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To explore how the gravitational force changes under specific conditions, let's begin by understanding the basic gravitational force formula and how each variable affects it. 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 given by Newton's law of universal gravitation:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where [tex]\( G \)[/tex] is the universal gravitational constant.
### Initial condition
Let's start with the initial condition where:
- [tex]\( m_1 \)[/tex] is 1 kg,
- [tex]\( m_2 \)[/tex] is 1 kg,
- [tex]\( r \)[/tex] is 1 meter.
Using the formula:
[tex]\[ F_{\text{initial}} = G \frac{m_1 \times m_2}{r^2} \][/tex]
[tex]\[ F_{\text{initial}} = G \frac{1 \times 1}{1^2} = G \][/tex]
### Modified condition
Now, let's consider the modified condition where:
- The mass of each object is doubled, hence [tex]\( m_1 \)[/tex] becomes 2 kg and [tex]\( m_2 \)[/tex] becomes 2 kg,
- The distance is made one fourth of the initial distance, so [tex]\( r \)[/tex] becomes [tex]\( \frac{1}{4} \)[/tex] meters.
Using the formula again:
[tex]\[ F_{\text{final}} = G \frac{m_1' \times m_2'}{r'^2} \][/tex]
where [tex]\( m_1' = 2 \, \text{kg} \)[/tex], [tex]\( m_2' = 2 \, \text{kg} \)[/tex], and [tex]\( r' = \frac{1}{4} \, \text{meters} \)[/tex].
Substituting these values in:
[tex]\[ F_{\text{final}} = G \frac{2 \times 2}{\left(\frac{1}{4}\right)^2} = G \frac{4}{\left(\frac{1}{4}\right)^2} = G \frac{4}{\frac{1}{16}} = G \times 4 \times 16 = 64G \][/tex]
### Comparison of Forces
Now, let's compare the final force [tex]\( F_{\text{final}} \)[/tex] with the initial force [tex]\( F_{\text{initial}} \)[/tex]:
[tex]\[ F_{\text{initial}} = G \][/tex]
[tex]\[ F_{\text{final}} = 64G \][/tex]
The change in force can be calculated by the ratio:
[tex]\[ \text{Change in Force} = \frac{F_{\text{final}}}{F_{\text{initial}}} = \frac{64G}{G} = 64 \][/tex]
### Conclusion
Therefore, under the described conditions, the gravitational force between the two objects increases by a factor of 64. The initial gravitational force is [tex]\( 6.6743 \times 10^{-11} \, \text{N} \)[/tex], the final gravitational force is [tex]\( 4.271552 \times 10^{-9} \, \text{N} \)[/tex], and the change in force is 64 times the initial force.
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where [tex]\( G \)[/tex] is the universal gravitational constant.
### Initial condition
Let's start with the initial condition where:
- [tex]\( m_1 \)[/tex] is 1 kg,
- [tex]\( m_2 \)[/tex] is 1 kg,
- [tex]\( r \)[/tex] is 1 meter.
Using the formula:
[tex]\[ F_{\text{initial}} = G \frac{m_1 \times m_2}{r^2} \][/tex]
[tex]\[ F_{\text{initial}} = G \frac{1 \times 1}{1^2} = G \][/tex]
### Modified condition
Now, let's consider the modified condition where:
- The mass of each object is doubled, hence [tex]\( m_1 \)[/tex] becomes 2 kg and [tex]\( m_2 \)[/tex] becomes 2 kg,
- The distance is made one fourth of the initial distance, so [tex]\( r \)[/tex] becomes [tex]\( \frac{1}{4} \)[/tex] meters.
Using the formula again:
[tex]\[ F_{\text{final}} = G \frac{m_1' \times m_2'}{r'^2} \][/tex]
where [tex]\( m_1' = 2 \, \text{kg} \)[/tex], [tex]\( m_2' = 2 \, \text{kg} \)[/tex], and [tex]\( r' = \frac{1}{4} \, \text{meters} \)[/tex].
Substituting these values in:
[tex]\[ F_{\text{final}} = G \frac{2 \times 2}{\left(\frac{1}{4}\right)^2} = G \frac{4}{\left(\frac{1}{4}\right)^2} = G \frac{4}{\frac{1}{16}} = G \times 4 \times 16 = 64G \][/tex]
### Comparison of Forces
Now, let's compare the final force [tex]\( F_{\text{final}} \)[/tex] with the initial force [tex]\( F_{\text{initial}} \)[/tex]:
[tex]\[ F_{\text{initial}} = G \][/tex]
[tex]\[ F_{\text{final}} = 64G \][/tex]
The change in force can be calculated by the ratio:
[tex]\[ \text{Change in Force} = \frac{F_{\text{final}}}{F_{\text{initial}}} = \frac{64G}{G} = 64 \][/tex]
### Conclusion
Therefore, under the described conditions, the gravitational force between the two objects increases by a factor of 64. The initial gravitational force is [tex]\( 6.6743 \times 10^{-11} \, \text{N} \)[/tex], the final gravitational force is [tex]\( 4.271552 \times 10^{-9} \, \text{N} \)[/tex], and the change in force is 64 times the initial force.
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