QJEEM18Phcf2b: JEE Main 2018 Physics - Class 11 JEEM18Phcf2b Solution: Take the mean distance of the moon and the sun from the earth…

Question

Verified Solution • 2025–26 Session moderate

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 :

Accepted Answer

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 R_earth

Given

m = 8

\times

10²² kg

M_s = 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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