What is the gravitational force between a [tex][tex]$70 \, \text{kg}$[/tex][/tex] physics student and her [tex][tex]$1 \, \text{kg}$[/tex][/tex] textbook at a distance of 1 meter?

- Object [tex]1\left(m_1\right): 70 \, \text{kg}[/tex] student
- Object [tex]2\left(m_2\right): 1 \, \text{kg}[/tex] textbook
- Distance [tex](r): 1 \, \text{meter}[/tex]

Use the gravitational force formula:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]

where:
[tex]\[ G = 6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \][/tex]



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.