document.write( "Question 1023909: THIS IS AN ALGEBRAIC FUNCTIONS VARIATIONS PROBLEM. There is also a constant k that is part of the equation I have to create. Can someone go through the logic for this problem? \r
\n" ); document.write( "\n" ); document.write( "Kepler's Third Law states that the square of time T required for a planet to complete one orbit around the sun (the period, that is, the length of one planetary year) is directly proportional to the cube of average distance d of the planet from the sun. For the planet Earth, assume d = 93 x 10^6 miles an T = 365 days. Find a) the period of Mars, given that Mars is approximately 1.5 times as distant from the sun as Earth; b) the average distance of Venus from the sun, given that the period of Venus is approximately 223 Earth days. \r
\n" ); document.write( "\n" ); document.write( "The units for Part a are in days and the units for part b are in miles.
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Algebra.Com's Answer #639406 by KMST(5328)\"\" \"About 
You can put this solution on YOUR website!
Stated in formulas, rather than words,
\n" ); document.write( "Kepler's Third Law states that
\n" ); document.write( "if \"T\"= time required for a planet to complete one orbit around the sun (the period, that is, the length of one planetary year), and
\n" ); document.write( "\"d\"= average distance of the planet from the sun,
\n" ); document.write( "then \"T%5E2=k%2Ad%5E3\" ,
\n" ); document.write( "where \"k\" is the proportionality constant.
\n" ); document.write( "So, for Earth, \"T%5BEarth%5D%5E2=k%2Ad%5BEarth%5D%5E3\"
\n" ); document.write( "For the planet Earth, assume \"d%5BEarth%5D\" = \"9%2A10%5E6\" \"miles\" an \"T%5BEarth%5D=+365\"\"days\" .
\n" ); document.write( "Since we are going to measure times in Earth days and distances in miles
\n" ); document.write( "(or in whatever units we want, but using the same units all along),
\n" ); document.write( "we do not need to keep writing the units with the calculations.
\n" ); document.write( "NOTE: I did not solve for \"k\" and I did not use the given average distance between Earth and the sun, because it would only complicate the calculations.
\n" ); document.write( "If your teacher insist that you do it, I am sorry for both of you.
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\n" ); document.write( "a) They tell us that \"d%5BMars%5D\"=\"approximately\"\"1.5%2Ad%5BEarth%5D\" ,
\n" ); document.write( "and according to Kepler's Third Law
\n" ); document.write( "\"T%5BMars%5D%5E2=k%2Ad%5BMars%5D%5E3\" , so
\n" ); document.write( "\"T%5BMars%5D%5E2=k%2A%281.5%2Ad%5BEarth%5D%29%5E3\"
\n" ); document.write( "\"T%5BMars%5D%5E2=k%2A1.5%5E3%2Ad%5BEarth%5D%5E3\"
\n" ); document.write( "\"T%5BMars%5D%5E2=T%5BEarth%5D%5E2%2A1.5%5E3\"
\n" ); document.write( "\"T%5BMars%5D%5E2=%28k%2Ad%5BEarth%5D%5E3%29%2A1.5%5E3\"
\n" ); document.write( "\"T%5BMars%5D%5E2=365%5E2%2A1.5%5E3\"
\n" ); document.write( "\"T%5BMars%5D%5E2=133225%2A3.375\"
\n" ); document.write( "\"T%5BMars%5D%5E2=449634.375\"
\n" ); document.write( "\"T%5BMars%5D=sqrt%28449634.375%29\"
\n" ); document.write( "\"T%5BMars%5D=about671\" (rounded).
\n" ); document.write( "So, the period of Mars is approximately 671 days.
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\n" ); document.write( "b) They tell us that \"T%5BVenus%5D=223\" ,
\n" ); document.write( "and according to Kepler's Third Law
\n" ); document.write( "\"T%5BVenus%5D%5E2=k%2Ad%5BVenus%5D%5E3\" , and \"T%5BVenus%5D%5E2=k%2Ad%5BVenus%5D%5E3\" ,
\n" ); document.write( "so with distances in miles this time
\n" ); document.write( "\"223%5E2=k%2Ad%5BVenus%5D%5E3\" , and \"365%5E2=k%2A%2893%2A10%5E6%29%5E3\" .
\n" ); document.write( "Dividing one equation by the other,
\n" ); document.write( "\"223%5E2%2F365%5E2=k%2Ad%5BVenus%5D%5E3%2F%28k%2A%2893%2A10%5E6%29%5E3%29\"
\n" ); document.write( "\"49729%2F133225=d%5BVenus%5D%5E3%2F%2893%5E3%2A10%5E18%29\"
\n" ); document.write( "\"49729%2F133225=d%5BVenus%5D%5E3%2F%28804357%2A10%5E18%29\"
\n" ); document.write( "\"49729%2A804357%2A10%5E18%2F133225=d%5BVenus%5D%5E3\"
\n" ); document.write( "\"49729%2A804357%2A10%5E18%2F133225=d%5BVenus%5D%5E3\"
\n" ); document.write( "\"300243%2A10%5E18=d%5BVenus%5D%5E3\"
\n" ); document.write( "\"root%283%2C300243%2A10%5E18%29=d%5BVenus%5D\"
\n" ); document.write( "\"d%5BVenus%5D=root%283%2C300243%29%2Aroot%283%2C10%5E18%29\"
\n" ); document.write( "\"d%5BVenus%5D=67%2A10%5E6\" (rounded).
\n" ); document.write( "So, the the average distance of Venus from the sun, is approximately 67 x 10^6 miles.
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