Take the mean distance of the moon and the sun from the earth to … - Vidyadip

MCQ

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

✓
$2$
B
$6$
C
$10^{-2}$
D
$0.6$

✓

Answer

Correct option: A.

$2$

a
As we know, Gravitational force of attraction,

$F = \frac{{GMm}}{{{R^2}}}$

${F_1} = \frac{{G{M_e}m}}{{r_1^2}}\,and\,{F_2} = \frac{{G{m_e}{M_s}}}{{r_2^2}}$

$\Delta {F_1} = \frac{{2G{M_e}m}}{{r_1^3}}\Delta {r_1}\,and\,\Delta {F_2} = \frac{{G{M_e}{M_s}}}{{r_2^3}}\Delta {r_2}$

$\frac{{\Delta {F_1}}}{{\Delta {F_2}}} = \frac{{m\Delta {r_1}}}{{r_1^3}}\frac{{r_2^3}}{{{M_s}\Delta {r_2}}} = \left( {\frac{m}{{{M_s}}}} \right)\left( {\frac{{r_2^3}}{{r_1^3}}} \right)\left( {\frac{{\Delta {r_1}}}{{\Delta {r_2}}}} \right)$

Using $\Delta {r_1} = \Delta {r_2} = 2{R_{earth}};m = 8 \times {10^{22}}\,kg;$

${M_s} = 2 \times {10^{30\,}}kg$

${r_1} = 0.4 \times {10^6}km\,and\,{r_2} = 150 \times {10^6}\,km$

$\frac{{\Delta {F_1}}}{{\Delta {F_2}}} = \left( {\frac{{8 \times {{10}^{22}}}}{{2 \times {{10}^{30}}}}} \right){\left( {\frac{{150 \times {{10}^6}}}{{0.4 \times {{10}^6}}}} \right)^3} \times 1 \cong 2$

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