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System Sem 1: Newton's Law of Gravitation

Question 3 of 10

Two shoes each have a mass of [tex][tex]$0.1 \, \text{kg}$[/tex][/tex]. If they are [tex][tex]$0.15 \, \text{m}$[/tex][/tex] apart, what is the gravitational force between them?

Given: [tex]G = 6.67 \times 10^{-11} \, \text{N} \cdot (\text{m}^2 / \text{kg}^2)[/tex]


Sagot :

Certainly! Let's solve this problem step-by-step using Newton's Law of Gravitation.

Newton's Law of Gravitation states that the gravitational force [tex]\( F \)[/tex] between two masses is given by the formula:

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

where:
- [tex]\( F \)[/tex] is the gravitational force between the two masses,
- [tex]\( G \)[/tex] is the gravitational constant,
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two objects,
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.

Given:
- The mass of the first shoe [tex]\( m_1 = 0.1 \, \text{kg} \)[/tex],
- The mass of the second shoe [tex]\( m_2 = 0.1 \, \text{kg} \)[/tex],
- The distance between the shoes [tex]\( r = 0.15 \, \text{m} \)[/tex],
- The gravitational constant [tex]\( G = 6.67 \times 10^{-11} \, \text{N} \cdot (\text{m}^2/\text{kg}^2) \)[/tex].

Now, substitute these values into the formula:

[tex]\[ F = 6.67 \times 10^{-11} \frac{0.1 \times 0.1}{0.15^2} \][/tex]

First, calculate [tex]\( 0.15^2 \)[/tex]:

[tex]\[ 0.15^2 = 0.0225 \][/tex]

Next, multiply the masses:

[tex]\[ 0.1 \times 0.1 = 0.01 \][/tex]

Now, compute the fraction:

[tex]\[ \frac{0.01}{0.0225} = 0.444444444\bar{4} \][/tex]

Finally, multiply by the gravitational constant [tex]\( G \)[/tex]:

[tex]\[ F = 6.67 \times 10^{-11} \times 0.444444444\bar{4} \][/tex]

[tex]\[ F \approx 2.964444444444445 \times 10^{-11} \, \text{N} \][/tex]

Therefore, the gravitational force between the two shoes is approximately:

[tex]\[ 2.964444444444445 \times 10^{-11} \, \text{N} \][/tex]