Let's solve the problem step-by-step.
1. Identify the known values:
- Initial gravitational force, [tex]\( F_{\text{initial}} = 6.02 \times 10^{-11} \, \text{N} \)[/tex]
- Final gravitational force, [tex]\( F_{\text{final}} = 4.2 \times 10^{-9} \, \text{N} \)[/tex]
- Change in distance, [tex]\( \Delta r = 0.06 \, \text{m} \)[/tex]
- Gravitational constant, [tex]\( G = 6.67430 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \)[/tex]
2. Recall the gravitational force formula:
[tex]\[
F = G \frac{m_1 m_2}{r^2}
\][/tex]
- Let [tex]\( r_1 \)[/tex] represent the initial distance.
- Let [tex]\( r_2 \)[/tex] represent the final distance.
3. Set up the equations using given forces:
[tex]\[
F_{\text{initial}} = G \frac{m_1 m_2}{r_1^2}
\][/tex]
[tex]\[
F_{\text{final}} = G \frac{m_1 m_2}{r_2^2}
\][/tex]
where [tex]\( r_2 = r_1 + \Delta r \)[/tex].
4. Eliminate the mass terms to find the relationship between distances and forces:
[tex]\[
F_{\text{final}} = G \frac{m_1 m_2}{(r_1 + \Delta r)^2}
\][/tex]
[tex]\[
F_{\text{final}} = G \frac{m_1 m_2}{r_1^2} \left( \frac{r_1^2}{(r_1 + \Delta r)^2} \right)
\][/tex]
Using [tex]\( F_{\text{initial}} = G \frac{m_1 m_2}{r_1^2} \)[/tex]:
[tex]\[
F_{\text{final}} = F_{\text{initial}} \left( \frac{r_1^2}{(r_1 + \Delta r)^2} \right)
\][/tex]
5. Rearrange to solve for [tex]\( r_1 \)[/tex]:
[tex]\[
\frac{F_{\text{final}}}{F_{\text{initial}}} = \frac{r_1^2}{(r_1 + \Delta r)^2}
\][/tex]
Taking the square root on both sides:
[tex]\[
\sqrt{\frac{F_{\text{final}}}{F_{\text{initial}}}} = \frac{r_1}{r_1 + \Delta r}
\][/tex]
Cross-multiplying to solve for [tex]\( r_1 \)[/tex]:
[tex]\[
r_1 = (r_1 + \Delta r) \sqrt{\frac{F_{\text{initial}}}{F_{\text{final}}}}
\][/tex]
6. Isolate [tex]\( r_1 \)[/tex] on one side:
[tex]\[
r_1 = (r_1 + 0.06) \sqrt{\frac{F_{\text{initial}}}{F_{\text{final}}}}
\][/tex]
Let:
[tex]\[
k = \sqrt{\frac{F_{\text{initial}}}{F_{\text{final}}}}
\][/tex]
Then:
[tex]\[
r_1 = (r_1 + 0.06) k
\][/tex]
[tex]\[
r_1 = r_1 k + 0.06 k
\][/tex]
[tex]\[
r_1 (1 - k) = 0.06 k
\][/tex]
7. Solve for [tex]\( r_1 \)[/tex]:
[tex]\[
r_1 = \frac{0.06 k}{1 - k}
\][/tex]
8. Substitute the fraction of the forces:
[tex]\[
k = \sqrt{\frac{6.02 \times 10^{-11}}{4.2 \times 10^{-9}}}
\][/tex]
Using the given results:
[tex]\[
k \approx \frac{1}{8.16} \text{ (approximation)}
\][/tex]
[tex]\[
r_1 \approx \frac{0.06}{\frac{1}{8.16} - 1}
\][/tex]
[tex]\[
r_1 \approx 0.00816 \, \text{m}
\][/tex]
Thus, the initial distance of separation between the two particles is approximately 0.00816 meters.
