rajen0201 said:
Gravity affects mass as per N/kg unit or not.
Yes, gravity accelerates bodies, but independent from their mass. It is the mass of the gravity source that matters.
rajen0201 said:
If yes, moons gravity acts accordingly for earth.
Yes, gravity doesn't make exceptions.
rajen0201 said:
Now, this reduced Earth gravity will reduce man weight on opposite side also.
Where does "this reduced Earth gravity" come from? In your spreadsheed you calculated the accelerations of Earth due to the gravity of Moon and vice versa. But this acceleration cannot reduce the gravity acting on a body on Earth or Moon because this body is affected by the gravity as well. With the same distance from the source of gravity it would be accelerated identically.
In order to reduce the weight of the body on Earth or Moon the acceleration needs to be different. As the mass of the the source of gravity is identical for all bodies (obviously) the gravitational acceleration can only be different due to different positions and the position of a person standing on the surface of Earth or Moon actually is different from the position of the center of mass of Earth or Moon.
Now let's do some math:
According to Newton's law of gravitation the gravitational acceleration in the field of a body located in the origin is
##a = - GM\cdot\frac{r}{{\left| r \right|^3 }}##
That's what you calculated in your spreadsheed and what
does not change the weight of the body on the surface. The weight of the body is affected by the tital acceleration = the difference of the acceleration of the body and the center of mass:
##\Delta a = a_{Body} - a_{CoM} = - GM\cdot\left( {\frac{{r_{Body} }}{{\left| {r_{Body} } \right|^3 }} - \frac{{r_{CoM} }}{{\left| {r_{CoM} } \right|^3 }}} \right)##
In case of small shperical bodies this can be approximated by
##\Delta a \approx grad\,a \cdot \Delta r = - \frac{GM}{{\left| r \right|^3 }}\cdot\left( {\Delta r - 3 \cdot \frac{{r \cdot \left( {r \cdot \Delta r} \right)}}{{r^2 }}} \right)##
This acceleration obviously depends on the direction of the displacement ##\Delta r## between the body and the center of mass. There are two special cases (which have already been discussed above) - the tidal acceleration on the near and fare sides:
##\Delta F_{||} \approx + 2 \cdot \frac{GM}{{\left| r \right|^3 }}\cdot\Delta r##
and the tidal acceleration at 90°:
##\Delta F_ \bot \approx - \frac{GM}{{\left| r \right|^3 }} \cdot \Delta r##
As you can see they have a different sign. On the near and far side it acts in the direction of the displacement (and therefore reduces the weight) and at 90° it acts against the displacement (and therefore increases the weight).
There is no uniform "reduced Earth gravity".