To determine the distance between Pluto and Charon given their gravitational force and masses, we can use Newton's Law of Universal Gravitation, which is given by the formula:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force between the objects,
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\(6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \)[/tex],
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two objects, and
- [tex]\( r \)[/tex] is the distance between the centers of the two objects.
Given:
[tex]\[ F = 3.61 \times 10^{18} \, \text{N} \][/tex]
[tex]\[ m_1 = 1.3 \times 10^{22} \, \text{kg} \, (\text{mass of Pluto}) \][/tex]
[tex]\[ m_2 = 1.6 \times 10^{21} \, \text{kg} \left(\text{mass of Charon}\right) \][/tex]
We aim to find the distance [tex]\( r \)[/tex] between Pluto and Charon. Rearrange the formula to solve for [tex]\( r \)[/tex]:
[tex]\[ r^2 = G \frac{m_1 m_2}{F} \][/tex]
Substitute the known values into the equation:
[tex]\[ r^2 = \left(6.67430 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2} \right) \left(\frac{1.3 \times 10^{22} \, \text{kg} \times 1.6 \times 10^{21} \, \text{kg}}{3.61 \times 10^{18} \, \text{N}}\right) \][/tex]
Perform the multiplication and division within the parentheses:
[tex]\[ r^2 = \left(6.67430 \times 10^{-11}\right) \left(\frac{2.08 \times 10^{43}}{3.61 \times 10^{18}}\right) \][/tex]
[tex]\[ r^2 = \left(6.67430 \times 10^{-11}\right) \left(5.76 \times 10^{24}\right) \][/tex]
[tex]\[ r^2 = 3.845580055401662 \times 10^{14} \, \text{m}^2 \][/tex]
To find [tex]\( r \)[/tex], take the square root of both sides:
[tex]\[ r = \sqrt{3.845580055401662 \times 10^{14}} \][/tex]
[tex]\[ r \approx 1.9610150574132934 \times 10^7 \, \text{m} \][/tex]
Approximately,
[tex]\[ r \approx 1.96 \times 10^7 \, \text{m} \][/tex]
None of the provided distances match the calculated distance exactly:
- [tex]\( 2.0 \times 10^7 \, \text{m} \)[/tex]
- [tex]\( 2.4 \times 10^{12} \, \text{m} \)[/tex]
- [tex]\( 3.8 \times 10^{14} \, \text{m} \)[/tex]
- [tex]\( 5.8 \times 10^{24} \, \text{m} \)[/tex]
Hence, the calculated distance [tex]\( r \approx 1.96 \times 10^7 \, \text{m} \)[/tex] does not exactly match any of the given options.
