Question

Take the mean distance of the moon and the sun from earth to be $4 \times {10^5}\,{\text{km}}$ and $150 \times {10^6}\,{\text{km}}$ respectively. Their masses are $8 \times {10^{22}}\,{\text{kg}}$ and $2 \times {10^{30}}\,{\text{kg}}$ respectively. The radius of the earth is. Let ${F_1}$ be the difference in the forces exerted by the moon at the nearest and farthest point on the earth and ${F_2}$ be difference in the force exerted by the sun at the nearest and farthest points on the earth. Then the number closest to $\dfrac{{{F_1}}}{{{F_2}}}$ is:
A. 0.01
B. 2
C. 0.6
D. 6

Answer 1

Use the expression for Newton’s law of gravitation. Using this formula, derive the expressions for the gravitational force of attraction between the earth and moon and the earth and sun at nearest and farthest points. Substitute these values in the given relation and calculate the required ratio.

Formula used:
The expression for Newton’s law of gravitation is
$F = \dfrac{{G{m_1}{m_2}}}{{{r^2}}}$ …… (1)
Here, $F$ is the force of attraction between the two objects, $G$ is the universal gravitational constant, ${m_1}$ and ${m_2}$ are masses of the two objects and $r$ is the distance between the centres of the two objects.

Complete step by step answer:
We have given that the mean distance between the moon and earth is $4 \times {10^5}\,{\text{km}}$ and the mean distance between the sun and earth is $150 \times {10^6}\,{\text{km}}$.
${d_1} = 4 \times {10^5}\,{\text{km}}$
$\Rightarrow{d_2} = 150 \times {10^6}\,{\text{km}}$
The masses of moon and the sun are $8 \times {10^{22}}\,{\text{kg}}$ and $2 \times {10^{30}}\,{\text{kg}}$ respectively.
${m_M} = 8 \times {10^{22}}\,{\text{kg}}$
$\Rightarrow{m_S} = 2 \times {10^{30}}\,{\text{kg}}$

Let ${F_{1N}}$ be the force of attraction between the moon and the earth at the nearest point on the earth and ${F_{1F}}$ be the force of attraction between the moon and the earth at the farthest point on the earth.Let ${d_1}$ be the nearest distance between the moon and the earth.Hence, the force ${F_1}$ becomes
${F_1} = {F_{1N}} - {F_{1F}}$
$\Rightarrow {F_1} = \dfrac{{G{m_E}{m_M}}}{{d_1^2}} - \dfrac{{G{m_E}{m_M}}}{{{{\left( {{d_1} + R} \right)}^2}}}$
Here, $R$ is the radius of the earth.

Let ${F_{2N}}$ be the force of attraction between the sun and the earth at the nearest point on the earth and ${F_{2F}}$ be the force of attraction between the sun and the earth at the farthest point on the earth.Let ${d_2}$ be the nearest distance between the sun and the earth.Hence, the force ${F_2}$ becomes
${F_2} = {F_{2N}} - {F_{2F}}$
$\Rightarrow {F_2} = \dfrac{{G{m_E}{m_S}}}{{d_2^2}} - \dfrac{{G{m_E}{m_S}}}{{{{\left( {{d_2} + R} \right)}^2}}}$

Let us now calculate the ratio $\dfrac{{{F_1}}}{{{F_2}}}$.
$\dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{\dfrac{{G{m_E}{m_M}}}{{d_1^2}} - \dfrac{{G{m_E}{m_M}}}{{{{\left( {{d_1} + R} \right)}^2}}}}}{{\dfrac{{G{m_E}{m_S}}}{{d_2^2}} - \dfrac{{G{m_E}{m_S}}}{{{{\left( {{d_2} + R} \right)}^2}}}}}$
$\Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{G{m_E}{m_M}\left[ {\dfrac{1}{{d_1^2}} - \dfrac{1}{{{{\left( {{d_1} + R} \right)}^2}}}} \right]}}{{G{m_E}{m_S}\left[ {\dfrac{1}{{d_2^2}} - \dfrac{1}{{{{\left( {{d_2} + R} \right)}^2}}}} \right]}}$
$\Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{{m_M}\left[ {\dfrac{{d_1^2 - d_1^2 + 2{d_1}R + {R^2}}}{{d_1^2{{\left( {{d_1} + R} \right)}^2}}}} \right]}}{{{m_S}\left[ {\dfrac{{d_2^2 - d_2^2 + 2{d_2}R - {R^2}}}{{d_2^2{{\left( {{d_2} + R} \right)}^2}}}} \right]}}$
$\Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{{m_M}\left[ {\dfrac{{2{d_1}R + {R^2}}}{{d_1^2{{\left( {{d_1} + R} \right)}^2}}}} \right]}}{{{m_S}\left[ {\dfrac{{2{d_2}R - {R^2}}}{{d_2^2{{\left( {{d_2} + R} \right)}^2}}}} \right]}}$

We know that $R < < {d_1}$ and $R < < {d_2}$. So we can neglect $R$.
$\Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{{m_M}\left[ {\dfrac{{2{d_1}R}}{{d_1^2d_1^2}}} \right]}}{{{m_S}\left[ {\dfrac{{2{d_2}R}}{{d_2^2d_2^2}}} \right]}}$
$\Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{{m_M}\left[ {\dfrac{1}{{d_1^3}}} \right]}}{{{m_S}\left[ {\dfrac{1}{{d_2^3}}} \right]}}$
$\Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{{m_M}}}{{{m_S}}} \times \dfrac{{d_2^3}}{{d_1^3}}$
$\Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{{m_M}}}{{{m_S}}} \times {\left( {\dfrac{{{d_2}}}{{{d_1}}}} \right)^3}$

Substitute $4 \times {10^5}\,{\text{km}}$ for ${d_1}$, $150 \times {10^6}\,{\text{km}}$ for ${d_2}$, $8 \times {10^{22}}\,{\text{kg}}$ for ${m_M}$ and $2 \times {10^{30}}\,{\text{kg}}$ for ${m_S}$ in the above equation.
$\Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{8 \times {{10}^{22}}\,{\text{kg}}}}{{2 \times {{10}^{30}}\,{\text{kg}}}} \times {\left( {\dfrac{{150 \times {{10}^6}\,{\text{km}}}}{{4 \times {{10}^5}\,{\text{km}}}}} \right)^3}$
$\Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = \dfrac{{8 \times {{10}^{22}}\,{\text{kg}}}}{{2 \times {{10}^{30}}\,{\text{kg}}}} \times {\left( {\dfrac{{150 \times {{10}^9}\,{\text{m}}}}{{4 \times {{10}^8}\,{\text{m}}}}} \right)^3}$
$\Rightarrow \dfrac{{{F_1}}}{{{F_2}}} = 2.1$
$\therefore \dfrac{{{F_1}}}{{{F_2}}} \approx 2$
Therefore, the number closest to $\dfrac{{{F_1}}}{{{F_2}}}$ is 2.

Hence, the correct option is B.

Note: The students should keep in mind that the radius of the earth is very small compared to the mean distance between the earth and moon and mean distance between the earth and sun. Hence, we have neglected or eliminated the terms including radius of the earth from the calculations and there will be no large difference in the final answer due to this elimination.