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Sagot :
Certainly! Let's work through this step by step. We will use Newton's law of gravitation which states that the gravitational force (F) between two masses is given by:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( G \)[/tex] is the gravitational constant ([tex]\( G = 6.67430 \times 10^{-11} \)[/tex] m[tex]\(^3\)[/tex] kg[tex]\(^{-1}\)[/tex] s[tex]\(^{-2}\)[/tex]),
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two bodies,
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
### Step 1: Initial Setup
Initially, we start with:
- [tex]\( m_1 = 1 \)[/tex] kg
- [tex]\( m_2 = 1 \)[/tex] kg
- [tex]\( r = 1 \)[/tex] m
Using these values, we can calculate the initial gravitational force:
[tex]\[ F_{\text{initial}} = G \frac{m_1 m_2}{r^2} \][/tex]
Substitute [tex]\( G \)[/tex], [tex]\( m_1 \)[/tex], [tex]\( m_2 \)[/tex], and [tex]\( r \)[/tex]:
[tex]\[ F_{\text{initial}} = 6.67430 \times 10^{-11} \frac{1 \times 1}{1^2} = 6.67430 \times 10^{-11} \, \text{N} \][/tex]
### Step 2: Changing Masses and Distance
Now, if we double the masses and halve the distance, we have:
- [tex]\( m_1' = 2 \times m_1 = 2 \times 1 = 2 \)[/tex] kg
- [tex]\( m_2' = 2 \times m_2 = 2 \times 1 = 2 \)[/tex] kg
- [tex]\( r' = \frac{r}{2} = \frac{1}{2} \)[/tex] m
Substituting these changed values back into the gravitational force formula, we get the new force ([tex]\( F_{\text{new}} \)[/tex]):
[tex]\[ F_{\text{new}} = G \frac{m_1' m_2'}{r'^2} \][/tex]
### Step 3: Calculating the New Force
Substitute [tex]\( G \)[/tex], [tex]\( m_1' \)[/tex], [tex]\( m_2' \)[/tex], and [tex]\( r' \)[/tex]:
[tex]\[ F_{\text{new}} = 6.67430 \times 10^{-11} \frac{2 \times 2}{(\frac{1}{2})^2} \][/tex]
First, compute the numerator and the denominator inside the fraction separately:
[tex]\[ F_{\text{new}} = 6.67430 \times 10^{-11} \frac{4}{(\frac{1}{2})^2} \][/tex]
[tex]\[ (\frac{1}{2})^2 = \frac{1}{4} \][/tex]
Now:
[tex]\[ F_{\text{new}} = 6.67430 \times 10^{-11} \frac{4}{\frac{1}{4}} \][/tex]
Simplify the fraction:
[tex]\[ \frac{4}{\frac{1}{4}} = 4 \times 4 = 16 \][/tex]
So:
[tex]\[ F_{\text{new}} = 6.67430 \times 10^{-11} \times 16 \][/tex]
Finally, calculate the product:
[tex]\[ F_{\text{new}} = 1.067888 \times 10^{-9} \, \text{N} \][/tex]
### Summary
- The initial gravitational force between the two bodies was:
[tex]\[ 6.67430 \times 10^{-11} \, \text{N} \][/tex]
- The new gravitational force, after doubling the masses and halving the distance, is:
[tex]\[ 1.067888 \times 10^{-9} \, \text{N} \][/tex]
Thus, the result demonstrates a significant increase in the gravitational force when both masses are doubled and the distance between them is halved.
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( G \)[/tex] is the gravitational constant ([tex]\( G = 6.67430 \times 10^{-11} \)[/tex] m[tex]\(^3\)[/tex] kg[tex]\(^{-1}\)[/tex] s[tex]\(^{-2}\)[/tex]),
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two bodies,
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
### Step 1: Initial Setup
Initially, we start with:
- [tex]\( m_1 = 1 \)[/tex] kg
- [tex]\( m_2 = 1 \)[/tex] kg
- [tex]\( r = 1 \)[/tex] m
Using these values, we can calculate the initial gravitational force:
[tex]\[ F_{\text{initial}} = G \frac{m_1 m_2}{r^2} \][/tex]
Substitute [tex]\( G \)[/tex], [tex]\( m_1 \)[/tex], [tex]\( m_2 \)[/tex], and [tex]\( r \)[/tex]:
[tex]\[ F_{\text{initial}} = 6.67430 \times 10^{-11} \frac{1 \times 1}{1^2} = 6.67430 \times 10^{-11} \, \text{N} \][/tex]
### Step 2: Changing Masses and Distance
Now, if we double the masses and halve the distance, we have:
- [tex]\( m_1' = 2 \times m_1 = 2 \times 1 = 2 \)[/tex] kg
- [tex]\( m_2' = 2 \times m_2 = 2 \times 1 = 2 \)[/tex] kg
- [tex]\( r' = \frac{r}{2} = \frac{1}{2} \)[/tex] m
Substituting these changed values back into the gravitational force formula, we get the new force ([tex]\( F_{\text{new}} \)[/tex]):
[tex]\[ F_{\text{new}} = G \frac{m_1' m_2'}{r'^2} \][/tex]
### Step 3: Calculating the New Force
Substitute [tex]\( G \)[/tex], [tex]\( m_1' \)[/tex], [tex]\( m_2' \)[/tex], and [tex]\( r' \)[/tex]:
[tex]\[ F_{\text{new}} = 6.67430 \times 10^{-11} \frac{2 \times 2}{(\frac{1}{2})^2} \][/tex]
First, compute the numerator and the denominator inside the fraction separately:
[tex]\[ F_{\text{new}} = 6.67430 \times 10^{-11} \frac{4}{(\frac{1}{2})^2} \][/tex]
[tex]\[ (\frac{1}{2})^2 = \frac{1}{4} \][/tex]
Now:
[tex]\[ F_{\text{new}} = 6.67430 \times 10^{-11} \frac{4}{\frac{1}{4}} \][/tex]
Simplify the fraction:
[tex]\[ \frac{4}{\frac{1}{4}} = 4 \times 4 = 16 \][/tex]
So:
[tex]\[ F_{\text{new}} = 6.67430 \times 10^{-11} \times 16 \][/tex]
Finally, calculate the product:
[tex]\[ F_{\text{new}} = 1.067888 \times 10^{-9} \, \text{N} \][/tex]
### Summary
- The initial gravitational force between the two bodies was:
[tex]\[ 6.67430 \times 10^{-11} \, \text{N} \][/tex]
- The new gravitational force, after doubling the masses and halving the distance, is:
[tex]\[ 1.067888 \times 10^{-9} \, \text{N} \][/tex]
Thus, the result demonstrates a significant increase in the gravitational force when both masses are doubled and the distance between them is halved.
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