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# The change in the value of the acceleration of earth towards sun, when the moon comes from the position of solar eclipse to the position in the other side of earth in line with sun is: (mass of moon $= 7.36 \times {10^{22}}kg$, radius of moon’s orbit $= 3.28 \times {10^8}m$)(A) $6.73 \times {10^{ - 2}}m/{s^2}$(B) $7.73 \times {10^{ - 3}}m/{s^2}$(C) $8.73 \times {10^{ - 4}}m/{s^2}$(D) $9.12 \times {10^{ - 5}}m/{s^2}$

Last updated date: 17th Apr 2024
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Hint: In the position of the solar eclipse, the moon will attract the earth towards the sun. When the moon moves to the other side, it will attract the earth away from the sun. The acceleration of the earth towards the moon would be directly proportional to the mass of the moon and inversely proportional to the orbital radius.
Formula used: In this solution we will be using the following formulae;
$a = G\dfrac{M}{{{r^2}}}$ where $a$ is the acceleration of an object near a gravitational body, $M$ is the mass of the gravitational body, and $r$ is the distance of the object to the gravitational body, and $G$ is the universal gravitational constant.

Complete Step-by-Step Solution:
Generally, it is common knowledge that the moon revolves around the earth due to the gravitational field of the earth attracting the moon. Nevertheless, the moon is also a gravitational body and the earth can act as an object to it.
The acceleration of the earth towards the moon can be given as
${a_e} = G\dfrac{{{M_m}}}{{{r^2}}}$ where $M$ is the mass of the moon, and $r$ is the distance of the moon to the earth (in this case the orbital radius), and $G$ is the universal gravitational constant.
Now, when the moon is on the other side, the acceleration will be in the opposite direction, hence the change in acceleration would be
$\Delta {a_e} = G\dfrac{{{M_m}}}{{{r^2}}} - \left( { - G\dfrac{{{M_m}}}{{{r^2}}}} \right) = 2\dfrac{{G{M_m}}}{{{r^2}}}$ (because acceleration is now in opposite direction relative to the sun)
Hence, by inserting all known values, we have
$\Delta {a_e} = 2\dfrac{{\left( {6.67 \times {{10}^{ - 11}}} \right)\left( {7.36 \times {{10}^{22}}} \right)}}{{{{\left( {3.28 \times {{10}^8}} \right)}^2}}}$
By computation, we have
$\Delta {a_e} = 9.12 \times {10^{ - 5}}m/{s^2}$

Hence, the correct option is D

Note: In the above, we assumed that the distance of the earth to the moon during the ellipse is the same as the distance on the other side. In actuality, this is not necessarily so, as the orbit of the moon is not a perfect circle. Also, although small, the earth actually shifts towards the moon as the moon is orbiting about it, allowing the earth itself to have an orbit about a point.