- [Voiceover] What does period and frequency mean?
The period is the number of seconds
it takes for a process to complete
an entire cycle, circle, or revolution.
So, if there's some repeating process,
the time it takes that process to reset is the period,
and it's measured in seconds.
The frequency is the number of cycles,
or circles, or revolutions completed in one second.
So, if there's some process that's repeating,
the number of times the process repeats
in one second would be the frequency.
This means it has units of one over second,
which is just called, the hertz.
And because the period and frequency are defined
in this inverse way as seconds per cycle
or cycles per second, each one
is just the inverse of the other.
In other words, the period is
just one over the frequency, and
the frequency is equal to one over the period.
One example of a repeating process
is an object going in a circle at a constant speed.
If this is the case, you can relate
the speed, the radius of the circle,
and the period of the motion since speed
is just distance per time, and the distance
the object travels in one cycle is two pi R
the circumference, the speed would
just be two pi R per the period,
or since one over the period is the frequency,
you could write the speed as two pi R times the frequency.
Since time is not a vector these quantities
are not vectors and they cannot be negative.
So, what's an example involving period
and frequency look like?
Let's say moon travels around a planet
in a circular orbit of radius R at a constant speed S.
And we wanna know what the period and frequency are
in terms of given quantities and
fundamental constants, so we'll use the relationship
between the speed, the period, and the frequency.
We know that for object in circular motion
the speed is two pi R over the period.
And that means the period here
would be equal to two pi R over the speed.
And since frequency is one over the period,
if we take one over this quantity
we just flip the top and bottom
and we get that this is the speed over two pi R.
But we can't leave our answer in terms of V.
We had to express this in terms of given quantities.
We were given S, so our answer for the period
has to be two pi R over S, and for frequency
it would be S over two pi R, which is C.
What is centripetal acceleration?
The centripetal acceleration of an object
is the acceleration that's causing
that object to go in a circle.
And it's important to note that
this centripetal acceleration always points
toward the center of the circle.
The formula to find the centripetal acceleration
is speed squared divided by the radius
of the circle the object is travelling in.
Even though this has a bit of an exotic formula
for the acceleration, it's still an acceleration,
so it still has units of meters per second squared,
and it is a vector, which means it does
have a direction, i.e toward the center of the circle.
But this centripetal acceleration
does not cause the object to speed up or slow down.
This centripetal acceleration is only changing
the direction of the velocity.
If the object going in the circle
is also speeding up or slowing down,
there's also gotta be a component
of the acceleration that's tangential to the circle,
in other words, if the object
is going in a circle and speeding up,
there's gotta be a component of acceleration
in direction of the velocity,
and if the object is slowing down,
there's gotta be a component of acceleration
in the opposite direction to the velocity.
So, centripetal acceleration changes the direction
of the velocity, and tangential acceleration changes
the magnitude or size of the velocity.
But this formula of V squared over R
is only giving you the magnitude
of the centripetal acceleration.
This does not account for any tangential acceleration.
So, what's an example problem
involving centripetal acceleration look like?
Let's say particle A is travelling in a circle
with a constant speed S and a radius R.
If particle B is travelling in a circle
with twice the speed of A and twice the radius of A,
what's the ratio of the acceleration
of particle A compared to particle B.
So, particle A is gonna have a centripetal acceleration
of the speed squared over the radius,
and particle B is also gonna have an acceleration
of the speed squared, but this speed
is twice as much of the speed of particle A,
and is travelling in a circle
with twice the radius of particle A.
When we square the two we'll get four over two,
gives us a factor of two times the speed
of A squared over the radius of A.
So, the ratio of the acceleration of particle A
compared to particle B is gonna be one half
since the acceleration of particle A
is half the acceleration of particle B.
Centripetal forces are not a new type of force,
centripetal forces are just one
of the any other forces that we've already met
that happen to be pointing towards the center
of the circle making an object travel in a circle.
So, for a moon going around the Earth,
gravity is the centripetal force.
For a yo-yo going around on a string,
the tension is the centripetal force.
For a skateboarder doing a loopty loop,
the normal force is the centripetal force.
And for a car going around a roundabout,
the static frictional force is the centripetal force.
And these forces still follow Newton's second law,
but using centripetal forces means you're
also going to have to use the expression
for the centripetal acceleration.
Now, if a force is directed radially inward
toward the center of the circle,
you would count that force as positive
since it points in the same direction
as the centripetal acceleration.
And if a force points radially out
from the center of the circle,
you would count that as a negative force.
And if force is directed tangential to the circle,
you wouldn't include it in this calculation at all.
You can include those forces in their own
Newton's second law equation,
but you wouldn't be using V squared over R
for that acceleration.
Those tangential forces change the speed of the object,
but the centripetal force changes
the direction of the object.
So, what's an example problem involving
centripetal forces look like?
Imagine a ball of mass M rolling over
the top of the hill of radius R at a speed S.
And we wanna know, at a top of the hill,
what's the magnitude of the normal force
exerted on the ball by the road.
So, we'll draw our force diagram.
There's gonna be an upward normal force
on the ball from the road, and there's
gonna be a downward force of gravity
on the ball from the Earth, and these two forces
are not going to be equal and opposite.
If they were equal and opposite
they would balance, and if the forces are balanced
the object would maintain its velocity
and keep travelling in a straight line.
But this ball doesn't travel in a straight line,
it starts accelerating downward.
So, this normal force is gonna have to be less
than the force of gravity.
To figure out how much less, we can use
Newton's second law with the formula
for centripetal acceleration.
The speed is S, the radius is R,
the force of gravity is gonna be a positive
centripetal force since it's directed toward
the center of the circle.
The normal force is gonna be a negative centripetal force
since it's directed radially away from
the center of the circle.
Then we divide by the mass, which,
if you solve this for normal force,
gives you the force of gravity
minus M, S squared over R, which makes sense
'cause this normal force has to be less
than the force of gravity.
Newton's universal law of gravity states that
all masses in the universe pull, i.e attract
every other mass in the universe
with gravitational force.
And this force is proportional
to each mass, and inversely proportional
to the square of the center to center distance
between the two masses.
In mathematical form, it just says
that the force of gravity is equal to big G,
a constant, which is 6.67 times 10 to the negative 11th
multiplied by each mass in kilograms,
and then divided by the center to center distance
between the two masses, in other words,
not the surface to surface distance,
but the center to center distance.
And even if these two objects have different
masses, the magnitude of the force
they exert on each other is gonna be the same.
This is illustrated by the formula
since you could swap these two masses
and you get the same number.
And it's also something we know from Newton's third law.
This force of gravity is a vector
and it has a direction, the direction is always such
that it attracts every other mass,
and since this is a force, the unit is in Newtons.
So, what's an example problem involving
Newton's universal law of gravity look like?
Let's say two masses, both of mass M,
exert a gravitational force F on each other.
If one of the masses is exchanged for a mass 3M
and the center to center distance between
the masses is tripled, what would
the new gravitational force be?
We know the gravitational force
is always big G times one of the masses multiplied
by the other mass divided by the center to center
distance squared.
So, the initial force between the two masses
would be big G M times M over R squared,
but the new force with the exchanged values would be
big G times 3M times M divided
by three times the radius squared.
The factor of three squared on the bottom
gives nine, and three divided by nine
is one over three times big G M M over R squared.
So, we can see that the force with the new values
1/3 of the force with the old values.
What's gravitational field mean?
The gravitational field is just another word
for the acceleration due to gravity near an object.
You can visualize a gravitational field
as vectors pointing radially in toward a mass.
All masses create a gravitational field that points
radially in toward them and dies off like
one over R squared the farther you get away from them.
So, the formula for the gravitational field little G
created by a mass M is big G times
the mass creating the field divided by the distance
from the center of the mass to the point where
you're trying to determine the value of the field.
And again, this value for the gravitational field
is going to be equal to the value
for the acceleration due to gravity
of an object placed at that point.
The gravitational field is a vector
since it has a direction, i.e. toward the center
of the object creating it.
And since gravitational field is equivalent
to acceleration due to gravity,
the units are meters per second squared,
but you could also write that as newtons per kilogram,
which is another way of thinking about what
gravitational field means.
Not only is it the acceleration due to gravity
of an object placed at that point,
but it's the amount of the gravitational force
exerted on a mass M placed at that point.
So, you could think of the gravitational field
as measuring the amount of gravitational force
per kilogram at a point in space,
which when rearranged gives you the familiar formula
that the force of gravity is just M times G.
So, what's an example problem involving
gravitational field look like?
Let's say a hypothetical planet X had three times
the mass of Earth and half the radius of Earth.
What would be the acceleration due to gravity
on planet X, i.e. the gravitational field
on planet X, in terms of the acceleration
due to gravity on Earth, which is GE.
So, we know that the gravitational field on Earth
has to be big G times mass of the Earth
over the radius of Earth squared,
which we're calling G sub E, and
the gravitational field on planet X would be
big G times three times the mass of the Earth
divided by half the radius of the Earth squared,
and when we square this factor of a half
we'll get 1/4, which is in the denominator,
so three divided by a 1/4 is 12 times big G
mass of the Earth over radius of the Earth squared,
and since this entire term here is the acceleration
due to gravity on Earth, the acceleration due
to gravity on planet X is gonna be 12 times
the acceleration due to gravity on Earth.
Sometimes when you're solving gravitational problems,
you'll be given the density instead of the mass.
The density is the amount of mass per volume
for a given material.
The symbol for density is the Greek letter rho,
and you can find it by taking the mass
divided by the volume.
So, the units of density are kilograms per meter queued.
And it's not a vector since it has no direction,
but it does let you solve for mass.
If you know the density you could say
that the mass is the density times the volume.
So, what's an example problem involving density look like?
Let's try the hypothetical planet problem again,
but this time instead of being told
that planet X has three times the mass of Earth,
let's say that planet X has three times
the density of Earth, and again, half the radius of Earth.
What would be the acceleration due to gravity
on planet X in terms of the acceleration
due to gravity on Earth GE.
We could write down the formula
for gravitational acceleration or gravitational field,
which is big G M over R squared,
but this time we don't know the mass,
we just know the density, so we want to rewrite
this formula in terms of density,
which we can do by rewriting the M as rho times V,
since density is mass per volume,
and mass is density times volume.
But we don't know the volume of this planet,
we just know the radius, so we need to rewrite volume
in terms of radius, which we could do
since planets are spherical and the volume of a sphere
is four thirds pi R cubed.
We can substitute this expression in
for the volume and finally get an expression
for the acceleration due to gravity of big G
times rho four thirds pi R cubed divided by R squared.
And we can cancel an R squared on the top and the bottom,
which leaves this little G as equaling big G
rho four thirds pi R.
So, the gravitational acceleration on Earth
would be big G rho of Earth four thirds pi
times the radius of Earth.
And the gravitational acceleration
on planet X would be big G times
the density of planet X, which is three times
the density of Earth times four thirds pi
times the radius of planet X, which is one half
the radius of Earth, which when we pull out
the three and factor of a half,
gives us three halves times the expression
for the acceleration due to gravity on Earth.
So, the gravitational acceleration
on planet X is gonna be three halves
the gravitational acceleration on planet Earth.
Gravitational orbits are just a special case
of centripetal acceleration where some object
is orbiting another object due
to the gravitational force.
And if that orbit is a circle,
we can relate the speed, the radius of the orbit,
and the larger mass to each other
using Newton's second law and centripetal acceleration.
You just plugin the acceleration
as the centripetal acceleration, V squared over R,
and since the centripetal force is the force of gravity,
you can plugin the expression for
the force of gravity as the centripetal force,
which is big G M M over the distance between them squared.
And since the mass of the orbiting object cancels
we get an expression that relates to the speed
of the orbiting object, the larger mass
that's pulling that object in,
and the center to center distance between the objects,
which, if we solve this for V,
gives us the square root of big G times the mass
pulling in the object divided by
the center to center distance between the objects.
Note that this formula does not depend on the mass
that's in orbit since that mass canceled out
in the calculation.
So, what's an example problem involving
gravitational orbits look like?
Well, imagine a space station of mass MS
is orbiting at an altitude of 3R
above a planet of mass MP and radius R
as seen in this diagram, and then imagine
a different space station of mass 3MS
is orbiting at an altitude of 2R above a planet
of mass 4MP and a radius of 2R as seen in this diagram,
and we wanna know, if the speed
of the space station of mass MS is V,
then in terms of V, what's the speed
of the space station of mass 3MS?
Well, we just showed that the speed
of an orbiting object is gonna be equal
to the square root of big G times
the mass of the larger object pulling in
the smaller object divided by the
center to center distance between the objects.
And since this formula doesn't involve the mass
of the orbiting object, it doesn't matter
that the objects have different masses,
but the mass of the planet can make a difference.
So, to get the speed of the space station MS,
we could say that it's the square root of Big G
the mass of planet P over the center to center distance,
which is not gonna be the radius
of the planet or the altitude,
it's gonna be the radius of the planet plus the altitude
since this has to be the center to center distance,
which in this case will be 3R plus R, which is 4R.
And now to get the speed of the space station
of mass 3MS, we'll use the same formula,
which is big G mass of the planet, which
in this case is 4MP divided by
the center to center distance,
which in this case would be 2R plus 2R,
and again that's 4R, and if we compare,
the only difference between these expressions
is that there's an extra factor of four within
this square root.
So, if we take that factor out, the square root
of four is two, we'd get two times the expression
for the speed of the space station MS.
So, the space station 3MS is travelling two times
the speed of the space station MS.