The table shows data for four planetary bodies. If your mass is [tex]68.05 \, \text{kg}[/tex], how much gravitational force would you experience on the surface of Mercury?

Newton's law of gravitation is [tex]F_{\text{gravity}} = \frac{G m_1 m_2}{r^2}[/tex]. The gravitational constant [tex]G[/tex] is [tex]6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2[/tex]. (For the purposes of calculating the gravitational force between a planet and an object on its surface, the distance [tex]r[/tex] is the radius of the planet.)

\begin{tabular}{|l|l|l|}
\hline
Planetary body & Mass (kg) & Radius (m) \\
\hline
Earth & [tex]5.97 \times 10^{24}[/tex] & [tex]6.37 \times 10^6[/tex] \\
\hline
Moon & [tex]7.35 \times 10^{22}[/tex] & [tex]1.74 \times 10^6[/tex] \\
\hline
Mars & [tex]6.42 \times 10^{23}[/tex] & [tex]3.39 \times 10^6[/tex] \\
\hline
Mercury & [tex]3.30 \times 10^{23}[/tex] & [tex]2.44 \times 10^6[/tex] \\
\hline
\end{tabular}

A. [tex]110 \, \text{N}[/tex]



Answer :

To determine the gravitational force you would experience on the surface of Mercury, we need to use Newton's law of gravitation.

The formula for calculating the gravitational force [tex]\( F_{\text{gravity}} \)[/tex] is:
[tex]\[ F_{\text{gravity}} = \frac{G \cdot m_1 \cdot m_2}{r^2} \][/tex]

Where:
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex]
- [tex]\( m_1 \)[/tex] is the mass of the object (your mass), [tex]\( 68.05 \, \text{kg} \)[/tex]
- [tex]\( m_2 \)[/tex] is the mass of Mercury, [tex]\( 3.30 \times 10^{23} \, \text{kg} \)[/tex]
- [tex]\( r \)[/tex] is the radius of Mercury, [tex]\( 2.44 \times 10^6 \, \text{m} \)[/tex]

Let's put these values into the formula step-by-step:

1. Identify the values:
- [tex]\( G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex]
- [tex]\( m_1 = 68.05 \, \text{kg} \)[/tex]
- [tex]\( m_2 = 3.30 \times 10^{23} \, \text{kg} \)[/tex]
- [tex]\( r = 2.44 \times 10^6 \, \text{m} \)[/tex]

2. Calculate [tex]\( r^2 \)[/tex]:
[tex]\[ r^2 = (2.44 \times 10^6 \, \text{m})^2 = 5.9536 \times 10^{12} \, \text{m}^2 \][/tex]

3. Plug the values into the formula:
[tex]\[ F_{\text{gravity}} = \frac{(6.67 \times 10^{-11}) \cdot (68.05) \cdot (3.30 \times 10^{23})}{5.9536 \times 10^{12}} \][/tex]

4. Calculate the numerator:
[tex]\[ 6.67 \times 10^{-11} \cdot 68.05 \cdot 3.30 \times 10^{23} = 1.4982835 \times 10^{14} \][/tex]

5. Calculate the force:
[tex]\[ F_{\text{gravity}} = \frac{1.4982835 \times 10^{14}}{5.9536 \times 10^{12}} = 251.58703137597416 \, \text{N} \][/tex]

Thus, the gravitational force you would experience on the surface of Mercury is approximately [tex]\( 251.59 \, \text{N} \)[/tex].