Question 109956
The given information is on a satellite that makes an orbit an hour. This means that it 
has a frequency of 24 orbits per day.
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Since it's frequency (f) varies inversely as {{{sqrt(r^3)}}} where r is the radius of the orbit, 
you can write the equation:
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{{{f = k/(sqrt(r^3))}}}
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where k is a constant of proportionality that adjusts for the units used etc.  
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Since for the given satellite you know that its frequency is 24 orbits per day and its radius
is 5080 km, you can substitute these values to get:
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{{{24 = k/(sqrt(5080^3))}}}
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Solve this for k by multiplying both sides by {{{sqrt(5080^3)}}} to eliminate the denominator.
With this multiplication you get:
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{{{24*sqrt(5080^3) = k}}}
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When you calculate {{{sqrt(5080^3)}}} you get 362072.5231. Multiplying this by 24 results
in this equation becoming:
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{{{8689740.555 = k}}}
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Now that we know the value of k we can substitute it into the equation for the frequency to
make it:
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{{{f = k/(sqrt(r^3))= 8689740.555/(sqrt(r^3))}}}
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Note that the way k was calculated requires that the frequency be specified in units of
orbits per day and the radius be specified in km in order for the equation to work.
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So now you have the equation:
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{{{f = 8689740.555/(sqrt(r^3))}}}
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You are given that a satellite makes 1 orbit per day and are asked to find the radius 
of its orbit. Substitute 1 for f and you have:
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{{{1 = 8689740.555/(sqrt(r^3))}}}
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Get rid of the denominator by multiplying both sides of this equation by {{{sqrt(r^3)}}}.
When you do that the equation becomes:
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{{{sqrt(r^3) = 8689740.555}}}
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Square both sides of this equation to get:
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{{{r^3 = 8689740.555^2 = 7.551159091*10^13}}}
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Solve for r by taking the cube root of both sides:
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{{{r = root(3,7.551159091*10^13) = 42267.30328}}}
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That rounds off to 42,300 km which is the answer you were given.
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Hope this helps you to see how to do the problem and aids you to identify the stuff that
gave you the difficulty.
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