Question

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 :

• A. 2
• B. $10^{-2}$
• C. 0.6
• D. 6

Answer 1

Answer: (A)

We have $\Delta F_{1}=$ difference in the forces exerted by the moon at the nearest and farthest points on the Earth. Therefore,
$F_{1}=\frac{G m_{ m } m_{ e }}{r_{1}^{2}}$
$\Rightarrow \Delta F_{1}=-\frac{G m_{ m } m_{ e }}{r_{1}^{3}} \Delta r_{1}$
where $m_{ m }$ is mass of moon, $m_{ e }$ is mass of Earth, $r_{1}$ is mean distance of the moon from the Earth. Now, $\Delta F_{2}=$ difference in the force exerted by the Sun at the nearest and farthest points on the Earth. Therefore,
$F_{2}=\frac{G m_{ c } m_{ s }}{r_{2}^{2}}$
$\Rightarrow \Delta F_{2}=-2 \frac{G m_{ c } m_{s}}{r_{2}^{3}} \Delta r_{2}$
where $m_{ s }$ is mass of Sun, $m_{ e }$ is mass of Earth, $r_{2}$ is mean distance of the Sun from the Earth. Thus,
$\frac{\Delta F_{1}}{\Delta F_{2}}=\frac{G m_{ m } m_{ e }}{G m_{ m } m_{ s }} \times \frac{r_{2}^{3}}{r_{1}^{3}} \times \frac{\Delta_{1}}{\Delta_{2}}$
Since, $\Delta r_{1}=\Delta r_{2}=2 R_{ e }$, therefore,
$\frac{\Delta F_{1}}{\Delta F_{2}}=\frac{8 \times 10^{22}}{2 \times 10^{30}} \times \frac{\left(150 \times 10^{6}\right)^{3}}{\left(0.4 \times 10^{6}\right)^{3}} \times \frac{2 R_{ e }}{2 R_{ e }}=2$