How the role of Mass is justified for below cases? (Earth/Moon, Jupiter/Io)

[A.T.]

As per equation g = GM/R^2, its value must be constant, but here, it is increasing.

This is the equation for a single gravity source. If you have multiple gravity sources, you have to combine their effects.

 

[rajen0201]

valuable lessons from about significant figure.

Note that, first off all, let me know if below is wrong. Gravity affects mass as per N/kg unit or not. If yes, moons gravity acts accordingly for earth. Now, this reduced Earth gravity will reduce man weight on opposite side also.

 

[russ_watters]

It's a part of the error but not the full problem...

That is not even mentioned in the spreadsheet.

Agreed. I'm just of the opinion that it is ok to do one thing at a time/tackle a problem in layers (if even necessary).

 

[rajen0201]

This is the equation for a single gravity source. If you have multiple gravity sources, you have to combine their effects.

All other objects in universe works to reduce the gravity of Earth and vice a Versa.

 

[Halc]

Note that, first off all, let me know if below is wrong.

OK, I'm letting you know.

Gravity affects mass as per N/kg unit or not.

This part is correct. Gravity can be expressed as acceleration. F=ma, so a=F/m, which is how you've chosen to express it. I'm not contesting that the moon affects my acceleration when it is overhead vs when it is not. I'm contesting that it affects my weight (outside of the tidal effects discussed).

moons gravity acts accordingly for earth. Now, this reduced Earth gravity will reduce man weight on opposite side also.

No reduction in Earth gravity results. Earth continues to contribute GM/r².

Weight is a combination of gravity effects from various sources, but also from acceleration effects. Your computation of weight only accounts for the former, not the latter. An astronaut going up in a Saturn-5 weighs maybe 10,000 Newtons, not because gravity changed, but because the acceleration of the Saturn-5 contributes far more than does gravity. You're not taking the acceleration of Earth into account in the weight calculation when computing the weight of this KG mass when the moon is overhead. Do all the calculations in the inertial frame of the barycenter of the Earth/moon system and the weight will come out correctly.

 

[ZapperZ]

Note that, first off all, let me know if below is wrong. Gravity affects mass as per N/kg unit or not. If yes, moons gravity acts accordingly for earth. Now, this reduced Earth gravity will reduce man weight on opposite side also.

This is a confusing jumble and it reflects your lack of understanding of vector addition. As I had ASKED you earlier, show me the free-body diagram of the situation that I showed in the diagram. There are two gravitational forces that are pulling on the person, and they act in the SAME direction. So how is the presence of the moon here caused the reduction in the gravitational force acting on that person?

You have a continuing habit of ignoring basic premises that you were asked to show. This is a common pattern in many of your questions, and it gets very frustrating when I am trying to make you realize that your problem goes down to something even more basic and fundamental here.

So please DRAW this FBD and show me why you think what you think. I'm trying to help you figure this out on your own, but I can only lead you to the water. You will have to take that last step and drink it.

If that continues to be ignored, then I no longer wish to waste any more of my time on this one.

Zz.

 

[Nugatory]

All other objects in universe works to reduce the gravity of Earth and vice a Versa.

This is simply incorrect, and repeating it isn't going to change that.
This thread is closed to new posts by you, and we strongly encourage you to draw the FBD and do teh vector addition recommended in post #41 above

 

[DrStupid]

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.

If yes, moons gravity acts accordingly for earth.

Yes, gravity doesn't make exceptions.

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".

 

[berkeman]

And the OP has left the building. Thanks everybody for trying to help him.