Question

The mass of the moon is $7.3\times {10}^{22}kg$ and the radius of the moon's orbit around the earth is $3.8\times {10}^{8}m$. Then, the change in the acceleration of the earth towards the sun from a total solar eclipse position to the total lunar eclipse position of the moon will be nearly

• A. $6.7\times {10}^{-5}m{s}^{-2}$
• B. $5.1\times {10}^{-5}m{s}^{-2}$
• C. $5.1\times {10}^{-4}m{s}^{-2}$
• D. zero

Answer 1

The correct option is A $6.7\times {10}^{-5}m{s}^{-2}$

The Gravitational force between sun and the earth is given as $F_{se}=G\frac{m_s{m}_e}{r_{se}^2}=6.67\times {10}^{-11}\frac{2\times {10}^{30}\times 6\times {10}^{24}}{(1.495\times {10}^{11}{)}^2}=35.811\times {10}^{21}N$
Similarly the force between the earth and the moon is given as $0.1987\times {10}^{21}N$
In the solar eclipse position the total force on the earth will the sum of the above two forces and the in the lunar eclipse position the total force will be the difference of the above two forces.
Thus we get the change in force as $\left(\right(35.811+0.1987)-(35.811-0.1987\left)\right)\times {10}^{21}N=4\times {10}^{20}N$.
Thus the change in acceleration is given as $a=\frac{F}{m}=\frac{4\times {10}^{20}}{6\times {10}^{24}}=6.67\times {10}^{-5}m{s}^{-2}$