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Step-by-Step Solution
Step 1: Understand the Problem
We need to compare the difference in gravitational forces (sometimes referred to as “tidal forces”) exerted by the Moon and by the Sun on the Earth, at the Earth's nearest and farthest points from each of these bodies. Symbolically:
ΔF1 = difference in force due to the Moon.
ΔF2 = difference in force due to the Sun.
We are asked to find the ratio (ΔF1 / ΔF2) and determine the numeric value closest to it.
Step 2: Write the General Expression for Tidal Force
The gravitational force on a mass Me (Earth) due to another mass m (Moon or Sun) separated by distance r is
$F = \frac{GM_{\mathrm{e}}\,m}{r^{2}}.$
However, the tidal force depends on how that gravitational pull changes over an extended body. If the radius of Earth is Re, then the difference in gravitational force between the nearest point and the farthest point from the external mass can be approximated by
$\Delta F \sim \frac{dF}{dr}\,\Delta r \;=\; \frac{2\,G\,M_{\mathrm{e}}\,m}{r^{3}} \,\Delta r,$
where Δr is effectively the diameter of the Earth (for tidal force difference from near side to far side).
Step 3: Express Differences for Moon and Sun
For the Moon, let:
$r_{1} =$ mean distance between Earth and Moon,
$\Delta F_{1} \approx \frac{2\,G\,M_{\mathrm{e}}\,m}{r_{1}^{3}} \,\Delta r_{1} \,.
For the Sun, let:
$r_{2} =$ mean distance between Earth and Sun,
$\Delta F_{2} \approx \frac{2\,G\,M_{\mathrm{e}}\,M_{s}}{r_{2}^{3}} \,\Delta r_{2} \,.
Assuming Δr1 = Δr2 = 2Re (the Earth's diameter), we can take this factor to be common when forming the ratio.
Step 4: Form the Required Ratio
The ratio of these differences is:
$\frac{\Delta F_{1}}{\Delta F_{2}} \;=\; \frac{\left(\frac{2\,G\,M_{\mathrm{e}}\,m}{r_{1}^{3}}\right)\Delta r_{1}}{\left(\frac{2\,G\,M_{\mathrm{e}}\,M_{s}}{r_{2}^{3}}\right)\Delta r_{2}}
\;=\;\left(\frac{m}{M_{s}}\right)\,\left(\frac{r_{2}^{3}}{r_{1}^{3}}\right)\,\left(\frac{\Delta r_{1}}{\Delta r_{2}} \right).$
Since Δr1 = Δr2, the ratio simplifies to:
$\frac{\Delta F_{1}}{\Delta F_{2}} \;=\;\left(\frac{m}{M_{s}}\right)\,\left(\frac{r_{2}^{3}}{r_{1}^{3}}\right).$
Step 5: Substitute the Given Values
From the problem statement:
Moon’s mass, $m = 8 \times 10^{22}\,\text{kg}.$
Sun’s mass, $M_{s} = 2 \times 10^{30}\,\text{kg}.$
Moon’s distance from Earth, $r_{1} = 0.4 \times 10^{6}\,\text{km}.$
Sun’s distance from Earth, $r_{2} = 150 \times 10^{6}\,\text{km}.$
Hence:
$\frac{\Delta F_{1}}{\Delta F_{2}} \;=\;
\left(\frac{8 \times 10^{22}}{2 \times 10^{30}}\right)
\times
\left(\frac{(150 \times 10^{6})^{3}}{(0.4 \times 10^{6})^{3}}\right).$
Step 6: Calculate the Numeric Value
First, simplify the ratio of masses:
$\frac{8 \times 10^{22}}{2 \times 10^{30}} = \frac{8}{2} \times 10^{-8} = 4 \times 10^{-8}.$
Next, simplify the distances' cubic ratio:
$\left(\frac{150 \times 10^{6}}{0.4 \times 10^{6}}\right)^{3}
= \left(\frac{150}{0.4}\right)^{3}
= (375)^{3}
= 375^{3}.
$
$375^{3} = 375 \times 375 \times 375.$ Numerically,
$375 \times 375 = 140625, \quad 140625 \times 375 \approx 5.2734 \times 10^{7}
\text{ (more precisely } 5.2734 \times 10^{7}\text{).}
$
So approximately,
$(375)^{3} \approx 5.3 \times 10^{7}.
$
Putting these together:
$\frac{\Delta F_{1}}{\Delta F_{2}}
\approx (4 \times 10^{-8}) \times (5.3 \times 10^{7})
= 4 \times 5.3 \times 10^{-1}
= 21.2 \times 10^{-1}
= 2.12.
$
This is close to 2, which matches the given correct answer.
Final Answer
The ratio $ \frac{\Delta F_{1}}{\Delta F_{2}} $ is approximately 2.
