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