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\markright{Chapter 3: Answers 3}
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\Bf{Chapter 3: Answers 3 \hfill Jack K. Cohen \hfill Colorado School of Mines}

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\item  Use \Mma\ and if you still don't see the light, ask your friends or the instructor for help.

\item Re the inverse square law.
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	\item \[W'(r) = -2 mg \frac{R^2}{r^3}. \] 

	\item W[R] and W'[R] are respectively:
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176. ft lbs-mass
-----------
      2
   sec

-0.000016835 lbs-mass
----------------
         2
      sec
	\end{verbatim}
	Note carefully that the second answer, though small, is \Em{not} 0!  If you got 0, you probably made the common mistake of substituting $R$ for $r$ before differentiating instead of afterwards as you should have done.
	
	\item 
	\begin{verbatim}
176. ft lbs-mass
-----------
      2
   sec

0
	\end{verbatim}

\item W[5500 miles] and W'[5500 miles] are respectively:
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91.2384 ft lbs-mass
--------------
        2
     sec

           -6
-6.28364 10   lbs-mass
-----------------
         2
      sec
	\end{verbatim}
Thus there is virtually no error in using the constant force model at the Earth's surface to compute weight.  While the correct derivative is small, replacing it by the approximation's identically zero derivative loses all information---if you really need derivative information, you should probably use the inverse square law.  Of course, if you need weights far away from the Earth's surface (as in the last part of the problem), you \Em{do} need the inverse square law to compute accurate weights.

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