Question

(T) The table shows the mean (average) distances d of the planets from the sun (taking the unit of measurement to be the distance from the earth to the sun) and their periods T (time of revolution in years).

(a) Fit a power model to the data.

(b) Kepler’s Third Law of Planetary Motion states that “The square of the period of revolution of a planet is proportional to the cube of its mean distance from the sun.”

Short answer

a. The power model of the data is \(T = 1.000431227{d^{1.49952875}}\).

b. The model corresponds to Kepler’s Third Law, \({T^2} = k{d^3}\).

Step-by-step solution

Fit a power model to the data

a)

Consider the values of\(d\)as\(x\)-coordinates and the values of\(T\)as\(y\)-coordinates.

The procedure to use the power regression calculator to obtain the power regression equation is as shown below:

1. Open the power regression calculator. Enter all the values of\(x\)in the\(x\)tab and\(y\)values in the\(y\)tab.
2. Select the “Calculate” button in the power regression equation calculator.

Obtain the power regression equation as shown below:

The power regression equation is\(T = 1.000431227{d^{1.49952875}}\).

Thus, the power model of the data is \(T = 1.000431227{d^{1.49952875}}\).

Kepler’s Third Law of planetary motion

b)

According to part (a), the power model is approximately\(T = {d^{1.5}}\).

Square both sides of the equation\(T = {d^{1.5}}\)as shown below:

\({T^2} = {d^3}\)

Thus, the model corresponds to Kepler’s Third Law, \({T^2} = k{d^3}\).