Answer :
To determine the mass of the planet, we can use the formula for the gravitational force:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force (in Newtons, [tex]\( N \)[/tex]),
- [tex]\( G \)[/tex] is the gravitational constant ([tex]\( 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex]),
- [tex]\( m_1 \)[/tex] is the mass of the satellite,
- [tex]\( m_2 \)[/tex] is the mass of the planet,
- [tex]\( r \)[/tex] is the radius of the orbit (in meters).
In this problem:
- The gravitational pull, [tex]\( F = 620 \, N \)[/tex],
- The mass of the satellite, [tex]\( m_1 = 450 \, kg \)[/tex],
- The orbital radius, [tex]\( r = 4.0 \times 10^6 \, m \)[/tex],
- The gravitational constant, [tex]\( G = 6.67 \times 10^{-11} \, N \cdot m^2 / kg^2 \)[/tex].
We need to find the mass of the planet, [tex]\( m_2 \)[/tex].
First, rearrange the formula 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]\[ m_2 = \frac{620 \, N \cdot (4.0 \times 10^6 \, m)^2}{6.67 \times 10^{-11} \, N \cdot m^2 / kg^2 \cdot 450 \, kg} \][/tex]
Calculate the value step by step:
1. Calculate [tex]\( r^2 \)[/tex]:
[tex]\[ (4.0 \times 10^6 \, m)^2 = 16 \times 10^{12} \, m^2 \][/tex]
2. Multiply [tex]\( F \)[/tex] by [tex]\( r^2 \)[/tex]:
[tex]\[ 620 \, N \cdot 16 \times 10^{12} \, m^2 = 9.92 \times 10^{15} \, N \cdot m^2 \][/tex]
3. Multiply [tex]\( G \cdot m_1 \)[/tex]:
[tex]\[ 6.67 \times 10^{-11} \, N \cdot m^2 / kg^2 \cdot 450 \, kg = 3.0015 \times 10^{-8} \, N \cdot m^2 / kg \][/tex]
4. Divide the result from step 2 by the result from step 3:
[tex]\[ m_2 = \frac{9.92 \times 10^{15}}{3.0015 \times 10^{-8}} \approx 3.305 \times 10^{23} \, kg \][/tex]
Thus, the approximate mass of the planet is:
[tex]\[ 3.305 \times 10^{23} \, kg \][/tex]
The answer is closest to option A:
[tex]\[ A. 3.3 \times 10^{23} \, kg \][/tex]
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( F \)[/tex] is the gravitational force (in Newtons, [tex]\( N \)[/tex]),
- [tex]\( G \)[/tex] is the gravitational constant ([tex]\( 6.67 \times 10^{-11} \, \text{N} \cdot \text{m}^2 / \text{kg}^2 \)[/tex]),
- [tex]\( m_1 \)[/tex] is the mass of the satellite,
- [tex]\( m_2 \)[/tex] is the mass of the planet,
- [tex]\( r \)[/tex] is the radius of the orbit (in meters).
In this problem:
- The gravitational pull, [tex]\( F = 620 \, N \)[/tex],
- The mass of the satellite, [tex]\( m_1 = 450 \, kg \)[/tex],
- The orbital radius, [tex]\( r = 4.0 \times 10^6 \, m \)[/tex],
- The gravitational constant, [tex]\( G = 6.67 \times 10^{-11} \, N \cdot m^2 / kg^2 \)[/tex].
We need to find the mass of the planet, [tex]\( m_2 \)[/tex].
First, rearrange the formula 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]\[ m_2 = \frac{620 \, N \cdot (4.0 \times 10^6 \, m)^2}{6.67 \times 10^{-11} \, N \cdot m^2 / kg^2 \cdot 450 \, kg} \][/tex]
Calculate the value step by step:
1. Calculate [tex]\( r^2 \)[/tex]:
[tex]\[ (4.0 \times 10^6 \, m)^2 = 16 \times 10^{12} \, m^2 \][/tex]
2. Multiply [tex]\( F \)[/tex] by [tex]\( r^2 \)[/tex]:
[tex]\[ 620 \, N \cdot 16 \times 10^{12} \, m^2 = 9.92 \times 10^{15} \, N \cdot m^2 \][/tex]
3. Multiply [tex]\( G \cdot m_1 \)[/tex]:
[tex]\[ 6.67 \times 10^{-11} \, N \cdot m^2 / kg^2 \cdot 450 \, kg = 3.0015 \times 10^{-8} \, N \cdot m^2 / kg \][/tex]
4. Divide the result from step 2 by the result from step 3:
[tex]\[ m_2 = \frac{9.92 \times 10^{15}}{3.0015 \times 10^{-8}} \approx 3.305 \times 10^{23} \, kg \][/tex]
Thus, the approximate mass of the planet is:
[tex]\[ 3.305 \times 10^{23} \, kg \][/tex]
The answer is closest to option A:
[tex]\[ A. 3.3 \times 10^{23} \, kg \][/tex]
