diff --git a/PhysicsGravitationLabReport/images/Force vs Distance Between Masses.pdf b/PhysicsGravitationLabReport/images/Force vs Distance Between Masses.pdf
new file mode 100644
index 0000000..6d04d36
Binary files /dev/null and b/PhysicsGravitationLabReport/images/Force vs Distance Between Masses.pdf differ
diff --git a/PhysicsGravitationLabReport/main.pdf b/PhysicsGravitationLabReport/main.pdf
index e9f1551..afecb4c 100644
Binary files a/PhysicsGravitationLabReport/main.pdf and b/PhysicsGravitationLabReport/main.pdf differ
diff --git a/PhysicsGravitationLabReport/main.tex b/PhysicsGravitationLabReport/main.tex
index c5278d6..8af9793 100644
--- a/PhysicsGravitationLabReport/main.tex
+++ b/PhysicsGravitationLabReport/main.tex
@@ -88,7 +88,7 @@ In the Principia, Newton asserted that every mass exerts an attractive force on
 This law states that the magnitude of the gravitational force between two masses is
 
 \begin{equation}
-F = G \frac{m_1 m_2}{r^2}
+F_g = G \frac{m_1 m_2}{r^2}
 \label{eq:NLUG}
 \end{equation}
 
@@ -449,7 +449,7 @@ Looking at Table \ref{tab:grav_equal}, this derived relationship can be verified
 
 
 
-\begin{figure}[h!] % h! = “here” placement
+\begin{figure}[H] % h! = “here” placement
     \centering
     \includegraphics[width=0.7\textwidth]{Force vs Mass m1 m2} % <-- your image file name
     \caption{$F_g$ vs mass of $m_1$ = $m_2$}
@@ -458,8 +458,90 @@ Looking at Table \ref{tab:grav_equal}, this derived relationship can be verified
 
 
 As suggested by equation \ref{eq:m1and2}, there is a proportional quadratic relationship between the masses of the objects and the resulting gravitational force (see Figure \ref{fig:m1m2graph}). 
-Therefore, equation \ref{eq:m1and2} is validated by the simulation.
+Therefore, equation \ref{eq:m1and2} is validated by the simulation.\\\\
 
+Moving onto Tables \ref{tab:grav_dist1} and \ref{tab:grav_dist2}, the output columns for the gravitational force appear identical. 
+Upon further inspection, the gravitational force seems to exhibit a dependency on the distance between the masses.
+A processed data table can be made combining Tables \ref{tab:grav_dist1} and \ref{tab:grav_dist2},
+showcasing the relationship between $F_g$ and $r:= |x_1-x_2|$.
+
+\begin{table}[H]
+\centering
+\caption{Gravitational force versus distance between two masses.}
+\label{tab:grav_vs_r}
+\renewcommand{\arraystretch}{1.3}
+
+\begin{tabularx}{0.8\textwidth}{
+    @{}
+    >{\centering\arraybackslash}X
+    >{\centering\arraybackslash}X
+    @{}
+}
+\toprule
+\textbf{Distance $r$ (m)} & \textbf{Gravitational Force $F_g$ (N)} \\
+\midrule
+2.00 & $1.67\times10^{-7}$ \\
+4.00 & $4.17\times10^{-8}$ \\
+6.00 & $1.85\times10^{-8}$ \\
+8.00 & $1.04\times10^{-8}$ \\
+10.00 & $6.67\times10^{-9}$ \\
+\bottomrule
+\end{tabularx}
+\end{table}
+
+This new relationship is graphed in the figure below:
+
+
+\begin{figure}[H] % h! = “here” placement
+    \centering
+    \includegraphics[width=0.7\textwidth]{Force vs Distance Between Masses} % <-- your image file name
+    \caption{$F_g$ vs $r= x_2 - x_1$}
+    \label{fig:rgraph}
+\end{figure}
+
+The shown trendline suggests a proprtional fit to $\frac{1}{r^2}$ as follows:
+
+\[
+F_g \propto \frac{1}{r^2},
+\]
+
+so 
+
+\begin{equation}
+F_g = k_4 \times \frac{1}{r^2}.
+\label{eq:finalEq}
+\end{equation}
+
+When combined with equation \ref{eq:m1and2}, NLUG pops out, and the proprtionality constant can be denotes as $G$:
+
+
+\begin{equation}
+F_g = G \frac{m_1 m_2}{r^2} \tag{\ref{eq:NLUG}}
+\end{equation}
+
+The final step is to solve for the proportionality constant, $G$. As the relationship has been proven, any data point can be used to solve for this constant.
+For simplicity, the control data point (used for setup) will be used, where $m_1$ = 100 kg, $m_2$ = 100 kg,  $r = x_2 - x_1 = 6\,\text{m} - 2\,\text{m} = 4\,\text{m}$.
+We have
+\[
+F_g = G \left( \frac{m_1 m_2}{r^2} \right) 
+= G \left( \frac{100\,\text{kg} \times 100\,\text{kg}}{4^2\,\text{m}^2} \right) 
+= 4.17 \times 10^{-8}\,\text{N}
+\]
+
+\[
+G = \frac{F_g \, r^2}{m_1 m_2} 
+= \frac{(4.17 \times 10^{-8}\,\text{N}) (4\,\text{m})^2}{100\,\text{kg} \times 100\,\text{kg}}
+= 6.67 \times 10^{-11}\,\text{N\,m}^2\text{/kg}^2
+\].
+
+Ergo, the data from the simulation can be used to derive NLUG and solve for $G$, the universal gravitational constant:
+
+
+\begin{equation}
+F_g = G \frac{m_1 m_2}{r^2}, \quad 
+G = 6.67 \times 10^{-11}\,\text{N\,m}^2\text{/kg}^2
+\label{eq:solvedNLUG}
+\end{equation}
 
 \section*{Error Analysis}