To find the mass of the Sun given the mass of the Earth, the distance between the Earth and the Sun, and the gravitational force between them, we can use Newton’s law of universal gravitation. The formula for 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:
[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]\( m_1 \)[/tex] is the mass of the first object (Earth),
- [tex]\( m_2\)[/tex] is the mass of the second object (Sun),
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
Given:
- [tex]\( F = 3.5 \times 10^{22} \, \text{N} \)[/tex],
- [tex]\( m_1 = 6 \times 10^{24} \, \text{kg} \)[/tex] (mass of Earth),
- [tex]\( r = 1.5 \times 10^{11} \, \text{m} \)[/tex],
- [tex]\( G = 6.7 \times 10^{-11} \, \text{N m}^2 \text{ kg}^{-2} \)[/tex].
We need to solve for [tex]\( m_2 \)[/tex]. Rearranging the equation to solve for [tex]\( m_2 \)[/tex]:
[tex]\[ m_2 = \frac{F \cdot r^2}{G \cdot m_1} \][/tex]
Substitute the known values into the equation:
[tex]\[ m_2 = \frac{(3.5 \times 10^{22} \, \text{N}) \cdot (1.5 \times 10^{11} \, \text{m})^2}{(6.7 \times 10^{-11} \, \text{N m}^2 \text{ kg}^{-2}) \cdot (6 \times 10^{24} \, \text{kg})} \][/tex]
First, calculate [tex]\( r^2 \)[/tex]:
[tex]\[ (1.5 \times 10^{11} \, \text{m})^2 = 2.25 \times 10^{22} \, \text{m}^2 \][/tex]
Now substitute this back into the equation:
[tex]\[ m_2 = \frac{(3.5 \times 10^{22} \, \text{N}) \cdot (2.25 \times 10^{22} \, \text{m}^2)}{(6.7 \times 10^{-11} \, \text{N m}^2 \text{ kg}^{-2}) \cdot (6 \times 10^{24} \, \text{kg})} \][/tex]
Next, calculate the numerator:
[tex]\[ 3.5 \times 10^{22} \, \text{N} \times 2.25 \times 10^{22} \, \text{m}^2 = 7.875 \times 10^{44} \, \text{N m}^2 \][/tex]
Now calculate the denominator:
[tex]\[ 6.7 \times 10^{-11} \, \text{N m}^2 \text{ kg}^{-2} \times 6 \times 10^{24} \, \text{kg} = 4.02 \times 10^{14} \, \text{N m}^2 \text{ kg}^{-1} \][/tex]
Finally, divide the numerator by the denominator to find [tex]\( m_2 \)[/tex]:
[tex]\[ m_2 = \frac{7.875 \times 10^{44}}{4.02 \times 10^{14}} = 1.9589552238805968 \times 10^{30} \, \text{kg} \][/tex]
Therefore, the mass of the Sun is approximately:
[tex]\[ m_2 \approx 1.96 \times 10^{30} \, \text{kg} \][/tex]
