Chapter 1.2, Problem 26E

a.

To determine

To fit:a power model to the given data on table.

Answer to Problem 26E

  $T=1.00253d^{1.49884}$ .

Explanation of Solution

Given:

The given table shows the mean distance of the planets from the sun and their periods:

Planet	Distance d	Periods T
Mercury	$0.387$	$0.241$
Venus	$0.723$	$0.615$
Earth	$1.000$	$1.000$
Mars	$1.523$	$1.881$
Jupiter	$5.203$	$11.861$
Saturn	$9.541$	$29.457$
Uranus	$19.190$	$84.008$
Neptune	$30.086$	$164.784$

Calculation:

Use technology to plot given points, to graph a curve.

The graph is:

  

Plot the given data and using a “graphing utility” get the power model $T=1.00253d^{1.49884}$ .

b.

To determine

To check: that the model corroboratesKepler’s third law.

Answer to Problem 26E

Yes.

Explanation of Solution

Given:

Kepler’s third law state that:

“The square of the period of revolution of a planet is proportional to the cube of its means distance from the sun”.

Calculation:

Power model from part (a):

  $T=1.00253d^{1.49884}$

For the square of T to be the proportional to the cube of d, thus the equation is :

  $T^2=k{d}^3\left\{\text{where }k\text{ is some constant}\right\}$

Taking the square root of both sides then:

  $\begin{array}{l}T=k{d}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ T=k{d}^{1.5}\end{array}$

Note that $1.5\approx 1.49953$ .

Therefore, the model corroborates keepers’ third law.