in this video we're looking at a uniform circular motion circular orbits problem and we're given a
little bit of context on this this is about the lunar reconnaissance orbiter or lro and this is a
spacecraft launched in 2009 to take topographical measurements and look for water ice on the moon so
we're told it's orbiting in a very nearly circular orbit around the moon with an average altitude of
about 90 kilometers above the surface and we're given data below the picture for the radius of the
moon and the mass of the moon now one of the first things we want to do here just so we don't run the
risk of forgetting is that with an altitude of 90 kilometers above the surface we have to add
that to the radius of the moon to get the total orbital radius so that gives us 1827 kilometers
incidentally that's about a five percent difference in the orbital radius compared to using
the radius of the moon itself so in part a we're asked to get the orbital period of the lro and the
point is not to quote some previous result here where the orbital mechanics problem was already
solved the point is to work from first principles and that means we start with a force diagram
so there's the force pointing to the center of curvature in other words the centripetal force
that causes the path of this orbiter to bend into a circular path and that force of course is caused
by gravity so we'll call it f g and it points to the center of the moon the magnitude of that
force is given by newton's law of gravitation and that's g big m for the mass of the moon little m
for the mass of the orbiter and then divided by r squared where that's the orbital radius
and we recall that g is approximately 6.67 times 10 to the negative 11 and the units there are
newton meters squared per kilogram squared okay so that force causes the centripetal
acceleration and before we put it all together i want to look at that centripetal acceleration
the formula most of us have memorized is v squared over r but we can rephrase this in terms of
the period which is going to be convenient for answering part a so we relate the speed to the
period by saying the speed is just the distance traveled divided by how long it takes and if we
do this on one orbit the distance traveled is 2 pi r or the circumference of that orbit the time
it takes is one period capital t and we're going to square that and then we still need to divide by
r so squaring that we get 4 pi squared r squared over t squared and then there's an
r in the denominator that's going to cancel out so we get 4 pi squared r over t squared
now it's time to put it all together and for this we use newton's second law so i can say that
f net equals ma for this orbiter and the net force on it there's only one force acting and that's the
force of gravity so that's gmm over r squared and then the mass of the orbiter is little m
and its acceleration can be written as 4 pi squared r over t squared now it turns out the mass
of the orbiter is irrelevant we see that canceling out of the algebra and what we're trying to solve
for here is the period so i need to multiply both sides by t squared divide by big m and g
multiply both sides by r squared and square root the result and i'll get t so again surviving in
the numerator on the right hand side i still have a 4 pi squared and there was an r there but when
i multiply by r squared that becomes an r cubed so that's in the numerator on the right hand side
then in the denominator on the right hand side well i had to divide by a big mg
where big m is the mass of the moon so we're ready to plug everything into this and we need
to convert to si units that means that radius the orbital radius should be expressed in meters
so it was 1827 kilometers we just have to tack on three zeros to get meters and we need to cube that
then our mass is 7.342 times 10 to the 22nd and big g 6.67 times 10 to the negative 11.
when we run the numbers on this we get about 7012 seconds and it's probably better to express this
in minutes so we divide by 60 and we end up with about 117 minutes all right so the lro goes around
in a little less than two hours the next part of the question asks us to get the speed of the lro
in its orbit and so we're just going back to this idea that happened in the second line
where we relate the speed to the orbital radius and the period again the speed is the distance
traveled divided by how long it took so that's one circumference divided by one period we plug in the
orbital radius in meters and the period in seconds and we're staying in si units here so that's seven
thousand and twelve seconds and when we run the numbers on this to three significant digits we
get 1640 meters per second and we're done if you enjoyed this video or at least found it useful
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