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The change in the value of acceleration of earth towards sun, when the moon comes from the position of solar eclipse to the position on the other side of earth in line with sun is :

(mass of the moon =7.36\times 10^{22}kg, radius of the moon's orbit =3.8\times 10^{8}kg

Option: 1

6.73\times10^{-5}\frac{m}{s^{2}}


Option: 2

6.73\times10^{-3}\frac{m}{s^{2}}


Option: 3

6.73\times10^{-2}\frac{m}{s^{2}}


Option: 4

6.73\times10^{-4}\frac{m}{s^{2}}


Answers (1)

best_answer

 

Let  be the gravitational force between earth and sun and  be the gravitational force between earth and moon.

Also, the acceleration of earth during solar and lunar eclipse can be derived from Newton’s second law, i.e.,
F=m \cdot a \rightarrow(1)

During solar eclipse, eqn (1) will be  

m \cdot a_{s}=F_{s}+F_{L} \rightarrow(2)

During lunar eclipse, eqn (1) will be  

m \cdot a_{L}=F_{s}-F_{L} \rightarrow(3)

So,

\begin{aligned} &m\left(a_{s}-a_{L}\right)=F_{s}+F_{L}-F_{s}+F_{L}\\ &\therefore m\left(a_{s}-a_{L}\right)=2 F_{L} \rightarrow(4) \end{aligned}

\left(a_{s}-a_{L}\right)=2 G \frac{M_{L}}{D^{2}} \rightarrow(5)

\begin{array}{l} \left(a_{s}-a_{L}\right)=2 \times 6.67 \times 10^{-11} \times \frac{7.36 \times 10^{22}}{\left(3.8 \times 10^{8}\right)^{2}} \\ \left(a_{s}-a_{L}\right)=6.799 \times 10^{-11+22-16} \\ \left(a_{s}-a_{L}\right)=6.799 \times 10^{-5} m / s ^{2} \end{array}

Posted by

Ajit Kumar Dubey

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