Take the mean distance of the moon and the sun from the earth to be $$0.4 \times 10^6$$ km and $$150 \times 10^6$$ km respectively. Their masses are $$8 \times 10^{22}$$ kg and $$2 \times 10^{30}$$ kg respectively. The radius of the earth is 6400 km. Let $$\Delta F_1$$ be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and $$\Delta F_2$$ be the difference in the force exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to $$\frac{\Delta F_1}{\Delta F_2}$$ is:

We begin with Newton’s universal law of gravitation, which gives the magnitude of the force exerted by a body of mass $$M$$ on a small test mass $$m$$ situated at a distance $$r$$ as

$$F \;=\; \dfrac{G\,M\,m}{r^{2}},$$

where $$G$$ is the gravitational constant.

The earth is a sphere of radius $$R = 6400\;\text{km}$$. For a celestial body whose centre is at a distance $$D$$ from the centre of the earth, the distance of that body from

• the nearest point on the earth’s surface is $$D - R$$,

• the farthest point on the earth’s surface is $$D + R$$.

Hence the forces on a unit test mass at these two points are

$$F_{\text{near}} = \dfrac{G\,M}{(D - R)^{2}}, \qquad F_{\text{far}} = \dfrac{G\,M}{(D + R)^{2}}.$$

The difference (which is what produces the tidal effect) is therefore

$$\Delta F \;=\; F_{\text{near}} \;-\; F_{\text{far}} \;=\; G\,M \!\left[ \dfrac{1}{(D - R)^{2}} \;-\; \dfrac{1}{(D + R)^{2}} \right].$$

Because the earth’s radius is very small compared with any celestial distance (i.e. $$R \ll D$$), we expand each denominator by writing

$$(D \pm R)^{-2} \;=\; D^{-2}\,(1 \pm \tfrac{R}{D})^{-2} \;\approx\; D^{-2}\,\bigl(1 \mp 2\tfrac{R}{D}\bigr),$$

using the binomial approximation $$(1 + x)^{n} \approx 1 + nx$$ for small $$x$$.

Substituting the expanded forms, we get

$$\dfrac{1}{(D - R)^{2}} \approx D^{-2}\!\left(1 + 2\frac{R}{D}\right), \qquad \dfrac{1}{(D + R)^{2}} \approx D^{-2}\!\left(1 - 2\frac{R}{D}\right).$$

Taking their difference yields

$$\dfrac{1}{(D - R)^{2}} - \dfrac{1}{(D + R)^{2}} \approx D^{-2}\!\Bigl(1 + 2\frac{R}{D} - 1 + 2\frac{R}{D}\Bigr) \;=\; 4\,\dfrac{R}{D^{3}}.$$

Hence

$$\boxed{\;\Delta F \approx 4\,G\,M\,\dfrac{R}{D^{3}}\;}.$$

Notice that $$G$$, the earth’s radius $$R$$ and the test mass have the same values for both the moon and the sun, so they will cancel when we form the required ratio. Writing

$$\Delta F_{1} \text{ (moon)} = 4\,G\,M_{\text{moon}}\dfrac{R}{D_{\text{moon}}^{\,3}},$$ $$\Delta F_{2} \text{ (sun)} = 4\,G\,M_{\text{sun }}\dfrac{R}{D_{\text{sun }}^{\,3}},$$

we obtain

$$\dfrac{\Delta F_{1}}{\Delta F_{2}} = \dfrac{M_{\text{moon}}}{M_{\text{sun}}} \left(\dfrac{D_{\text{sun}}}{D_{\text{moon}}}\right)^{\!3}.$$

Now we substitute the numerical data (the units are all consistent because every distance is in kilometres and every mass in kilograms):

$$M_{\text{moon}} = 8 \times 10^{22}, \qquad M_{\text{sun}} = 2 \times 10^{30},$$

$$D_{\text{moon}} = 0.4 \times 10^{6}, \qquad D_{\text{sun}} = 150 \times 10^{6}.$$

First the mass ratio:

$$\dfrac{M_{\text{moon}}}{M_{\text{sun}}} = \dfrac{8 \times 10^{22}}{2 \times 10^{30}} = 4 \times 10^{-8}.$$

Next the distance ratio:

$$\dfrac{D_{\text{sun}}}{D_{\text{moon}}} = \dfrac{150 \times 10^{6}}{0.4 \times 10^{6}} = \dfrac{150}{0.4} = 375.$$

Its cube is

$$375^{3} = (375)^{2}\times 375 = 140\,625 \times 375 = 52\,734\,375.$$

Finally, combining everything,

$$\dfrac{\Delta F_{1}}{\Delta F_{2}} = \bigl(4 \times 10^{-8}\bigr)\; \times 52\,734\,375 = 2.109375 \; \approx \; 2.$$

Thus the number closest to $$\dfrac{\Delta F_{1}}{\Delta F_{2}}$$ is $$2$$.

Hence, the correct answer is Option A.