The change in the value of acceleration of th | vedclass

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(A) During a solar eclipse,the sun and moon are on the same side of the earth. During a lunar eclipse,the moon and the sun are on opposite sides of th

Vedclass · Accepted Answer

$6.73 	imes 10^{-5} \ m/s^2$

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(A) During a solar eclipse,the sun and moon are on the same side of the earth. During a lunar eclipse,the moon and the sun are on opposite sides of the earth.Let $F_S$ be the gravitational force between the earth and the sun,and $F_L$ be the gravitational force between the earth and the moon.According to Newton's second law,the net force on the earth is $F = m_e a$,where $m_e$ is the mass of the earth.During a solar eclipse,the forces from the sun and moon act in the same direction on the earth:$m_e a_S = F_S + F_L$  --- $(1)$During a lunar eclipse,the forces act in opposite directions:$m_e a_L = F_S - F_L$  --- $(2)$The change in acceleration is $\Delta a = a_S - a_L$. Subtracting $(2)$ from $(1)$:$m_e (a_S - a_L) = (F_S + F_L) - (F_S - F_L) = 2F_L$Since $F_L = G \frac{m_e M_L}{D^2}$,where $M_L$ is the mass of the moon and $D$ is the orbital radius:$m_e \Delta a = 2 G \frac{m_e M_L}{D^2} \implies \Delta a = \frac{2 G M_L}{D^2}$Substituting the values: $G = 6.67 	imes 10^{-11} \ N \cdot m^2/kg^2$,$M_L = 7.36 	imes 10^{22} \ kg$,$D = 3.8 	imes 10^8 \ m$:$\Delta a = \frac{2 	imes 6.67 	imes 10^{-11} 	imes 7.36 	imes 10^{22}}{(3.8 	imes 10^8)^2} = \frac{98.1776 	imes 10^{11}}{14.44 	imes 10^{16}} \approx 6.8 	imes 10^{-5} \ m/s^2$.The closest option is $6.73 	imes 10^{-5} \ m/s^2$.

The change in the value of acceleration of the earth towards the sun,when the moon moves from the position of a solar eclipse to the position on the other side of the earth in line with the sun,is: (mass of the moon $= 7.36 \times 10^{22} \ kg$,radius of the moon's orbit $= 3.8 \times 10^8 \ m$).

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