Gravitation

JEE Main level, single correct, medium

27 of 223

Question 27 of 223

Take the mean distance of the moon and the sun from the earth to be 0.4×106 km0.4 \times 10^6 \mathrm{~km} and 150×106 km150 \times 10^6 \mathrm{~km} respectively. Their masses are 8×1022 kg8 \times 10^{22} \mathrm{~kg} and 2×2 \times 1030 kg10^{30} \mathrm{~kg} respectively. The radius of the earth is 6400 km6400 \mathrm{~km}. Let ΔF1\Delta F_1 be the difference in the forces exerted by the moon at the nearest and farthest points on the earth and ΔF2\Delta \mathrm{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 ΔF1ΔF2\frac{\Delta F_1}{\Delta F_2} is:

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Answer: (A) 22

As we know, Gravitational force of attraction,
F=GMmR2\mathrm{F}=\frac{\mathrm{GMm}}{\mathrm{R}^2}
 F1=GMemr12\mathrm{~F}_1=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{r}_1^2} and F2=GMeMsr22\mathrm{F}_2=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{M}_{\mathrm{s}}}{\mathrm{r}_2^2}
ΔF1=2GMemr13Δr1\Delta \mathrm{F}_1=\frac{2 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{r}_1^3} \Delta \mathrm{r}_1 and ΔF2=GMeMsr23Δr2\Delta \mathrm{F}_2=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{M}_{\mathrm{s}}}{\mathrm{r}_2^3} \Delta \mathrm{r}_2
Using
Δr1=Δr2=2Rearth ;m=8×1022 kg\Delta \mathrm{r}_1=\Delta \mathrm{r}_2=2 \mathrm{R}_{\text {earth }} ; \mathrm{m}=8 \times 10^{22} \mathrm{~kg};
Ms=2×1030 kg\mathrm{M}_{\mathrm{s}}=2 \times 10^{30} \mathrm{~kg}
r1=0.4×106 km2\mathrm{r}_1=0.4 \times 10^6 \mathrm{~km}^2 and r2=150×106 km\mathrm{r}_2=150 \times 10^6 \mathrm{~km}
ΔF1ΔF2=(8×10222×1030)(150×1060.4×106)3×1≅2\frac{\Delta \mathrm{F}_1}{\Delta \mathrm{F}_2}=\left(\frac{8 \times 10^{22}}{2 \times 10^{30}}\right)\left(\frac{150 \times 10^6}{0.4 \times 10^6}\right)^3 \times 1 \cong 2

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