Gravity as we've said, 
is the force, that mostly governs the 
behavior an the motions of celestial 
objects. 
And so Newton's theory is going to be the 
cornerstone of everything we do from now 
on, and it behooves us to go back and 
fill in some details and see what more we 
can get out of this wonderful 
formulation. 
And so, in this clip we're going to work 
through some more detailed calculations, 
applying Newton's universal law of 
gravity and along the way learn some 
things, both about astronomy and about 
gravity and even some math. 
And so, let's proceed, we have things to 
do. 
First thing you have to do is reconcile 
our story about the universality of 
gravity with the rather simple picture 
that we've been using so far. 
So let's understand, when I am standing 
in the vicinity of the earth, of the 
actual gravitational force that the earth 
exerts on me tells me Newton's theory, is 
the sum total of gravitational attraction 
of every little bit of earth. 
because gravity is universal, every atom 
of earth attracts every atom of me and I 
need to add them all up. Now I'm pretty 
small, 
but I need to figure out, how to add up 
the some total of gravitational 
attraction from all different parts of 
the earth. 
And so Newton showed us the way to do 
this along the way inventing, theory of 
multiple integrals to do it, but what he 
showed and we're not going to repeat that 
calculation for that reason, is that if 
you just take a round spherically 
symmetric shell of mass, so imagine just 
a crust of the earth, with an empty 
inside. 
Then the total force it exerts on someone 
standing over here at some arbitrary 
point outside well, if this mass of the 
shell is N, and my distance from the 
center of the shell is R, then the force 
that the shell exerts on a person over 
here is GMM over R squared, and it's 
directed towards the center of the shell 
as it had to be by symmetry. 
The force is the same, therefore as would 
have been created by eliminating the 
shell and replacing it with a mass N 
located at the center of the circle. 
Now if I on the other hand ask the same 
question, and this will become important 
to us later, about what is the force on 
an observer here? 
Well here again, some parts of the shell, 
attract you this way. 
Some parts attract you this way. 
The result for the forces striking, the 
results for the force is that it's 
absolutely zero. 
There is no force applied by a spherical 
shell anywhere on the inside of it. 
We don't tend to make too many 
measurements on the inside of earth but 
we are on the outside. 
And what all of this tells me is that if 
I consider earth as a collection of 
concentric shells and I consider the full 
gravitational effect of the earth to be 
the sum of all the gravitational effects 
of each concentric shell, then this has 
the effect, as far as I am concerned of 
replacing the whole earth by an object 
with a mass equal to mass of the earth 
located at the center. 
And, this simplifies our calculation a 
great deal. 
Let's see an example of how we use this. 
So, let's go back to figuring out my 
weight. My mass you will recall, 
hopefully hasn't changed much, was 59 
Kilos. 
And so to compute the force on me, 
standing on the surface of the earth That 
means my distance from that imaginary 
mass at the center of the earth is the 
radius of the earth. 
I will use the astrological symbol for 
earth to signify things like the mass of 
the earth and the radius of the earth 
because it's simple to produce in tech. 
And so the force on me is given by 
Newton's, expression, GMM over R squared, 
where I've replaced the Earth by an 
object with the same mass as Earth 
sitting at its centre, plugging in the 
numbers for G for the mass of the Earth 
and for its radius in metres squared. 
I find not surprisingly that when I 
multiply all of these things, 
guess what? 
I found our old friend, 
MG This object is what we call the 
gravitational acceleration on the surface 
of earth. 
And if you wish, then measuring the 
gravitational constant as Cavendish did, 
measuring the radius of the earth, which 
you can do geometrically, this is the way 
to determine the earth's mass. 
We've computed the gravitational 
acceleration on earth from Newton's 
formula 
We know, that as I get farther away, from 
the center of earth, or as I get higher 
up beyond, into and beyond the 
atmosphere, then the denominator in 
Newton's formula becomes larger. 
That R squared in the denominator, my 
distance from the center of the Earth 
grows, and gravity weakens. 
It's not that gravity does not exist of 
course beyond Earth, 
but it is weakened with distance. 
I want to get a quantitative estimate of 
this. 
So let H be my altitude above the ground. 
This is what Newton's formula predicts 
for the force of gravity on an object of 
mass MR, say, me. 
I'm going to rewrite this in a useful 
way. 
I'm going to rewrite this as. 
G times the mass of earth divided by R of 
radius of the earth squared. 
Well I've divided and multiplied, by the 
radius of the earth squared so here I 
need to put, radius of the earth squared, 
over, my actual distance, from the center 
of the earth squared, and then I'll 
remember MR. 
And then I recognize this combination, as 
what we previously called G. 
When H is zero this, quantity is one and 
I'm back to the surface. 
So, in the second line, I'm, rewriting 
this, as, 
my weight on the surface of the Earth 
times dimensionless number that indicates 
how much its decreased and then I 
simplify this dividing numerator and 
denominator by the radius of the Earth to 
get this expression. 
And the point I want to apply here is 
that I want to see, 
This is an exact expression. 
I want to see if I can simplify for cases 
where my height above the earth is not 
very large compared to the radius of the 
earth. 
And to the rescue comes of course none 
another than Newton who derived the 
following important expression. 
It is perhaps the only application of 
calculus we will use broadly in this 
class. 
So, I won't derive it, 
but it's a fact that if you take this 
combination, one plus x to the power, 
and if x is much smaller than one in 
absolute value, X can be negative or 
positive, then one plus X to the power A 
is, approximately given by one plus AX, 
the smaller X and the smaller A, the 
closer the approximation is, and because 
I'm not proving it, here's a few graphs 
that demonstrate this. 
this is the X axis, X is running from 
zero to a tenth. 
This is one plus x to the a for various 
values of a indicated here running from 
five to minus four, and you see that as 
long as x is not too large, this 
approximation is valid and this will 
allow us to simplify many, many 
calculations in the class. 
So, it's one, 
result of calculus that I want you to 
accept from me, and you're welcome to 
test it, 
Just as I have shown you in this graph 
you can get some more examples. 
So at the moment, let's apply it to this 
example. 
So here, H over R, the radius of the 
earth is X, which I'm going to assume is 
not too large. 
Remember a tenth, gives me an altitude of 
about, 700 kilometers, and A is negative 
two. And so the approximation yields one 
plus X to the minus two is about one 
minus 2X. 
And this gives me the approximate amount 
by which the, force of gravity decreases 
as I 
elevate myself to an altitude H above the 
surface of the ground. 
Of course when H is zero I reproduce my 
weight on the ground, and this is valid 
so long as your height is negligible or 
small compared to the radius of the 
earth. 
It supplies what we've learned to 
understanding potential energy and 
gravity. 
We've already had an expression for 
potential energy, mgh, but this was valid 
as long as we assume the force of gravity 
was constantly mg. 
We saw that as you get farther from the 
center of earth, the force decreases 
every height gain costs less energy and 
therefore taking that into account, we 
find an exact expression. 
This is the exact expression for 
gravitational potential energy, Newtonian 
gravity. 
These two do not look the same, and I 
didn't derive this because it requires 
some calculus. 
Let me convince you that they're not as 
different as they appear. 
Another opportunity to use some skills 
we've developed. 
So, let's write the potential energy 
using the correct expression at an 
altitude h, above the earth's surface. 
So I write. 
For distance from the earth's center, r 
plus h. 
And, as always, I rewrite this by scaling 
it to a known quantity. 
And, that would be this one. This is your 
potential energy on the surface of the 
earth. 
And, then I need to write this 
dimensionless quantity, which gives me 
the scaling. 
And again, I factored, I divide numerator 
and denominator by r to get -g and m over 
r times one plus h over r inverse. 
And another opportunity to apply Newton's 
formula, A is negative one, X is, your 
altitude, if your altitude is smaller 
than the earth's radius significantly, 
then Newton's formula applies. 
That replaces this by one minus H over R, 
and I will do the calculation, in my head 
and on that one, it's easy to do. 
It's this constant, number. 
And then minus H over R gives me a 
positive contribution here, an extra 
factor of R in the denominator, 
GM over R squared. 
and H, and I recognize two things here. 
One is this is independent of your 
altitude. It's just a constant, 
and this object here is G. 
So the difference from, between this 
exact ob, quantity and our approximation 
for H that is significantly smaller than 
the earth's radius is just a constant 
shift. 
We said that a constant shift is 
irrelevant and so this formula is as 
valid as this for H much smaller than the 
radius of earth and then at higher 
altitudes this becomes valid we can see 
that 
Mathematical calculation graphically 
here, this would be mgh. 
This is the exact potential energy. 
the graph shows us two salient 
properties. 
One is the potential energy is laways 
negative, that's just a choice of our 
constant that we added. 
It becomes closer to zero, it increases 
towards zero as one gets farther from 
earth. 
So, it approaches zero at infinite 
distance from earth near Earth it 
increasesm mgh and minus GMM over R are 
not the same. 
But if one shifts this down by an 
appropriate constant, then one sees that 
as long as H is much smaller than the 
radius of the Earth you get a good 
approximation. 
The statement oft heard that there is no 
gravity in space, you've seen is patently 
false. 
The earth's gravitational attraction 
extents throughout the universe. 
But it does weaken with distance. 
And so, perhaps the correct aversion 
that's times one here's, is that 
astronauts in the space station are 
weightless because the gravitational 
force of earth is weaker at larger 
distances from the center of the Earth. 
here we see Stephen Hawking weightless. 
no, no, no, the space station. 
as we have said, orbits at an altitude of 
400 kilometers. 
So, we can compute the force of 
gravitation, we can compute the weight of 
an object, say, myself, or even Hawking 
at an altitude of H above the earth. 
We found that the approximate version for 
the force of gravity for altitudes H 
smaller than the radius of the earth is 
this formula. 
This makes it easy to plug in the 
numbers, and you find that in the space 
station your wait is in fact 87 percent. 
In other words, the force with which the 
earth attracts to is 87 percent of what 
it would be on earth. 
Why then are astronauts weightless in the 
space station? 
Why is Hawking weightless in this 
picture? 
Stephen Hawking is far too important to 
humanity to allow him to travel to space 
and take the risks of real space travel. 
Hawking is weightless because, in fact he 
is accelerating towards the ground with 
an acceleration that is essentially g. 
When you are in free fall, there is no 
gravity, and there is a beautiful demo 
that explains this. 
We said that all you have to do is to be 
weightless is be in free fall. 
So what I'm going to do in this 
demonstration is I'm jumping off the box. 
I'm jumping in ultra slow motion, because 
we form a video over the high speed 
camera, and I've slowed it down further. 
And as I fall, to demonstrate that I'm 
really weightless, I'm dropping this 
little white ball. 
And what you see is that as I drop the 
ball, of course the ball's falling. 
So am I. 
To my eyes the ball is hovering in front 
of my face and it's not moving. 
And it won't stop falling in concert with 
me until my feet hit the ground. 
You can tell exactly the moment when my 
feet hit the ground because that is the 
moment when the ball is no longer level 
with my nose. 
Well, I hope you found that amusing. 
What we saw is that when you're freely 
falling, 
you, you experience no gravity. 
A ball you drop well however in front of 
your face. 
Had I thrown that ball up it would have 
relatively to my nose, moved up at a 
constant velocity, and this is important 
because in space there are no floors for 
your feet to hit and so everything is in 
free fall. For example, the earth as it 
orbits the sun in its circular motion is 
in free fall under the gravitational 
influence of the sun which means that 
when studying the motions of objects near 
if we can completely ignore the sun's 
gravity. 
We are all in free fall towards the sun. 
Really? 
Almost. 
The reason we're only almost all in free 
fall towards the sun is because we're all 
actually held together by staying on 
earth. 
So we're all falling towards the sun 
with, one constant acceleration, and that 
could not possibly be the free fall 
acceleration towards the sun, at all the 
positions we are because at different 
positions, the free fall acceleration 
towards the sun is different, depending 
on your distance from and direction to 
the sun. 
These differences, in the gravitational 
free fall acceleration at different 
points of an extended object, will lead 
to forces, tidal forces, we'll see why we 
call them tidal forces, that account for 
the differences in gravitational 
acceleration between different parts of 
an object. 
These differences manifest themselves in 
left over gravitational forces. 
Like all gravitational forces, they are 
proportional to your mass. 
So the tidal force on a refrigerator is 
more than the tidal force on a tennis 
ball. 
And they are given by your mass times the 
difference between the gravitational 
acceleration towards, say the sun, where 
you are, 
and the acceleration with which as a 
denizen of Earth you have to the R 
falling onto the Sun. 
That would be the gravitational 
acceleration towards the Sun, as it 
applies at the centre of the Earth. 
This sounds a little bit confusing and 
like many things can be clarified by a 
good calculation. 
Let's turn to that. 
Let's do this and let's do the 
calculation. 
so in this image we have the earth over 
here. 
The sun I'm going to assume is way over 
there in the left hand corner and way 
outside the image. 
And because the sun's gravitation applies 
at force to the earth given by FG equals 
G times the mass of the sun, 
divided by the distance to the sun 
squared, times the mass of the earth. 
The entire Earth is accelerating to the 
left towards the sun with an acceleration 
given by, this is equal to the mass of 
the Earth times the acceleration of the 
Earth and I can, as always with gravity, 
cancel the factors of the mass of the 
Earth and so what I find. 
Is that the entire earth is accelerating, 
towards the sun with, this acceleration, 
where, the distance to the sun I'm going 
to take as the, always, as the center to 
center distance, because that is what 
Newton taught us to do, with spherically 
symmetric objects. 
Now, the important thing is, the entire 
earth is accelerating to the left with 
this acceleration, and because this is 
the acceleration of free fall, gravity 
would cancel except. 
But if, if you are sitting at this point 
on earth, you are a little bit nearer the 
sun at the center of the earth by a very 
small amount. 
This is 6,400 kilometers, the distance of 
the sun is 150,000,000 kilometers, but 
it's a difference nonetheless so the 
gravitational acceleration towards the 
sun here is a little bit larger. 
And so the gravitational acceleration 
towards the sun at this point is larger 
because you are closer to the sun. 
Indeed, the denominator for computing 
here should be decreased by the radius of 
the earth. 
Similarly, the gravitational acceleration 
on the far side of the sun will be 
smaller than the value in the center of 
the earth because you're at a larger 
distance by the radius of the earth. 
Now, the radius of the earth is of course 
much smaller than the distance to the 
sun. 
This gives us a wonderful opportunity to 
apply our approximation that we learned 
from Newton. 
We do the usual steps. 
We write here. 
Pull out d squared of shortcut. 
The next step we will get one minus r, 
radius of the earth over distance to the 
sun to the power minus two. 
X is the small quantity radius of the 
earth divided by distance to the sun, a 
is negative two and the result is gn over 
d squared, 
which is, simply A. 
And then times one, plus two, R over D. 
And so, this is our approximate value, 
valid because the earth is much smaller, 
than the distance to the sun, for, the 
value of the acceleration, of gravity on 
the near side to the sun, 
and we can make a similar, calculation 
over here. 
The difference will be, this plus sign 
will turn into a minus sign when we 
multiply it by minus two. 
So, we can write, the acceleration over 
here, in a similar form. 
Now this allows us, that suggests, 
rewriting, this is a sum of two terms. 
The acceleration of with which the Earth 
approaches, and then an extra little bit 
so an object over here has a free-fall 
acceleration towards the sun that is 
larger than this one by the amount of 
this little yellow arrow which signifies 
A times 2R over D, and similarly an 
object on the far side of the sun has a 
gravitational acceleration toward the sun 
that is less than the acceleration with 
which the earth moves toward the sun by 
approximately the same amount. 
To this order of approximation. 
Now, one can apply a similar logic to 
these positions on Earth. 
Their distance from the Sun to first 
order, is the same as that of the center 
of the Earth. 
So, one asks, why is the gravitational 
acceleration here, different from that at 
the center of the Earth? 
Well, that is because to a very small 
extent the direction from here to the Sun 
is not the same as the direction from 
here to the Sun, as these arrows and 
again, a grossly exaggerated version 
demonstrate, this would have the Sun 
about here. 
The Sun should be very far, so this angle 
should be tiny, but again I can rewrite 
these as the gravitational acceleration 
at the center of the earth, plus a small 
correction. 
And so, the net result is that the 
gravitational acceleration with which the 
earth is moving towards the sun, that 
part of the sun's gravity is cancelled by 
the earth's motion. 
But the leftovers, 
these yellow little differences, are not 
cancelled. 
So if I put those differences back in 
their place, I see that the net result is 
that the Sun exerts a small force upward 
away from Earth, 
on this side of the Earth, 
and a small force upward away from the 
Earth on this side of the Earth. 
And a small force downward on this side 
of the Earth. 
In other words, the Sun, is doing its 
best to smoosh the Earth into this, oval 
that is elongated in the direction of the 
sun. 
And so of course the Earth is kind of a 
rigid object. 
The sun cannot manage to deform the 
Earth, but the Earth is under some kind 
of tidal stress. 
Now before we get too excited, let's 
figure out how strong this force is that 
we've calculated. 
Again, the magnitude of the force depends 
on what it's acting on. 
These are all accelerations, so imagine a 
one kilo rock and figure out the net 
force that this has. 
Or better yet. 
Compute an acceleration and compare it to 
the acceleration of gravity so we know, 
is this rock essentially rendered, 
Weightless as it lifted off the earth by 
these tidal forces. 
So we can do the calculation. 
We have here the. 
All the quantities we need. 
The title acceleration due to the sun is 
the difference. 
That little yellow difference, which was 
2GM over D squared times R over D. 
The salient point is that because I have 
2GM over D square times R over D, this 
decreases with distance from the sun like 
the third power of distance. 
Remember the gravitational force itself 
decreases like the second power. 
And I want to compare this acceleration 
to something that I know. 
So I will do the usual scaling trick that 
I like to use. 
Try to pull out a factor over here of 
The mass of the earth divided by the 
radius of the earth squared. 
I need didn't have the mass of the earth 
I had the mass of the sun so I get one 
factor here. 
I had the radius of the earth. 
Divided by the distance to the Sun cubed. 
So, I get a scaling factor, notice I have 
an acceleration that is known, this is 
our friend g. 
These are dimensionless factors, so I am 
relating an acceleration to an 
acceleration. 
And, plugging in the ratio of the Sun's 
mass to the earth, and the cube of the 
ratio of the earth's radius to the 
distance to the Sun, I find that the 
acceleration, the tidal acceleration due 
to the Sun is very small. 
It's five parts in a 100 million, 
the acceleration of gravity. 
So no it is not about to rip the earth 
apart. 
This is. 
The, the grav, tidal acceleration due to 
the sun. 
What else might exert a tidal influence 
on the earth? 
Well, the second 
Object that we might imagine would be the 
moon. 
Now the moon is much less massive than 
the sun but it is closer and so because 
of this, third power of the distance one 
might imagine that the tidal force 
generated by the moon might be comparable 
to that generated by the sun and indeed a 
simple scaling argument. 
Tells us that, to get the tidal 
acceleration due to this moon relative to 
that from the sun, I multiply by the 
ratio of the masses, and the ratio of the 
distances to the, power negative three. 
Plugging all the numbers in, I find, that 
in fact, the tidal acceleration due to 
the moon is approximately twice, that due 
to the sun. 
So the dominant tidal effects on earth, 
are. 
Lunar tides, rather than solar tides. 
but. 
Solar tides are not negligible. 
They are a 50% effect. 
The net result is a combination of the 
two. 
Let's take a look at what that does. 
So here we see the earth and the moon of 
course, drawn way too close to earth to 
be to scale. 
And we see the elongated shape into which 
the moon is attempting to draw the earth 
again, 
Exaggerated way out of proportion. 
Remember though that while the Earth is 
rigid and unable to respond to this tidal 
pull except very slightly, because it's a 
rigid rocky object, there is a soft 
component of Earth, namely the water. 
So the surface of the oceans will deform 
in this direction. 
Ever so slightly. 
again not to this extent but there will 
be high tide on this side of the earth 
and a low, high tide on this side of the 
earth. 
And low tide at these points, because of 
the, tidal information that we discussed. 
And of course, as the moon orbits the 
earth, the 
tidal bulge will point towards the Moon 
and as the Earth rotates, once a day, 
each point will, each point on Earth will 
appear, have two high tides and two low 
tides. 
And these because you pass through these 
two points and these once a day. 
And, because of the Moon's motion along 
the celestial sphere, tides will repeat 
not every 24 hours, but every 24 hours 
and 44 minutes, 
because the moon advances by 44 minutes 
relative to the sun, remember. 
And so, indeed if you look up the table 
of tides you'll find the tide repeats 
every 24 hours and 44 minutes. 
and, and this is because they are 
associated to the moon. 
But, there, there is another effect, that 
two effects that come into play. 
One is. 
We can include the sun. 
So, in this picture, the sun is way off 
to the right. 
The sun is attempting to extend the earth 
as the horizontally extended oblong. 
Where as the moon, attempts to draw the 
earth to whatever direction points 
towards the moon. 
When the moon is full. 
Earth sun and moon are in line and the 
sun's deformation, the moon's deformation 
are in exactly the same direction. 
We get very, very exaggeratedly high 
tides, they are called spring tides. 
When we have a quarter moon, the moon is 
elongating the earth this way, and the 
sun this way, and the tides are 
diminished, we call them neap tides. 
Tides are very weak at quarter moon, and 
of course when we get a new moon, the 
earth and the sun are elongating the 
earth in the same direction and tides are 
stronger. 
So, tides are strongest at new moon and 
at full moon. 
Those are the spring tides, Tides are 
weakest at quarter moon. 
Those are the neap tides, and Newton's 
theory makes easy sense of all that. 
And then there's another effect that we 
can take into account. 
The earth rotates. 
What the Earths rotation does is as the 
moon creates a bulge of water the 
rotation of the Earth, which is much 
faster than the moons orbit by a factor 
of 30 remember, drives this bulge so that 
water that was raised here is actually, 
over here. 
This means. 
farther to the east. 
This means that the high tide bulge is a 
little bit to the east of the moon, which 
means that high tide where you are is a 
little bit later than the time the moon 
crosses your meridian. 
it also has other implications which 
we'll discuss momentarily. 
But these are the three effects that this 
demonstration shows very convincingly. 
To summarize what we've seen, the moon, 
Attempts to deform, the earth. 
It succeeds in deforming, the old water 
in the oceans, so there's a bulge facing 
the moon. 
As the earth rotates the bulge moves 
around the earth, so, everybody gets two 
high tides, and two low tides, once 
every, 24 hours and 44 minutes so the 
time between, one high tide and the next, 
is in fact about twelve and a half hours 
rather than twelve. 
The earth's rotation drags the bulge to 
the east, so therefore it lags the moon 
the. 
Like everything else in that image the 
lag was way exaggerated. 
In fact it turns out to be about three 
degrees or twelve minutes. 
Now in addition to the Moon the Sun 
exerts a tidal force directed towards the 
Sun with about half as, the strength as 
that of the Moon. 
At full or new Moon, these act in concert 
creating, creating the intense tides we 
call spring tides, which have nothing to 
do with spring. 
Whereas at quarter Moon in either 
direction, the Sun's deformation 
counteracts the Moon's deformation and 
you get the relatively weak leap tides. 
Pull. 
Something we understand, even more, 
consequences of tidal forces. 
the moon is not able to deformed the 
earth, but when the moon formed a long 
time ago, much closer to the earth, we'll 
see, and molten. 
it was soft. 
And closer to earth. 
And the tidal force of earth. 
Remember at a closer distance with a mass 
bigger than the mass of the Moon you get 
more intense tidal forces. 
The tidal forces of Earth deform the Moon 
so that when it solidified it froze with 
a permanent bulge. 
The Moon is not quite round. 
Now the tidal forces act to keep this 
bulge always aligned in the direction of 
earth. 
We call this tidal locking. 
And this is the reason why, as I told you 
a long time ago, when we look up at the 
moon, we always see the same side of the 
moon. 
There is a near side and a far side to 
the moon, where the locks, the moon's 
spin period about its axis to its orbital 
period about the earth is tidal locking, 
and this is a phenomenon we'll see 
before. 
Moreover, since the tidal bulge on earth 
is being dragged to the east of the moon, 
the tidal force from the moon attempting 
to align the tidal bulge with the moon, 
is actually trying to slow down the 
rotation of the earth. 
Think about it differently, 
there is friction between the title bulge 
and the rotating earth, because the title 
bulge cannot keep pace with the rotating 
earth. 
It's held, locked three degrees away from 
the moon. 
What this is doing is it's slowing down 
of the earth rotation? 
Yes. 
The Earth's rotation is slowing down. 
But, and, but, but, Roland, what about 
conservation of angular momentum? 
Well angular momentum is conserved. 
The way it's conserved is that, that same 
bulge acts on the moon, to in in fact, 
sling it faster along its trajectory, so 
that the moon is being accelerated, and 
as it gains angular momentum, it's 
actually moving to higher orbits. 
So the moon is actually receding from 
earth. 
This is an observation first argued 
strongly by George Darwin, son of the 
famous Charles Darwin in 1898, 
and was the reasoning behind his theory 
that, when the moon formed, it was indeed 
much nearer earth, and as we will see 
that coincides with modern theories we 
think that the moon formed much closer to 
the earth. 
And, these were conjunctures until the 
Apollo missions left, reflectors on the 
moon enabling us to make a very precise 
measurement of the distance to the moon. 
And indeed we can now verify 
experimentally that the moon is receding 
form earth. 
So remember it is now exactly the angular 
size of the sun in the sky. 
In a few millennia total solar eclipses 
maybe a thing of the past. 
So we worked pretty hard, but we're 
getting somewhere. 
Understanding universal laws has given us 
a unifying understanding of many, many 
phenomena. 
So, we understand tidal forces. 
And remember, in space. 
Nothing is held up. 
Everything then is in free fall. 
Everything is an orbit around everything 
else. 
And so tidal forces are extremely 
important. 
this is very nicely evident in this 
beautiful image of M51, the whirlpool 
nebula. 
And what we see is that these are in 
fact, two galaxies that are in the 
process of colliding. 
Galactic collisions are slow and rather 
delicate but we see that there's a tail. 
Of stars from this galaxy being dragged 
out towards it's neighbor by the 
gravitational force. 
And now that we understand tidal forces, 
it's not surprising that there is an 
equal tail being dragged out by the same 
tidal forces because these are in free 
fall about each other. 
And the other side of the galaxy, tidal 
forces are approximately symmetric. 
These are falling in because the 
gravitational acceleration is larger than 
that with which the galaxy is falling, 
these because it's smaller. 
The net result is the galaxies being 
stretched in both directions. 
So understanding the uses and 
applications of F equals MA is a very 
powerful tool and we're going to continue 
to apply it as we go along. 
And what we're going to turn to next is 
understanding a little bit more about the 
forces in nature beyond gravity. 
What else is there? 
And something about the M, something 
about the structure of the matter upon 
which these forces act. 
This will be important for understanding 
the structure and workings of various 
astronomical phenomena. 
And so that's what we'll turn to in the 
next clip.