Take the mean distance of the moon and the sun from the Earth to be $0.4 \times 10^{6} k m$ 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 $∆ F_{1}$ be the difference in the forces exerted by the moon at the nearest and farthest points on the Earth and $∆ F_{2}$ the difference in the force exerted by the sun at the nearest and farthest points on the Earth. Then, the number closest to $\frac{∆ F_{1}}{∆ F_{2}}$ is

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Detailed Solution

$F_{1} = \frac{G M_{e} M_{m}}{{r_{1}}^{2}} and F_{2} = \frac{G M_{e} M_{s}}{{r_{2}}^{2}} \Rightarrow ∆ F_{1} = \frac{2 G M_{e} M_{m}}{{r_{1}}^{3}} ∆ r_{1} a n d ∆ F_{2} = \frac{2 G M_{e} M_{s}}{{r_{2}}^{3}} ∆ r_{2} \Rightarrow \frac{∆ F_{1}}{∆ F_{2}} = \frac{M_{m} ∆ r_{1}}{{r_{1}}^{3}} \frac{{r_{2}}^{3}}{M_{s} ∆ r_{2}} = (\frac{M_{m}}{M_{s}}) (\frac{{r_{2}}^{3}}{{r_{1}}^{3}}) (\frac{∆ r_{1}}{∆ r_{2}})$

Using $∆ r_{1} = ∆ r_{2} = 2 R_{E a r t h}, M_{m} = 8 \times 10^{22} kg, M_{s} = 2 \times 10^{30} kg r_{1} = 0.4 \times 10^{6} km, r_{2} = 150 \times 10^{6} km .$