Answered

Two masses are 4.62 m apart. Mass 1 is 84.2 kg and Mass 2 is 28.4 kg. What is the gravitational force between the two masses?

[tex]\[
\begin{array}{c}
\vec{F} = G \frac{m_1 m_2}{r^2} \\
G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \\
\vec{F} = [?] \times 10^{[?]} \, \text{N}
\end{array}
\][/tex]



Answer :

To find the gravitational force [tex]\(\vec{F}\)[/tex] between two masses, we use Newton's Law of Universal Gravitation, given by the formula:

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

where:
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( G = 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex],
- [tex]\( m_1 \)[/tex] is the mass of the first object,
- [tex]\( m_2 \)[/tex] is the mass of the second object,
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.

Given:
- [tex]\( m_1 = 84.2 \, \text{kg} \)[/tex],
- [tex]\( m_2 = 28.4 \, \text{kg} \)[/tex],
- [tex]\( r = 4.62 \, \text{m} \)[/tex].

Step-by-step solution:
1. Substitute the given values into the formula:

[tex]\[ \vec{F} = 6.67 \times 10^{-11} \cdot \frac{84.2 \cdot 28.4}{4.62^2} \][/tex]

2. Calculate the product of the masses:

[tex]\[ m_1 \cdot m_2 = 84.2 \, \text{kg} \times 28.4 \, \text{kg} = 2391.28 \, \text{kg}^2 \][/tex]

3. Calculate the square of the distance:

[tex]\[ r^2 = 4.62 \, \text{m} \times 4.62 \, \text{m} = 21.3444 \, \text{m}^2 \][/tex]

4. Substitute these results back into the formula:

[tex]\[ \vec{F} = 6.67 \times 10^{-11} \, \frac{2391.28}{21.3444} \][/tex]

5. Compute the division inside the parentheses:

[tex]\[ \frac{2391.28}{21.3444} \approx 112.037 \][/tex]

6. Finally, multiply by the gravitational constant:

[tex]\[ \vec{F} \approx 6.67 \times 10^{-11} \, \times 112.037 \][/tex]

7. Calculate the result:

[tex]\[ \vec{F} \approx 7.472609958583983 \times 10^{-9} \, \text{N} \][/tex]

Therefore, the gravitational force [tex]\( \vec{F} \)[/tex] between the two masses is approximately:

[tex]\[ \vec{F} \approx 7.47 \times 10^{-9} \, \text{N} \][/tex]

So, we have:

[tex]\[ \vec{F} \approx 7.47 \times 10^{-9} \, \text{N} \][/tex]