જ્યારે ચંદ્ર સૂર્ય ગ્રહણથી પૃથ્વી અને સૂર્ય ને જોડતી રેખામાં બીજી બાજુએ આવે ત્યારે પૃથ્વીના સૂર્ય તરફના પ્રવેગમાં કેટલો ફેરફાર થાય? (ચંદનું દળ  \(= 7.36 \times 10^{22}\,kg,\) ચંદ્રની ત્રિજ્યા \(= 3 .8 \times 10^8\,m\) ).

So the gravitational force acting on earth due to sun will add up with the gravitational force acting on earth due to moon during solar eclipse and the gravitational forces will subtract with each other during lunar eclipse.

Let \(F_S\) be the gravitational force between earth and sun and \(F_L\) 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, 

\(F=ma\)........(1)

During solar eclipse, eqn (1) will be

\(m.{a_s} = {F_s} + {F_L} \to \left( 2 \right)\)

During lunar ecllpse, eqn \((1)\) will be

\(m.{a_L} = {F_s} - {F_L} \to \left( 3 \right)\)

Here, \(a_s\) and \(a_L\) are the acceleration of earth towards sun during solar and lunar ecllpse respectlvely and \(m\) is the mass of earth.

So the change in acceleration of earth toward sun from solar to lunar ecllpse perlod, can be found, can be found by subtracting eqn \((2)\) from eqn \((3)\)

\(m\left( {{a_s} - {a_L}} \right) = {F_s} + {F_L} - {F_s} + {F_L}\)

\(\therefore m\left( {{a_s} - {a_L}} \right) = 2{F_L} \to \left( 4 \right)\)

Here,

\({F_L} = G\frac{{m.{M_L}}}{{{D^2}}}\)

Where \(G\) is the gravltational constant, \(m\) is the mass of earth, \(ML\) is the mass of moon, and \(D\) is the distance between moon and earth.

\(D\) is equal to the orbital radlus of moon.

So the eqn \((4)\) becomes,

\(\left( {{a_s} - {a_L}} \right) = 2G\frac{{{M_L}}}{{{D^2}}} \to \left( 5 \right)\)

Now, substitute the glven values in eqn \((5)\),

\({M_L} = 7.36 \times {10^{22}}\,kg\)

\(D = 3.8 \times {10^8}\,m\)

Gravltational constant value, \(G = 6.67 \times {10^{ - 11}}N.{m^2}.k{g^{ - 2}}\)

\(\left( {{a_s} - {a_L}} \right) = 2 \times 6.67 \times {10^{ - 11}} \times \frac{{7.36 \times {{10}^{22}}}}{{{{\left( {3.8 \times {{10}^8}} \right)}^2}}}\)

\(\left( {{a_s} - {a_L}} \right) = 6.799 \times {10^{ = 11 + 22 - 16}}\)

\(\left( {{a_s} - {a_L}} \right) = 6.799 \times {10^{ - 5}}m/{s^2}\)