Question 218: Gravitation

Question

Take the mean distance of the moon and the sun from the earth to be $$0.4 	imes {10^6}$$ km and $$150 	imes {10^6}$$ km respectively. Their masses are $$8 	imes {10^{22}}$$ kg and $$2 	imes {10^{30}}$$ 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 :

Accepted Answer

As gravitational force of attraction,
F = $${{GMm} \over {{R^2}}}$$
$$	herefore\,\,\,\,$$ Force of attraction berween earth and moon
F
1
= $${{G{M_e}m} \over {r_1^2}}$$
Force of attraction between earth and sun,
F
2
= $${{GMeMs} \over {r_2^2}}$$
$$	herefore\,\,\,\,$$ $$\Delta $$F
1
= $$-$$ $${{2G{M_e}m} \over {r_1^3}}$$ $$\Delta $$r
1
$$\Delta $$F
2
= $$-$$  $${{2GMe\,Ms} \over {r_2^3}}$$ $$\Delta $$r
2
$$	herefore\,\,\,\,$$ $${{\Delta {F_1}} \over {\Delta {F_2}}}$$ = $${{m\Delta {r_1}} \over {r_1^3}} 	imes {{r_2^3} \over {Ms\,\Delta {r_2}}}$$
= $$  \left( {{m \over {Ms}}} ight)\left( {{{r_2^3} \over {r_1^3}}} ight)\left( {{{\Delta {r_1}} \over {\Delta {r_2}}}} ight)$$
$$\Delta {r_1} = \Delta {r_2}$$ = diameter of the earth = 2 R
earth
Given
m = 8 $$ 	imes $$ 10
22
kg
M
s
= 2 $$ 	imes $$ 10
30
kg
r
1
= 0.4 $$ 	imes $$ 10
6
km
r
2
= 150 $$ 	imes $$ 10
6
km
$$	herefore\,\,\,\,$$ $${{\Delta {F_1}} \over {\Delta {F_2}}} = \left( {{{8 	imes {{10}^{22}}} \over {2 	imes {{10}^{30}}}}} ight){\left( {{{150 	imes {{10}^6}} \over {0.4 	imes {{10}^6}}}} ight)^3} 	imes 1$$
=  2

Answer

$$2$$

Answer

$${10^{ - 2}}$$

Answer

$$0.6$$

Answer

$$6$$

Gravitation

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