Answer :
To find the gravitational force between a 70 kg physics student and her 1 kg textbook at a distance of 1 meter, we can use Newton's law of universal gravitation. The 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]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( G \)[/tex] is the gravitational constant, approximately equal to [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex],
- [tex]\( m_1 = 70 \)[/tex] kg (mass of the student),
- [tex]\( m_2 = 1 \)[/tex] kg (mass of the textbook),
- [tex]\( r = 1 \)[/tex] meter (distance between the student and the textbook).
Now, let's plug in the values:
[tex]\[ F = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \times \frac{70 \, \text{kg} \times 1 \, \text{kg}}{(1 \, \text{m})^2} \][/tex]
Simplifying the expression inside the parentheses:
[tex]\[ F = 6.67430 \times 10^{-11} \times \frac{70}{1^2} \][/tex]
Since [tex]\( 70 \)[/tex] divided by [tex]\( 1^2 \)[/tex] is simply [tex]\( 70 \)[/tex]:
[tex]\[ F = 6.67430 \times 10^{-11} \times 70 \][/tex]
Finally, multiplying these values yields:
[tex]\[ F \approx 4.672010 \times 10^{-9} \, \text{N} \][/tex]
Therefore, the gravitational force between the 70 kg student and the 1 kg textbook at a distance of 1 meter is approximately [tex]\( 4.672010 \times 10^{-9} \)[/tex] newtons.
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( G \)[/tex] is the gravitational constant, approximately equal to [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex],
- [tex]\( m_1 = 70 \)[/tex] kg (mass of the student),
- [tex]\( m_2 = 1 \)[/tex] kg (mass of the textbook),
- [tex]\( r = 1 \)[/tex] meter (distance between the student and the textbook).
Now, let's plug in the values:
[tex]\[ F = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \times \frac{70 \, \text{kg} \times 1 \, \text{kg}}{(1 \, \text{m})^2} \][/tex]
Simplifying the expression inside the parentheses:
[tex]\[ F = 6.67430 \times 10^{-11} \times \frac{70}{1^2} \][/tex]
Since [tex]\( 70 \)[/tex] divided by [tex]\( 1^2 \)[/tex] is simply [tex]\( 70 \)[/tex]:
[tex]\[ F = 6.67430 \times 10^{-11} \times 70 \][/tex]
Finally, multiplying these values yields:
[tex]\[ F \approx 4.672010 \times 10^{-9} \, \text{N} \][/tex]
Therefore, the gravitational force between the 70 kg student and the 1 kg textbook at a distance of 1 meter is approximately [tex]\( 4.672010 \times 10^{-9} \)[/tex] newtons.