Question 876382
{{{t}}}= time a planet takes to revolve once around the sun (in Earth years).
{{{d}}}= distance from the sun to the planet (in millions of km).
{{{t=f(d)=kd^("3 / 2")}}}
 
a) For Earth, {{{t=1}}} and {{{d=150}}}
{{{1=k*150^("3 / 2")}}}
{{{1/150^("3 / 2")=k}}}
{{{k=1/750sqrt(6)}}}--->{{{highlight(k=sqrt(6)/4500)}}}
The approximate value is {{{highlight(k=0.00054433)}}}
{{{highlight(t=f(d)=0.00054433d^("3 / 2"))}}}
 
b) For Mercury, {{{d=58}}} , so
{{{t=f(58)=0.00054433*58^("3 / 2")=0.240}}}
It takes Mercury {{{highlight(0.240)}}} Earth years to revolve once around the sun.
 
c) I had not heard the term "irrational function" before, but then again, I am not a fan of calling anything names.
{{{t=f(d)=0.00054433d^("3 / 2")}}} is not a rational function,
because rational functions are quotients of polynomials, and {{{t=f(d)=0.00054433d^("3 / 2")}}} is an exponential function.
It does have a restricted domain because we expect {{{d>0}}} ,
and {{{d<0}}} makes {{{d^("3 / 2")=sqrt(d^3)}}} undefined.