Answer :
To find the distance between Pluto and Charon, we can use the gravitational force formula:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
Where:
- [tex]\( F \)[/tex] is the gravitational force between the two masses,
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-1} \)[/tex],
- [tex]\( m_1 \)[/tex] is the mass of Pluto, [tex]\( 1.3 \times 10^{22} \, \text{kg} \)[/tex],
- [tex]\( m_2 \)[/tex] is the mass of Charon, [tex]\( 1.6 \times 10^{21} \, \text{kg} \)[/tex],
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
We need to rearrange this formula to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{\frac{G m_1 m_2}{F}} \][/tex]
First, substitute the known values into the rearranged formula:
[tex]\[ r = \sqrt{\frac{6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-1} \times 1.3 \times 10^{22} \, \text{kg} \times 1.6 \times 10^{21} \, \text{kg}}{3.61 \times 10^{18} \, \text{N}}} \][/tex]
Simplify the expression inside the square root:
[tex]\[ r = \sqrt{\frac{6.67430 \times 1.3 \times 1.6 \times 10^{-11 + 22 + 21}}{3.61 \times 10^{18}}} \][/tex]
Combine the constants:
[tex]\[ r = \sqrt{\frac{13.877392 \times 10^{32}}{3.61 \times 10^{18}}} \][/tex]
Now, divide the exponents:
[tex]\[ r = \sqrt{3.84494155 \times 10^{14}} \][/tex]
Finally, take the square root:
[tex]\[ r \approx 19610150.574132934 \, \text{m} \][/tex]
Thus, the distance between Pluto and Charon is approximately [tex]\( 1.96 \times 10^7 \, \text{m} \)[/tex].
Looking at the provided multiple-choice answers, the closest option is:
[tex]\[ 2.0 \times 10^7 \, \text{m} \][/tex]
So the correct answer is:
[tex]\[ 2.0 \times 10^7 \, \text{m} \][/tex]
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
Where:
- [tex]\( F \)[/tex] is the gravitational force between the two masses,
- [tex]\( G \)[/tex] is the gravitational constant, [tex]\( 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-1} \)[/tex],
- [tex]\( m_1 \)[/tex] is the mass of Pluto, [tex]\( 1.3 \times 10^{22} \, \text{kg} \)[/tex],
- [tex]\( m_2 \)[/tex] is the mass of Charon, [tex]\( 1.6 \times 10^{21} \, \text{kg} \)[/tex],
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
We need to rearrange this formula to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{\frac{G m_1 m_2}{F}} \][/tex]
First, substitute the known values into the rearranged formula:
[tex]\[ r = \sqrt{\frac{6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-1} \times 1.3 \times 10^{22} \, \text{kg} \times 1.6 \times 10^{21} \, \text{kg}}{3.61 \times 10^{18} \, \text{N}}} \][/tex]
Simplify the expression inside the square root:
[tex]\[ r = \sqrt{\frac{6.67430 \times 1.3 \times 1.6 \times 10^{-11 + 22 + 21}}{3.61 \times 10^{18}}} \][/tex]
Combine the constants:
[tex]\[ r = \sqrt{\frac{13.877392 \times 10^{32}}{3.61 \times 10^{18}}} \][/tex]
Now, divide the exponents:
[tex]\[ r = \sqrt{3.84494155 \times 10^{14}} \][/tex]
Finally, take the square root:
[tex]\[ r \approx 19610150.574132934 \, \text{m} \][/tex]
Thus, the distance between Pluto and Charon is approximately [tex]\( 1.96 \times 10^7 \, \text{m} \)[/tex].
Looking at the provided multiple-choice answers, the closest option is:
[tex]\[ 2.0 \times 10^7 \, \text{m} \][/tex]
So the correct answer is:
[tex]\[ 2.0 \times 10^7 \, \text{m} \][/tex]
