Gravitation

JEE Main level, single correct, medium

93 of 223

Question 93 of 223

Take the mean distance of the moon and the sun from the earth to be 0.4×106km0.4\times {10}^{6}km and 150×106km150\times {10}^{6}km, respectively. Their masses are 8×10228\times {10}^{22} kgkg and 2×10302\times {10}^{30} kgkg, respectively. The radius of the earth is 6400km6400km. Let ΔF1\Delta {F}_{1} be the difference in the forces exerted by the moon at the nearest and farthest point on the earth, and ΔF2\Delta {F}_{2} be the difference in the forces exerted by the sun at the nearest and farthest points on the earth. Then, the number closest to ΔF1ΔF2\dfrac{\Delta {F}_{1}}{\Delta {F}_{2}} is,

Show answer and solution

Answer: (C) 22

F1=GMemr13,F2=GMeMsr22{F}_{1}=\dfrac{G{M}_{e}m}{{r}_{1}^{3}},{F}_{2}=\dfrac{G{M}_{e}{M}_{s}}{{r}_{2}^{2}}

ΔF1=(−)2GMemr13Δr1,ΔF2=(−)2GMeMsr22Δr2\Delta {F}_{1}= (-)\dfrac{2G{M}_{e}m}{{r}_{1}^{3}}\Delta {r}_{1},\Delta {F}_{2}=(-)\dfrac{2G{M}_{e}{M}_{s}}{{r}_{2}^{2}}\Delta {r}_{2}

ΔF1ΔF2=mΔr1r13r23MsΔr2=(mMs)(r23r13)(Δr1Δr2)\dfrac{\Delta {F}_{1}}{\Delta {F}_{2}}=\dfrac{m\Delta {r}_{1}}{{r}_{1}^{3}}\dfrac{{r}_{2}^{3}}{{M}_{s}\Delta {r}_{2}}=\left(\dfrac{m}{{M}_{s}}\right)\left(\dfrac{{r}_{2}^{3}}{{r}_{1}^{3}}\right)\left(\dfrac{\Delta {r}_{1}}{\Delta {r}_{2}}\right)

Using
Δr1=Δr2=2Rearth\Delta {r}_{1}=\Delta {r}_{2}=2{R}_{earth}

m=8×1022kgm=8\times {10}^{22}kg

Ms=2×1030kgMs=2\times {10}^{30}kg

r1=0.4×106km{r}_{1}=0.4\times {10}^{6}km

r2=150×106km{r}_{2}=150\times {10}^{6}km

We get
ΔF1ΔF2=2\dfrac{\Delta {F}_{1}}{\Delta {F}_{2}}=2.

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