Answer
See the detailed answer below.
Work Step by Step
a) This part is a math review part, we need to find $y$ in terms of $x$ where $x=\log u$ and $y=\log \nu$.
We got the following formula to work with;
$$\nu^p=Cu^q$$
Taking the logarithm
$$\log [\nu^p]=\log[ Cu^q]$$
$$p\log\nu=\log C+\log u^q$$
$$p\;\overbrace{\log\nu}^{y}=\log C+q\;\overbrace{\log u}^{x} $$
$$py=qx+\log C$$
Thus,
$$\boxed{y=\left[\dfrac{q}{p}\right]x+\dfrac{\log C}{p}}$$
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b) From the boxed formula above, the shape of the graph is a straight line since $q$, $p$, and $\log C$ are constants.
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c) According to the straight line formula $y=mx+b$ where $m$ is the slope and $b$ is the $y$-axis intercept of the graph, the slope here of this boxed formula is
$${\rm Slope}=\dfrac{q}{p}$$
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d) The formula of the given line is
$$\log T=1.5\log r-9.264\tag 1$$
where $\log T=y$ and $\log r=x$
$$y=1.5x-9.264$$
where $\dfrac{q}{p}=1.5$, and $\dfrac{\log C}{p}= 9.264$
Noting that the distance between the planet and the sun is given by
$$\dfrac{GM \color{red}{\bf\not}m}{r^2}= \color{red}{\bf\not}m\dfrac{v^2}{r}$$
where $M$ is the sun's mass, and $m$ is the planet mass.
Recall that the speed of the planet around the sun is given by $v=2\pi r/T$;
$$\dfrac{GM }{r^{ \color{red}{\bf\not}2}}=\dfrac{2^2\pi^2 r^2}{ \color{red}{\bf\not}rT^2}$$
$$\dfrac{GM }{r }=\dfrac{4\pi^2 r^2}{ T^2}$$
Hence,
$$ T^2 =\dfrac{4\pi^2 r^3}{ M G}$$
Taking the logarithm for both sides;
$$2\log T=\log \left(\dfrac{4\pi^2 r^3}{ M G}\right)$$
$$2 \log T= \log r^3+\log \dfrac{4\pi^2 }{ M G} $$
$$2 \log T=3\log r+ \log \left( \dfrac{4\pi^2 }{ M G} \right)$$
$$ \log T=\frac{3}{2}\log r+ \dfrac{\log \left( \dfrac{4\pi^2 }{ M G} \right)}{2}$$
From (1), the boxed formula, and this previous formula, we can see that
$$\dfrac{\log \left( \dfrac{4\pi^2 }{ M G} \right)}{2}=\dfrac{\log C}{p}=-9.264$$
$$ \log \left( \dfrac{4\pi^2 }{ M G} \right) =2(-9.264)$$
Hence,
$$ \dfrac{4\pi^2 }{ M G} =10^{-18.528}$$
Thus,
$$M=\dfrac{4\pi^2 }{G\times 10^{-18.528}}=\dfrac{4\pi^2 }{(6.67\times 10^{-11})\times 10^{-18.528}}$$
$$M=\color{red}{\bf 1.99634\times10^{30}}\;\rm kg$$
which is the same mass of the sun in table 13.2.
