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AP Physics 1 review of Centripetal Forces | Physics | Khan Academy

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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
  • 12:33 - 12:35
    times the radius of Earth.
  • 12:35 - 12:37
    And the gravitational acceleration
  • 12:37 - 12:39
    on planet X would be big G times
  • 12:39 - 12:41
    the density of planet
    X, which is three times
  • 12:41 - 12:44
    the density of Earth times four thirds pi
  • 12:44 - 12:47
    times the radius of planet
    X, which is one half
  • 12:47 - 12:49
    the radius of Earth,
    which when we pull out
  • 12:49 - 12:51
    the three and factor of a half,
  • 12:51 - 12:53
    gives us three halves times the expression
  • 12:53 - 12:56
    for the acceleration
    due to gravity on Earth.
  • 12:56 - 12:57
    So, the gravitational acceleration
  • 12:57 - 12:59
    on planet X is gonna be three halves
  • 12:59 - 13:02
    the gravitational
    acceleration on planet Earth.
  • 13:02 - 13:04
    Gravitational orbits
    are just a special case
  • 13:04 - 13:07
    of centripetal acceleration
    where some object
  • 13:07 - 13:08
    is orbiting another object due
  • 13:08 - 13:10
    to the gravitational force.
  • 13:10 - 13:11
    And if that orbit is a circle,
  • 13:11 - 13:14
    we can relate the speed,
    the radius of the orbit,
  • 13:14 - 13:16
    and the larger mass to each other
  • 13:16 - 13:19
    using Newton's second law
    and centripetal acceleration.
  • 13:19 - 13:21
    You just plugin the acceleration
  • 13:21 - 13:23
    as the centripetal
    acceleration, V squared over R,
  • 13:23 - 13:26
    and since the centripetal
    force is the force of gravity,
  • 13:26 - 13:28
    you can plugin the expression for
  • 13:28 - 13:30
    the force of gravity as
    the centripetal force,
  • 13:30 - 13:34
    which is big G M M over the
    distance between them squared.
  • 13:34 - 13:37
    And since the mass of the
    orbiting object cancels
  • 13:37 - 13:39
    we get an expression
    that relates to the speed
  • 13:39 - 13:41
    of the orbiting object, the larger mass
  • 13:41 - 13:42
    that's pulling that object in,
  • 13:42 - 13:46
    and the center to center
    distance between the objects,
  • 13:46 - 13:47
    which, if we solve this for V,
  • 13:47 - 13:51
    gives us the square root
    of big G times the mass
  • 13:51 - 13:53
    pulling in the object divided by
  • 13:53 - 13:55
    the center to center
    distance between the objects.
  • 13:55 - 13:57
    Note that this formula
    does not depend on the mass
  • 13:57 - 14:00
    that's in orbit since
    that mass canceled out
  • 14:00 - 14:01
    in the calculation.
  • 14:01 - 14:03
    So, what's an example problem involving
  • 14:03 - 14:04
    gravitational orbits look like?
  • 14:04 - 14:07
    Well, imagine a space station of mass MS
  • 14:07 - 14:10
    is orbiting at an altitude of 3R
  • 14:10 - 14:13
    above a planet of mass MP and radius R
  • 14:13 - 14:16
    as seen in this diagram, and then imagine
  • 14:16 - 14:18
    a different space station of mass 3MS
  • 14:18 - 14:21
    is orbiting at an altitude
    of 2R above a planet
  • 14:21 - 14:26
    of mass 4MP and a radius of
    2R as seen in this diagram,
  • 14:26 - 14:28
    and we wanna know, if the speed
  • 14:28 - 14:31
    of the space station of mass MS is V,
  • 14:31 - 14:33
    then in terms of V, what's the speed
  • 14:33 - 14:35
    of the space station of mass 3MS?
  • 14:35 - 14:37
    Well, we just showed that the speed
  • 14:37 - 14:38
    of an orbiting object is gonna be equal
  • 14:38 - 14:41
    to the square root of big G times
  • 14:41 - 14:43
    the mass of the larger object pulling in
  • 14:43 - 14:45
    the smaller object divided by the
  • 14:45 - 14:48
    center to center distance
    between the objects.
  • 14:48 - 14:50
    And since this formula
    doesn't involve the mass
  • 14:50 - 14:52
    of the orbiting object, it doesn't matter
  • 14:52 - 14:54
    that the objects have different masses,
  • 14:54 - 14:56
    but the mass of the planet
    can make a difference.
  • 14:56 - 14:58
    So, to get the speed of
    the space station MS,
  • 14:58 - 15:01
    we could say that it's
    the square root of Big G
  • 15:01 - 15:04
    the mass of planet P over the
    center to center distance,
  • 15:04 - 15:06
    which is not gonna be the radius
  • 15:06 - 15:08
    of the planet or the altitude,
  • 15:08 - 15:11
    it's gonna be the radius of
    the planet plus the altitude
  • 15:11 - 15:14
    since this has to be the
    center to center distance,
  • 15:14 - 15:17
    which in this case will
    be 3R plus R, which is 4R.
  • 15:17 - 15:19
    And now to get the speed
    of the space station
  • 15:19 - 15:22
    of mass 3MS, we'll use the same formula,
  • 15:22 - 15:24
    which is big G mass of the planet, which
  • 15:24 - 15:26
    in this case is 4MP divided by
  • 15:26 - 15:28
    the center to center distance,
  • 15:28 - 15:31
    which in this case would be 2R plus 2R,
  • 15:31 - 15:33
    and again that's 4R, and if we compare,
  • 15:33 - 15:35
    the only difference
    between these expressions
  • 15:35 - 15:37
    is that there's an extra
    factor of four within
  • 15:37 - 15:39
    this square root.
  • 15:39 - 15:41
    So, if we take that factor
    out, the square root
  • 15:41 - 15:44
    of four is two, we'd get
    two times the expression
  • 15:44 - 15:46
    for the speed of the space station MS.
  • 15:46 - 15:50
    So, the space station 3MS
    is travelling two times
  • 15:50 - 15:53
    the speed of the space station MS.
Title:
AP Physics 1 review of Centripetal Forces | Physics | Khan Academy
Description:

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Video Language:
English
Team:
Khan Academy
Duration:
15:54

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