction Lift” (Johnson),
74

air, weight of,
60
–61

air drag,
43
,
51

maglev trains and,
164

pendulums and,
52

Air Force Cambridge Research Laboratories (AFCRE),
193
–94

air pressure,
59
–65,
71

barometer and,
63
,
66
–67

on Earth’s surface,
63
–64

measuring,
61
–63

wind and,
64

Albee, Edward,
123

alpha decay,
183

alternating current (AC),
158
,
161

aluminum,
151

Amazon Star (Bellatrix),
237

amber,
126

American Physical Society,
213

American Science and Engineering (ASE),
17
,
193
–94

Ampère, André-Marie,
138
,
154
,
166

ampere (amp),
138
,
141

Ampère’s law,
166

amplitude,
118

energy and,
105

of pendulums,
52
–54,
106

of sound waves,
104
–6

of tuning fork,
106

AM radio band,
191
–92

Andromeda galaxy,
4
,
28