91影视

First it is going to use the same function as in exercise 39. And then, substitute the data points$\left(\log \left(a\right),lgg\left(D\right)\right)$into the function $log\left(D\right)=c_0+c_{1}log\left(a\right)$.

Then, it has a linear system that needs to solve.

$\left[\begin{array}{cc}1& 0.387\\ 1& 1.000\\ 1& 5.023\\ 1& 19.19\\ 1& 39.53\end{array}\right]\left[\begin{array}{c}{c}_0\\ c_1\end{array}\right]=\left[\begin{array}{c}0.241\\ 1.000\\ 11.86\\ 84.04\\ 248.6\end{array}\right]$

Now, it is going to find the least solution using the formulas below.

$$\begin{gathered} {c^{*}}_0=\frac{\left({}_i{a^2}_i\right)\left({}_i{b}_i\right)-\left({}_i{a}_i\right)\left({}_i{a}_i{b}_i\right)}{n\left({}_i{a^2}_i\right)-{\left({}_i{a}_i\right)}^2} \\ =\frac{\left(1959.098\right)\left(345.741\right)-\left(65.310\right)\left(11502.686\right)}{5\left(1959.098\right)-{\left(65.310\right)}^2} \\ =\frac{-73900}{5530} \\ 鈮�-13.363 \\ {c}_1^{*}=\frac{n\left({}_i{a}_i{b}_i\right)-\left({}_i{a}_i\right)\left({}_i{b}_i\right)}{n\left({}_i{a^2}_i\right)-{\left({}_i{a}_i\right)}^2} \\ =\frac{5\left(11502.686\right)-\left(65.310\right)\left(345.741\right)}{5\left(1959.098\right){\left(65.310\right)}^2} \\ =\frac{-34.933}{5530} \\ 鈮�6.317 \end{gathered}$$

$鈭�\log \left(D\right)=-13.363+6.317\log \left(a\right)$

Convert the logarithm function to an exponent equation it gets,

$$\begin{gathered} 鈬�D={10}^{-13.363+6.317log\left(a\right)} \\ 鈬�D={10}^{13.363).{10}^{6.317log\left(a\right)}} \\ 鈬�D=(0.0000000000000433)a^{6.317} \\ 鈬�k=0.0000000000000433{}^{^}n=6.317 \end{gathered}$$

So, the formula of the third keeper鈥檚 law is $D^2=a^3$. And if it does square of both side then we get$D=a^{1.5}$ and that鈥檚 why the value of k is close to 1.

Hence, the formula $D=a^{1.5}$, that is approximating, and it is the keeper鈥檚 third law of planetary. And the value of k is close to 1.