Gravitational Field due to Point Masses

To determine the gravitational field \(\vec{g}\) produced by the array of masses at point P we will have to use the superposition principle: the gravitational field generated at point P will be the sum of the gravitational field generated due to mass 1 \(m_1=2000\,\text{kg}\) and mass 2 \(m_2=1000\,\text{kg}\), explicitly

\begin{equation}\label{g}
\vec{g}=\vec{g}_1+\vec{g}_2.
\end{equation}

The gravitational field produced by \(m_1\) can be calculated as

\begin{equation}\label{g1}
\vec{g}_1=-\frac{Gm_1}{r_1}\vec{r}_1,
\end{equation}

where \(G= 6.67430 \times 10^{-11}\, \text{m}^3\, \text{kg}
^{-1}\,\text{s}^{-2}\) is the Newtonian constant of gravitation, \(\vec{r}_1\) is the distance vector between \(m_1\) and P and \(r_1\) is the magnitude of the distance vector. For the gravitational field generated by the mass \(m_2\) we obtain an equivalent equation

\begin{equation}\label{g2}
\vec{g}_2=-\frac{Gm_2}{r_2}\vec{r}_2,
\end{equation}

where in this case \(\vec{r}_2\) is the distance vector between \(m_2\) and P and \(r_2\) is the magnitude of the distance vector. Using the results of \eqref{g1} and \eqref{g2} into equation \eqref{g} we get

\begin{equation}\label{gg}
\vec{g}=-\frac{Gm_1}{r_1}\vec{r}_1-\frac{Gm_2}{r_2}\vec{r}_2.
\end{equation}

The distance vectors are the ones shown in the following figure.

The distance vector can be easily calculated by subtraction of the coordinates of point P \((x_P,y_P)\) and the coordinates of the masses \(m_1\), \((x_1,y_1)\) and \(m_2\), \((x_2,y_2)\). Explicitly

\begin{equation}\label{vr1}
\vec{r}_1=(x_P-x_1)\,\hat{\textbf{i}}+(y_P-y_1)\,\hat{\textbf{i}},
\end{equation}

and

\begin{equation}\label{vr2}
\vec{r}_2=(x_P-x_2)\,\hat{\textbf{i}}+(y_P-y_2)\,\hat{\textbf{i}}.
\end{equation}

From the equations above, it follows that the magnitudes are

\begin{equation}\label{r1}
r_1=\sqrt{(x_P-x_1)^2+(y_P-y_1)^2},
\end{equation}

and

\begin{equation}\label{r2}
r_2=\sqrt{(x_P-x_2)^2+(y_P-y_2)^2}.
\end{equation}

We can then write the component along the X axis of the gravitational field from equation \eqref{gg} using the expressions given in \eqref{vr1}, \eqref{vr2}, \eqref{r1} and \eqref{r2}, namely

\begin{equation}
g_x\,\hat{\textbf{i}}=-\frac{G m_1(x_P-x_1)}{\sqrt{(x_P-x_1)^2+(y_P-y_1)^2}}\,\hat{\textbf{i}}-\frac{G m_2(x_P-x_2)}{\sqrt{(x_P-x_2)^2+(y_P-y_2)^2}}\,\hat{\textbf{i}},
\end{equation}

which numerically is

\begin{equation}
g_x\,\hat{\textbf{i}}=-\frac{(6.67430 \times 10^{-11}\, \text{m}^3\, \text{kg}
^{-1}\,\text{s}^{-2})(2000\,\text{kg})(0.8\,\text{m}-(-1\,\text{m}))}{\sqrt{(0.8\,\text{m}-(-1\,\text{m}))^2+(0.6\,\text{m}-1\,\text{m})^2}}\,\hat{\textbf{i}}-\frac{(6.67430 \times 10^{-11}\, \text{m}^3\, \text{kg}
^{-1}\,\text{s}^{-2})(1000\,\text{kg})(0.8\,\text{m}-0.5\,\text{m})}{\sqrt{(0.8\,\text{m}-0.5\,\text{m})^2+(0.6\,\text{m}-(-0.7\,\text{m}))^2}}\,\hat{\textbf{i}},
\end{equation}

\begin{equation}
g_x\,\hat{\textbf{i}}\approx -1.45\times 10^{-7}\,\text{m/s}^2\,\hat{\textbf{i}}.
\end{equation}

We can now write the component along the Y axis of the gravitational field from equation \eqref{gg} using the expressions given in \eqref{vr1}, \eqref{vr2}, \eqref{r1} and \eqref{r2}, namely

\begin{equation}
g_y\,\hat{\textbf{j}}=-\frac{G m_1(y_P-y_1)}{\sqrt{(x_P-x_1)^2+(y_P-y_1)^2}}\,\hat{\textbf{j}}-\frac{G m_2(y_P-y_2)}{\sqrt{(x_P-x_2)^2+(y_P-y_2)^2}}\,\hat{\textbf{j}},
\end{equation}

which numerically is

\begin{equation}
g_y\,\hat{\textbf{i}}=-\frac{(6.67430 \times 10^{-11}\, \text{m}^3\, \text{kg}
^{-1}\,\text{s}^{-2})(2000\,\text{kg})(0.6\,\text{m}-1\,\text{m})}{\sqrt{(0.8\,\text{m}-(-1\,\text{m}))^2+(0.6\,\text{m}-1\,\text{m})^2}}\,\hat{\textbf{j}}-\frac{(6.67430 \times 10^{-11}\, \text{m}^3\, \text{kg}
^{-1}\,\text{s}^{-2})(1000\,\text{kg})(0.6\,\text{m}-(-0.7\,\text{m}))}{\sqrt{(0.8\,\text{m}-0.5\,\text{m})^2+(0.6\,\text{m}-(-0.7\,\text{m}))^2}}\,\hat{\textbf{j}},
\end{equation}

\begin{equation}
g_y\,\hat{\textbf{j}}\approx -3.61\,\times 10^{-8}\,\text{m/s}^2\,\hat{\textbf{j}}.
\end{equation}

b) The force exerted on a particle of mass \(m=200\,\text{kg}\) placed on point P can be calculated as the product of \(m\) and the gravitational field, explicitly

\begin{equation}\label{fg}
\vec{F}=m\vec{g}.
\end{equation}

The component of the force along the X axis will then be, according to \eqref{fg}

\begin{equation}
F_x\,\hat{\textbf{i}}=mg_x\,\hat{\textbf{i}},
\end{equation}

which numerically is

\begin{equation}
F_x\,\hat{\textbf{i}}=(200\,\text{kg})(-1.45\times 10^{-7}\,\text{m/s}^2)\,\hat{\textbf{i}},
\end{equation}

\begin{equation}
F_x\,\hat{\textbf{i}}=-2.9\times 10^{-5}\,\text{N}\,\textbf{i}.
\end{equation}

The component of the force along the Y axis will be, according to \eqref{fg}

\begin{equation}
F_y\,\hat{\textbf{j}}=mg_y\,\hat{\textbf{j}},
\end{equation}

which numerically is

\begin{equation}
F_y\,\hat{\textbf{i}}=(200\,\text{kg})(-3.61\times 10^{-8}\,\text{m/s}^2)\,\hat{\textbf{j}},
\end{equation}

\begin{equation}
F_y\,\hat{\textbf{j}}=-7.22\times 10^{-6}\,\text{N}\,\textbf{j}.
\end{equation}

The forces are really small due to the weak nature of gravity when working with masses on the non-planetary scale.