\begin{tabular}{|l|l|}
\hline Mass 1 & Mass 2 \\
\hline [tex]$m_1=30 \, \text{kg}$[/tex] & [tex]$m_2=30 \, \text{kg}$[/tex] \\
\hline Distance \\
\hline [tex]$r=2.0 \, \text{m}$[/tex] & \\
\hline
\end{tabular}

Calculate the gravitational force in this scenario.

[tex]\[ F = \square \, \text{N} \][/tex]



Answer :

To determine the gravitational force exerted between two masses, we utilize Newton's law of universal gravitation. This law states that the gravitational force [tex]\( F \)[/tex] between two masses [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] separated by a distance [tex]\( r \)[/tex] is given by:

[tex]\[ F = G \frac{m_1 \cdot m_2}{r^2} \][/tex]

where:
- [tex]\( G \)[/tex] is the gravitational constant, approximately [tex]\( 6.67430 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex],
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses in kilograms,
- [tex]\( r \)[/tex] is the distance between the centers of the two masses in meters.

In this problem, we have:
- [tex]\( m_1 = 30 \, \text{kg} \)[/tex],
- [tex]\( m_2 = 30 \, \text{kg} \)[/tex],
- [tex]\( r = 2.0 \, \text{m} \)[/tex].

First, we calculate the gravitational force [tex]\( F \)[/tex] in Newtons (N):

[tex]\[ F = \left( 6.67430 \times 10^{-11} \right) \frac{30 \cdot 30}{2.0^2} \][/tex]
[tex]\[ F = \left( 6.67430 \times 10^{-11} \right) \frac{900}{4} \][/tex]
[tex]\[ F = \left( 6.67430 \times 10^{-11} \right) \cdot 225 \][/tex]
[tex]\[ F = 1.5017174999999998 \times 10^{-8} \, \text{N} \][/tex]

Next, since we want the force in nanoNewtons (nN), we must convert Newtons to nanoNewtons. Remember that [tex]\( 1 \, \text{N} = 10^9 \, \text{nN} \)[/tex]:

[tex]\[ F \, (\text{nN}) = 1.5017174999999998 \times 10^{-8} \, \text{N} \times 10^9 \, \text{nN} / \text{N} \][/tex]
[tex]\[ F \, (\text{nN}) = 15.017174999999998 \, \text{nN} \][/tex]

Thus, the gravitational force between the two masses is:

[tex]\[ F = 15.017174999999998 \, \text{nN} \][/tex]

So, in this scenario, the force is [tex]\( 15.017174999999998 \, \text{nanoNewtons} \)[/tex].