Newton's Third Law of Motion: There are three laws of motion developed by Isaac
Newton. The third one deals with what happens when an object exerts a force on another
object. For instance, consider your fist smashing into a thing wall. It might be possible
that you punch a hole in the wall. Yet it is also possible that your fist is in a lot of pain-if not outright broken--from the impact. According to our everyday common sense, the
fist would be exerting the force, the wall would be receiving the force. So, what causes
the bones in the fist to break? Newton's Third Law of Motion answers that by saying that
there's no such thing as "receiving" a force--when an object exerts a force on a second
object, the second object exerts just as much force on the first object. This can lead to
some situations that make absolutely no sense at all on the surface. For instance,
consider a small car pushing a big truck up a hill. How can the car be pushing the big
truck up the hill if the force exerted by the car on the truck is equal but opposite to the
force the truck exerts on the car? The answer is that those aren't the only forces involved
in the problem. In fact, since the whole system is the car plus the truck, the force the car
exerts on the truck and the force the truck exerts on the car are purely internal forces
within the system. It takes a force external to the system to cause the system to move.
What's pushing the car-truck system up the hill is that the car is pushing off of the
ground--the ground then responds by pushing on the car, according to Newton's Third
Law of Motion, and thus providing the force external to the car-truck system necessary to
get the system moving.
Force of Gravity in Free-Fall: Let's apply Newton's Second Law of Motion to free-fall.
If the downward acceleration of a falling object of mass m is g, then the gravitational
force acting on the mass is Fgravity = mg. Thus, the more massive the object is, the greater
the gravitational force acting on it. However, that doesn't impart a greater acceleration
because a heavier mass is intrinsically more difficult to accelerate than a lighter mass.
The result is that no matter what the mass is, the resulting accelerations in free-fall are
equal.
Newton's Law of Universal Gravitation: Suppose you were to perform experiments to
determine the acceleration due to gravity on a freely-falling object both at sea level and at
the top of Mount Everest? Careful experiments would tell you that g = 9.80 m/s2 at sea
level but g = 9.77 m/s2 on top of Mount Everest. Apparently, therefore, g is not only not
constant, but it decreases the further you get from sea level. To what value would it
decrease by the time you were as far out as the Moon? Isaac Newton thought about it in
the mid-17th century. He was ultimately able to show that the law obeyed by the
gravitational force applies throughout the solar system and later physicists were able to
show that Newton's Law of Gravitation applies to the known universe. In Newton's Law
of Universal Gravitation, it turns out that the force due to gravity decreases with the
square of the distance between the two masses. Thus, out at the distance of the Moon, the
acceleration you would feel due to Earth's gravitational attraction is g = 0.003 m/s2.
Beyond that, the acceleration due to the Earth becomes even more negligible. Newton
derived the following equation:
Fgravity  m
GM
r2
where m is the mass of the object in question, M is the mass of the body to which the first
object is attracted (e.g., the Earth), r is the distance from the center of the first body to the
center of the second body, and G is known as the "gravitational constant of the universe.
G is equal to 6.67 x 10-11 N·m2/kg2, which means that it's a very small number. This is
indicative of the fact that as fundamental forces go, gravity is very weak. Note also that r
doesn't represent the distance from the surface of either body but from the centers. Thus,
at sea level on Earth, r is not equal to zero but is, instead, equal to the radius of the Earth.
We can compare the equation above with the equation we previously had for the
gravitational force, Fgravity = mg. When we compare these, we find g = MG/r2. If we plug
in the radius of the Earth, 6.37 x 106 m and the mass of the Earth, 5.98 x 1024 kg, we get g
= 9.83 m/s2. This agrees very well with the measured value of 9.80 m/s2 (when the
effects of the Earth's rotation are included, we find that theory and experiment are in
excellent agreement).
The "inverse-square law" is very common in physics. Aside from gravity, other
phenomena that are affected this way include electromagnetism and intensity (sound,
light and radiation). The inverse-square law of gravity gives rise to extreme gravitational
"gradients" found near black holes. Suppose we have a black hole with the mass of the
Sun; its event horizon would then have a radius of 2954 m (almost 3 km). Suppose you
are standing straight up just beyond the event horizon (your feet are 2955 m away from
the center of the black hole). If you are 1.8 m tall (about 5'11") then your head is at about
2956.8 m away from the center of the black hole. Let's say your mass is 80 kg (about 177
pounds). Then the difference in the gravitational forces you feel
GmM 6.67 10 11  80 1.99 1030
Ffeet 

 1.22 1015 N
2
2
r
2955
11
GmM 6.67 10  80 1.99 1030
Fhead 

 1.211015 N
2
2
r
2956.8
F  Ffeet  Fhead  1.5 1012 N.
Admittedly, we made some approximations for the masses involved, but even if this
calculation is off by a factor of 100, that would still leave a force difference of about 10 10
N between your head and feet. This is equivalent to having a mass of 10 10/9.8  109 kg
tied and dangling from your feet--a modern aircraft carrier. As you inch closer to the
event horizon, this differential gets even more extreme. This is what we mean by a sharp
gradient (like a steep hill) and this is the effect of the inverse-square law.