use what you know about newtons law of universal to predict the values for his missing variables.
system $f_g$ $m_1$ $m_2$ r
kepler - 92 $2.7×10^{26}n$ $2.3×10^{27}kg$ $5.6×10^{30}k$ $5.7×10^{10}m$
gliese - 597 $8.03×10^{24}n$ choose $3.03×10^{31}kg$ $1.93×10^{11}m$
kepler - 731 choose $1.48×10^{26}kg$ $6.69×10^{28}kg$ $1.17×10^{11}m$
kepler - 1059 $5.98×10^{22}n$ $1.54×10^{36}kg$ $6.68×10^{30}kg$ choose

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Accepted Answer

# Explanation:
## Step1: Recall Newton's law of universal gravitation formula
Newton's law of universal gravitation is given by $F_g = G\frac{m_1m_2}{r^2}$, where $G = 6.67\times 10^{- 11}\text{ N}\cdot\text{m}^2/\text{kg}^2$.

## Step2: Solve for $m_1$ in the Gliese - 597 system
We need to find $m_1$ for the Gliese - 597 system. Rearranging the formula for $m_1$ gives $m_1=\frac{F_gr^2}{Gm_2}$.
Substitute $F_g = 8.03\times 10^{24}\text{ N}$, $r = 1.93\times 10^{11}\text{ m}$, $G = 6.67\times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$, and $m_2 = 3.03\times 10^{31}\text{ kg}$ into the formula:
$$
\begin{align*}
m_1&=\frac{(8.03\times 10^{24})\times(1.93\times 10^{11})^2}{6.67\times 10^{-11}\times3.03\times 10^{31}}\
&=\frac{8.03\times10^{24}\times3.7249\times 10^{22}}{2.02041\times 10^{21}}\
&=\frac{29.919947\times 10^{46}}{2.02041\times 10^{21}}\
&\approx1.48\times 10^{26}\text{ kg}
\end{align*}
$$

## Step3: Solve for $F_g$ in the Kepler - 731 system
Substitute $m_1 = 1.48\times 10^{26}\text{ kg}$, $m_2 = 6.69\times 10^{28}\text{ kg}$, $r = 1.17\times 10^{11}\text{ m}$, and $G = 6.67\times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$ into $F_g = G\frac{m_1m_2}{r^2}$:
$$
\begin{align*}
F_g&=6.67\times 10^{-11}\times\frac{1.48\times 10^{26}\times6.69\times 10^{28}}{(1.17\times 10^{11})^2}\
&=6.67\times 10^{-11}\times\frac{9.9012\times 10^{54}}{1.3689\times 10^{22}}\
&=6.67\times 10^{-11}\times7.2329\times 10^{32}\
&\approx4.83\times 10^{22}\text{ N}
\end{align*}
$$

## Step4: Solve for $r$ in the Kepler - 1059 system
Rearrange $F_g = G\frac{m_1m_2}{r^2}$ to $r=\sqrt{\frac{Gm_1m_2}{F_g}}$. Substitute $F_g = 5.98\times 10^{22}\text{ N}$, $m_1 = 1.54\times 10^{36}\text{ kg}$, $m_2 = 6.68\times 10^{30}\text{ kg}$, and $G = 6.67\times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$ into the formula:
$$
\begin{align*}
r&=\sqrt{\frac{6.67\times 10^{-11}\times1.54\times 10^{36}\times6.68\times 10^{30}}{5.98\times 10^{22}}}\
&=\sqrt{\frac{6.67\times1.54\times6.68\times 10^{55}}{5.98\times 10^{22}}}\
&=\sqrt{\frac{68.77\times 10^{55}}{5.98\times 10^{22}}}\
&=\sqrt{11.5\times 10^{33}}\
&=\sqrt{1.15\times 10^{34}}\
&\approx1.07\times 10^{17}\text{ m}
\end{align*}
$$

# Answer:
- For Gliese - 597: $m_1 = 1.48\times 10^{26}\text{ kg}$
- For Kepler - 731: $F_g\approx4.83\times 10^{22}\text{ N}$
- For Kepler - 1059: $r\approx1.07\times 10^{17}\text{ m}$

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QUESTION IMAGE

Question

Explanation:

Step1: Recall Newton's law of universal gravitation formula

Step2: Solve for \(m_1\) in the Gliese - 597 system

Step3: Solve for \(F_g\) in the Kepler - 731 system

Step4: Solve for \(r\) in the Kepler - 1059 system

Answer: