Welcome back.
So where I left off, I was hopefully trying to give you
an intuition of this inward acceleration.
In this case, I have a rock spinning around, and I'm
holding it on a string.
And I want to think about what the inward acceleration is
dependent on.
And so we talked a little bit about its velocity in these
multiple red arrows.
These are potential velocity vectors.
So the point I was trying to make is, if the velocity
vector is bigger, to change its direction requires even
more acceleration.
Because the change in the x direction is bigger, and then
the change in the y direction is bigger.
Because for example, right here, it's y-- I'll draw it in
a different color-- it's y velocity vector.
And I'll do it for the big velocity.
Its y velocity vector was here.
And then now its y velocity vector is almost essentially
the whole vector.
So when the velocity increases, and let's say this
is the case of the big arrow.
If the velocity was less, like the small arrow, its velocity
of vectors was here, and it would have to go to here.
So I don't know if it's completely obvious, but the
change in velocity, for example in the y direction
over this course of time, has to be more.
And likewise, the change in x direction over this, and this
might be more obvious, because here this is the x velocity
vector if the velocity was less.
And this is the x velocity vector if it was more.
And that's going to 0.
Between when the dot goes from here to here.
Here it has no x component.
So clearly, I would have had to decelerate in the x
direction more, if we have a larger velocity vector.
I'm either giving you an intuition, or I'm confusing
you, but I'll keep going.
So velocity clearly-- the greater the velocity, the more
I'm going to have to pull in on this one.
The more the inward acceleration has to be to keep
this thing going in the circle.
And not only does the acceleration have to be bigger
just to get that velocity vector down, but you've got to
realize when, between this point and this point, the
faster I go, the less time it's going to take me to get
from here to here.
So not only am I going to have to change the velocity more, I
have to do that in less time.
So the velocity is affecting this acceleration that I need
to pull in two ways.
So the higher the velocity, I have to change more of the
velocity, and I have to do it in less time.
And it actually turns out that the acceleration is
proportional to the velocity squared.
And then there's another term, and I won't
prove it in this video.
I'm going to prove it in another video.
And that might be out of your scope just now, because it
requires some calculus.
But I want to give you an intuition.
So we know that the acceleration is dependent on
the velocity squared.
And I want to make that point of why it's not just velocity,
because there's two ways that the velocity is affecting how
much of an acceleration I need.
One, is just the magnitude of the velocity.
The more that is, the more I have to accelerate to change
this direction.
And the second is, between this point and this point, I
have less time to change that velocity.
So the velocity affects it in two ways.
And that's where you get an intuition for the v squared.
The other thing that matters is actually the radius.
So let's say this is the radius.
And the acceleration is inversely
proportional to the radius.
And why is that?
Well, it's kind of like the second, I guess we can call
it, velocity argument I gave. The bigger the radius.
So let's say that, let's look at this rate and let's look at
another radius that is bigger.
Let's say that this is another radius.
Let's say if the string was longer and
this was another case.
Let's says this was the case.
But the velocity was the same.
So this is the case with a larger-- so let's say it's the
same velocity vector.
So we have the same thing at play, to go from
there to say, there.
We have to change the velocity the same magnitude.
As we had to change the velocity from here to here.
But one thing changes.
This distance is longer.
So if this velocity is the same as this velocity, but
this distance is longer, you have more time
to change the velocity.
So the larger the radius, the more time you have to change
the velocity.
And so the less acceleration you need.
Hopefully that makes sense.
Let me repeat it.
Because I think it's nice to have the intuition.
Between this point and this point, unless you have a
smaller radius, if I want to change the acceleration vector
from that to that, I have however long it takes to go
this distance.
I have that long to change its velocity vector.
But if the radius was a little longer.
And I had the same velocity, so here the velocity is that.
And then here, the velocity is that.
I have a little bit more time to change its velocity.
And remember, acceleration is just change in velocity over
change in time.
And so if I have the same change in velocity, but I have
more time to do it with.
This is a bigger number.
The required acceleration is less.
So maybe I've given you an intuition.
And I will prove this for sure.
Because this is kind of a very touchy, very soft way, and not
very rigorous way of proving that the required acceleration
to make something going in a circle is v squared over r.
But accept that as a bit of faith, and I'll prove it.
Especially once you learn your calculus, and you can watch
that video.
And we could do a couple of problems. And this is probably
what you care about anyway.
But I think it's nice to have that intuition.
So once again, let's say we're in deep space.
Let's do a little orbit problem.
Let say that this is Earth.
I will do Earth in blue.
Let's say that is Earth.
And I have a satellite that is going around Earth.
And I'm not going to calculate its-- this is a satellite
that's going around Earth.
And I'm going to do a whole other video on actually
figuring out the gravitational force depending on how far
away you are from an object.
But let's just assume that its gravitational force is similar
to what it is on the surface of Earth.
And that is a big assumption, or let's assume this is a
different planet and I'm going to give you the gravitational
acceleration.
Let's say that the planet, and maybe this isn't Earth, is
pulling inwards.
It always pulls towards its center.
And I'll do that in a different color.
That it's pulling inwards at 30 feet per second.
My question is, what does the velocity of the satellite have
to be for the object to stay in orbit?
Well, we could just apply the formula and then maybe we can
have a little intuition.
So acceleration is equal-- oh, I'm sorry.
Let's say that the object is 6,000 feet from the center.
And when we learn gravity, we'll learn actually that it's
the distance from the radius that you're going around is
actually relative to the center.
Well, actually it's always going to be,
because this is a circle.
So, the radius.
This is the center of that circle.
You don't have to worry about that.
It has nothing to do with gravity.
OK, so the acceleration is v squared over r.
Acceleration, we know is 30 feet per second.
We want to figure out the velocity, and then the radius,
I said, is 6,000 feet.
So let's ignore the units for now.
This could make things complex.
Multiply both sides of this equation times 6,000, so 3
times 6 is 18.
You'll have four 0's.
2, 3, 4.
So v squared is equal to 18,000.
Everything's in feet per second.
And so what's v?
So v is equal to the square root of 180,000.
Let me get the calculator out, because I haven't memorized my
perfect squares that high.
1, 2, 3.
Take the square root, so 424.
So the required velocity is 424 feet per second.
So if the satellite is at that velocity, it will go in
perfect orbit around this planet.
Now what happens if this object's
velocity is a little slower.
What happens-- let me do it in a different color so you know
what I'm talking about.
What happens if its velocity is a little
bit less than that?
Let's say its is 300 feet per second.
Well, then it's not going to be able to travel far enough
tangent to the circle to-- well, it'll
essentially just do this.
It'll always be getting a little bit closer.
And it'll spiral in and it'll hit the planet.
And another way to think about what the required velocity is,
you have to go fast enough so that you're always falling in.
But every time you fall in, you want to go in this case,
the right enough so that you're an equal distant--
you're still the same distance away from the center.
So that your distance from the center never changes if you're
going fast enough.
And then what happens if its velocity is too fast?
What happens in the case where its velocity is
600 feet per second?
Well, in that case, the object will have enough velocity to
actually escape the orbit of the planet.
But anyway, that's enough about that.
I'm actually out of time, so I'll see
you in the next video.