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.