SOLUTION: Hi everyone. I am stuck on another problem which I have been working on since yesterday morning, and although I have the answer from the text. I have absolutely no idea how they h

Algebra ->  Proportions -> SOLUTION: Hi everyone. I am stuck on another problem which I have been working on since yesterday morning, and although I have the answer from the text. I have absolutely no idea how they h      Log On


   

Question 109956This question is from textbook Mathematics for Technicians
: Hi everyone.
I am stuck on another problem which I have been working on since yesterday morning, and although I have the answer from the text. I have absolutely no idea how they have came about it, any assistance would be greatly appreciated. Cheers D
The question is as follows:-
The frequency of revolution, n (i.e. the number of orbits per unit of time), of a satellite above the earth varies inversely as sqrt r^3, where r is the radius of orbit. Given that a satellite whose radius of orbit is 5080 km orbits the earth each hour, find:
The radius of orbit for a satellite that orbits the earth once per day (correct to the nearest 100 km)
It gives me an answer of 42,300 km. This is however how I have gone about trying to solve this problem: ((sqrt5080^3)*(1/24))/(24/1)= 628.598. I have tried various different combinations but am still struggling to get the correct answer.
Again your help and assistance would be greatly appreciated. Thank you. Cheers D This question is from textbook Mathematics for Technicians

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
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.
.
Since it's frequency (f) varies inversely as sqrt%28r%5E3%29 where r is the radius of the orbit,
you can write the equation:
.
f+=+k%2F%28sqrt%28r%5E3%29%29
.
where k is a constant of proportionality that adjusts for the units used etc.
.
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:
.
24+=+k%2F%28sqrt%285080%5E3%29%29
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Solve this for k by multiplying both sides by sqrt%285080%5E3%29 to eliminate the denominator.
With this multiplication you get:
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24%2Asqrt%285080%5E3%29+=+k
.
When you calculate sqrt%285080%5E3%29 you get 362072.5231. Multiplying this by 24 results
in this equation becoming:
.
8689740.555+=+k
.
Now that we know the value of k we can substitute it into the equation for the frequency to
make it:
.
f+=+k%2F%28sqrt%28r%5E3%29%29=+8689740.555%2F%28sqrt%28r%5E3%29%29
.
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.
.
So now you have the equation:
.
f+=+8689740.555%2F%28sqrt%28r%5E3%29%29
.
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:
.
1+=+8689740.555%2F%28sqrt%28r%5E3%29%29
.
Get rid of the denominator by multiplying both sides of this equation by sqrt%28r%5E3%29.
When you do that the equation becomes:
.
sqrt%28r%5E3%29+=+8689740.555
.
Square both sides of this equation to get:
.
r%5E3+=+8689740.555%5E2+=+7.551159091%2A10%5E13
.
Solve for r by taking the cube root of both sides:
.
r+=+root%283%2C7.551159091%2A10%5E13%29+=+42267.30328
.
That rounds off to 42,300 km which is the answer you were given.
.
Hope this helps you to see how to do the problem and aids you to identify the stuff that
gave you the difficulty.
.