Answer :
To determine the gravitational force between you and your textbook, we will apply Newton's law of gravitation, which states that the gravitational force [tex]\( F_{\text{gravity}} \)[/tex] between two masses is given by:
[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 first object (in this case, your mass, [tex]\( 72 \, \text{kg} \)[/tex]).
- [tex]\( m_2 \)[/tex] is the mass of the second object (in this case, the mass of the textbook, [tex]\( 3.7 \, \text{kg} \)[/tex]).
- [tex]\( r \)[/tex] is the distance between the centers of the two objects (in this case, [tex]\( 0.33 \, \text{m} \)[/tex]).
Let's substitute the given values into the formula:
[tex]\[ F_{\text{gravity}} = \frac{6.67 \times 10^{-11} \cdot 72 \cdot 3.7}{0.33^2} \][/tex]
First, calculate the denominator:
[tex]\[ 0.33^2 = 0.1089 \][/tex]
Next, substitute this value back into the equation:
[tex]\[ F_{\text{gravity}} = \frac{6.67 \times 10^{-11} \cdot 72 \cdot 3.7}{0.1089} \][/tex]
Now, perform the multiplication in the numerator:
[tex]\[ 6.67 \times 10^{-11} \cdot 72 \cdot 3.7 = 1.779816 \times 10^{-8} \][/tex]
Then, divide by [tex]\( 0.1089 \)[/tex]:
[tex]\[ F_{\text{gravity}} = \frac{1.779816 \times 10^{-8}}{0.1089} \][/tex]
By performing the division, we get:
[tex]\[ F_{\text{gravity}} \approx 1.63 \times 10^{-7} \, \text{N} \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{1.63 \times 10^{-7} \, \text{N}} \][/tex]
The option corresponding to this value is:
B. [tex]\(1.63 \times 10^{-7} \, \text{N}\)[/tex]
[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 first object (in this case, your mass, [tex]\( 72 \, \text{kg} \)[/tex]).
- [tex]\( m_2 \)[/tex] is the mass of the second object (in this case, the mass of the textbook, [tex]\( 3.7 \, \text{kg} \)[/tex]).
- [tex]\( r \)[/tex] is the distance between the centers of the two objects (in this case, [tex]\( 0.33 \, \text{m} \)[/tex]).
Let's substitute the given values into the formula:
[tex]\[ F_{\text{gravity}} = \frac{6.67 \times 10^{-11} \cdot 72 \cdot 3.7}{0.33^2} \][/tex]
First, calculate the denominator:
[tex]\[ 0.33^2 = 0.1089 \][/tex]
Next, substitute this value back into the equation:
[tex]\[ F_{\text{gravity}} = \frac{6.67 \times 10^{-11} \cdot 72 \cdot 3.7}{0.1089} \][/tex]
Now, perform the multiplication in the numerator:
[tex]\[ 6.67 \times 10^{-11} \cdot 72 \cdot 3.7 = 1.779816 \times 10^{-8} \][/tex]
Then, divide by [tex]\( 0.1089 \)[/tex]:
[tex]\[ F_{\text{gravity}} = \frac{1.779816 \times 10^{-8}}{0.1089} \][/tex]
By performing the division, we get:
[tex]\[ F_{\text{gravity}} \approx 1.63 \times 10^{-7} \, \text{N} \][/tex]
Thus, the correct answer is:
[tex]\[ \boxed{1.63 \times 10^{-7} \, \text{N}} \][/tex]
The option corresponding to this value is:
B. [tex]\(1.63 \times 10^{-7} \, \text{N}\)[/tex]