To determine at which stage the gravitational force between the Earth and a star is the greatest, let's consider each stage separately and use the gravitational force formula:
[tex]\[
F = G \frac{m_1 m_2}{r^2}
\][/tex]
Where:
- [tex]\( F \)[/tex] is the gravitational force
- [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] is the mass of the Earth, [tex]\( 5.97 \times 10^{24} \, \text{kg} \)[/tex]
- [tex]\( m_2 \)[/tex] is the mass of the star which varies with each stage
- [tex]\( r \)[/tex] is the distance between the Earth and the star (assuming [tex]\( r = 1 \)[/tex] for simplicity)
We need to calculate [tex]\( F \)[/tex] for each stage:
### Stage 1
[tex]\[
m_2 = 1.8 \times 10^{30} \, \text{kg}
\][/tex]
[tex]\[
F_1 = G \frac{(5.97 \times 10^{24}) (1.8 \times 10^{30})}{1^2} = 7.17220278 \times 10^{44} \, \text{N}
\][/tex]
### Stage 2
[tex]\[
m_2 = 2.7 \times 10^{27} \, \text{kg}
\][/tex]
[tex]\[
F_2 = G \frac{(5.97 \times 10^{24}) (2.7 \times 10^{27})}{1^2} = 1.07583042 \times 10^{42} \, \text{N}
\][/tex]
### Stage 3
[tex]\[
m_2 = 5.0 \times 10^{25} \, \text{kg}
\][/tex]
[tex]\[
F_3 = G \frac{(5.97 \times 10^{24}) (5.0 \times 10^{25})}{1^2} = 1.99227855 \times 10^{40} \, \text{N}
\][/tex]
### Stage 4
[tex]\[
m_2 = 4.5 \times 10^{23} \, \text{kg}
\][/tex]
[tex]\[
F_4 = G \frac{(5.97 \times 10^{24}) (4.5 \times 10^{23})}{1^2} = 1.79305069 \times 10^{38} \, \text{N}
\][/tex]
From the calculated forces:
- [tex]\( F_1 = 7.17220278 \times 10^{44} \, \text{N} \)[/tex]
- [tex]\( F_2 = 1.07583042 \times 10^{42} \, \text{N} \)[/tex]
- [tex]\( F_3 = 1.99227855 \times 10^{40} \, \text{N} \)[/tex]
- [tex]\( F_4 = 1.79305069 \times 10^{38} \, \text{N} \)[/tex]
The greatest gravitational force occurs in Stage 1.
Therefore, the gravitational force between Earth and the star is the greatest at Stage 1.
