JEE MAIN - Physics (2018 - 15th April Morning Slot - No. 17)
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 $${{\Delta {F_1}} \over {\Delta {F_2}}}$$ is :
$$2$$
$${10^{ - 2}}$$
$$0.6$$
$$6$$
Explanation
As gravitational force of attraction,
F = $${{GMm} \over {{R^2}}}$$
$$\therefore\,\,\,\,$$ Force of attraction berween earth and moon
F1 = $${{G{M_e}m} \over {r_1^2}}$$
Force of attraction between earth and sun,
F2 = $${{GMeMs} \over {r_2^2}}$$
$$\therefore\,\,\,\,$$ $$\Delta $$F1 = $$-$$ $${{2G{M_e}m} \over {r_1^3}}$$ $$\Delta $$r1
$$\Delta $$F2 = $$-$$ $${{2GMe\,Ms} \over {r_2^3}}$$ $$\Delta $$r2
$$\therefore\,\,\,\,$$ $${{\Delta {F_1}} \over {\Delta {F_2}}}$$ = $${{m\Delta {r_1}} \over {r_1^3}} \times {{r_2^3} \over {Ms\,\Delta {r_2}}}$$
= $$ \left( {{m \over {Ms}}} \right)\left( {{{r_2^3} \over {r_1^3}}} \right)\left( {{{\Delta {r_1}} \over {\Delta {r_2}}}} \right)$$
$$\Delta {r_1} = \Delta {r_2}$$ = diameter of the earth = 2 Rearth
Given
m = 8 $$ \times $$ 1022 kg
Ms = 2 $$ \times $$ 1030 kg
r1 = 0.4 $$ \times $$ 106 km
r2 = 150 $$ \times $$ 106 km
$$\therefore\,\,\,\,$$ $${{\Delta {F_1}} \over {\Delta {F_2}}} = \left( {{{8 \times {{10}^{22}}} \over {2 \times {{10}^{30}}}}} \right){\left( {{{150 \times {{10}^6}} \over {0.4 \times {{10}^6}}}} \right)^3} \times 1$$
= 2
F = $${{GMm} \over {{R^2}}}$$
$$\therefore\,\,\,\,$$ Force of attraction berween earth and moon
F1 = $${{G{M_e}m} \over {r_1^2}}$$
Force of attraction between earth and sun,
F2 = $${{GMeMs} \over {r_2^2}}$$
$$\therefore\,\,\,\,$$ $$\Delta $$F1 = $$-$$ $${{2G{M_e}m} \over {r_1^3}}$$ $$\Delta $$r1
$$\Delta $$F2 = $$-$$ $${{2GMe\,Ms} \over {r_2^3}}$$ $$\Delta $$r2
$$\therefore\,\,\,\,$$ $${{\Delta {F_1}} \over {\Delta {F_2}}}$$ = $${{m\Delta {r_1}} \over {r_1^3}} \times {{r_2^3} \over {Ms\,\Delta {r_2}}}$$
= $$ \left( {{m \over {Ms}}} \right)\left( {{{r_2^3} \over {r_1^3}}} \right)\left( {{{\Delta {r_1}} \over {\Delta {r_2}}}} \right)$$
$$\Delta {r_1} = \Delta {r_2}$$ = diameter of the earth = 2 Rearth
Given
m = 8 $$ \times $$ 1022 kg
Ms = 2 $$ \times $$ 1030 kg
r1 = 0.4 $$ \times $$ 106 km
r2 = 150 $$ \times $$ 106 km
$$\therefore\,\,\,\,$$ $${{\Delta {F_1}} \over {\Delta {F_2}}} = \left( {{{8 \times {{10}^{22}}} \over {2 \times {{10}^{30}}}}} \right){\left( {{{150 \times {{10}^6}} \over {0.4 \times {{10}^6}}}} \right)^3} \times 1$$
= 2
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