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

[rajen0201]

TL;DR

Refer attachment. The distance between Io moon to Jupiter and our Moon to Earth is nearly the same. also, diameter, density & gravity are nearly the same. The ratio of gravity for Earth/Moon is 80.4 times vs ratio of gravity for Jupiter/Io is 20276.7. Also, Jupiter has reduced Io’s gravity by 0.68m/s2 vs Earth has reduced Moon’s gravity by 0.0027m/s2. If Io escapes its absolute gravity will be 1.8+0.68=2.48m/s2 vs Moon’s absolute gravity is 1.62+0.0027=1.6227m/s2.

How the role of Mass justified for the above cases in Newton's gravitation laws & in Einstein's GR?

 

[Dale]

What do you mean by justifying the role of mass? Also, what are you talking about reducing gravity and absolute gravity?

Also, please put your calculations in LaTeX in your post. I am not downloading and opening an excel spreadsheet.

 

[Halc]

Earth has reduced Moon’s gravity by 0.0027m/s2.

This seems to refer to the moon's centripetal acceleration due to Earth's gravity. It is a mistake to word this as a reduction in the moon's gravity. If I were to place a weight scale anywhere on the moon, it would read my weight as say 150N anywhere on the moon, and this would not change at all if the moon happened not to be in orbit about anything. Ditto with IO. There's no reduction of gravity going on.

There are tidal effects which may be significant, but 0.0027m/s² is not a measure of tidal acceleration or force.

 

[rajen0201]

reducing gravity and absolute gravity?

we know Moon gravity reduces the effect of Earth's surface gravity . the same term is used for Earth gravity reduces the effect of moon surface gravity.
I termed absolute gravity for Moons when they are not in the influence of other object's gravity. Trust that I am not familiar with LaTeX. PDF attached.

 

[rajen0201]

It is a mistake to word this as a reduction in the moon's gravity

Note that, On Earth's surface, Moon's attraction effect reduces our weight by 3.31E-5 N/kg. We can directly say Moon reduces Earth gravity in a similar way. I hope you understand what I want to say.

 

[Halc]

Note that, On Earth's surface, Moon's attraction effect reduces our weight by 3.31E-5 N/kg.

It does not. Please show your work.

I get less than 1.2E-6 due to tidal forces on nearside:
GM_m/(R-r_e)² - GM_m/R²
To compute the same on the moon, switch all the m and e subscripts.

 

[russ_watters]

It does not. Please show your work.

I get less than 1.2E-6 due to tidal forces:

...and that's positive or negative (or zero) depending on where you/the moon is.

 

[russ_watters]

Note that, On Earth's surface, Moon's attraction effect reduces our weight by 3.31E-5 N/kg. We can directly say Moon reduces Earth gravity in a similar way. I hope you understand what I want to say.

I'm guessing you didn't account for the fact that they orbit each other.

 

[Halc]

...and that's positive or negative (or zero) depending on where you/the moon is).

Interesting to compute the tiny added weight at half-moon (that causes low tides), but the figure I posted is expressed as a positive reduction in weight on either near or far side. The far side figure is just over 1.1E-6.

Is "positive reduction" a contradiction?
Anyway, it's the way the OP expressed the value when asserting the 3.31E-5 figure.

 

[russ_watters]

Interesting to compute the tiny added weight at half-moon (that causes low tides), but the figure I posted is a positive reduction in weight on either near or far side. The far side figure is just over 1.1E-6.

Is "positive reduction" a contradiction? ;)
Anyway, it's the way the OP expressed the value when asserting the 3.31E-5 figure.

Dang, I fell into the OP's trap; yeah, tidal force is always negative (up) - it's the OP's idea I suspect (stationary bodies) that would have positive or negative forces.

 

[Dale]

we know Moon gravity reduces the effect of Earth's surface gravity

No. We don’t know that. It is wrong.

Note that, On Earth's surface, Moon's attraction effect reduces our weight by 3.31E-5 N/kg. We can directly say Moon reduces Earth gravity in a similar way. I hope you understand what I want to say.

No. This is wrong. Or at least it is not right everywhere on the surface. This will change in magnitude and direction on the surface

 

[rajen0201]

1.2E-6

You are talking about tidal force that is equal to gravity force at near side-gravity force at center of earth.
I have pt the value as per g=GM/r2 for moon.

 

[A.T.]

Dang, I fell into the OP's trap; yeah, tidal force is always negative (up) - it's the OP's idea I suspect (stationary bodies) that would have positive or negative forces.

Not sure if we mean the same by "tidal force", but it is "up" at the near and far side, while being "down" at 90° to that.

 

[rajen0201]

No. We don’t know that. It is wrong.

No. This is wrong. Or at least it is not right everywhere on the surface. This will change in magnitude and direction on the surface

how?

 

[Dale]

how?

At one point on the surface the moon is “up”, on the opposite side it is “down”, and in between it is “horizontal”.

 

[russ_watters]

Not sure if we mean the same by "tidal force", but it is "up" at the near and far side, while being "down" at 90° to that.

Fair enough -- I didn't think the separation angle created at 90 degrees was significant enough for a "down" force to be significant/notable.

 

[Dale]

@rajen0201 also, what do you mean by justifying the role of mass? You still haven’t explained that. Your OP is incomprehensible as is.

 

[Halc]

You are talking about tidal force that is equal to gravity force at near side-gravity force at center of earth.
I have pt the value as per g=GM/r2 for moon.

Yes, that gets the value for centripetal acceleration at some distance from mass M. Centripetal acceleration has no direct effect on your weight on the moon since both you and the moon accelerate identically. Any weight you lose by having Earth overhead is completely canceled by the moon also accelerating towards you from below. Only tidal effects have a nonzero effect on weight since it computes the difference in respective centripetal accelerations of you and the moon.

Fair enough -- I didn't think the separation angle created at 90 degrees was significant enough for a "down" force to be significant/notable.

It's tiny compared to the 'up' of say farside, but it's not zero. At a point on Earth where moon is exactly at horizon, the effect is zero. At halfway, the moon is just below the horizon.

 

[sophiecentaur]

we know Moon gravity reduces the effect of Earth's surface gravity

I think the OP is adding (vectorially) the effects of Earth and Moon gravity vectors for a position with the Moon overhead. So, in the Antipodes, the attractive forces would add so weight would appear to be greater. But, as has been mentioned, this ignores the small matter of the mutual Orbit.
Perhaps the OP should read about Lagrange Points.

 

[ZapperZ]

how?

How is the gravitational force on the person "reduced" here by the moon?

Zz.

 

[rajen0201]

@rajen0201 also, what do you mean by justifying the role of mass? You still haven’t explained that. Your OP is incomprehensible as is.

Suppose moon/Io is not there, both Earth/Jupiter will have higher gravity than current values against their masses. How current theories address this issue?

 

[sophiecentaur]

Suppose moon/Io is not there, both Earth/Jupiter will have higher gravity than current values against their masses. How current theories address this issue?

1. Have you read all the replies on the thread?
2. Do you have any citations for your idea?

 

[A.T.]

Suppose moon/Io is not there, both Earth/Jupiter will have higher gravity than current values against their masses. How current theories address this issue?

Are you asking about how gravity from different sources is combined? In Newtonian Gravity it's just vector addition. In GR it's complicated.

 

[Dale]

Suppose moon/Io is not there, both Earth/Jupiter will have higher gravity than current values against their masses. How current theories address this issue?

First, the claim isn’t true. Second in general the gravitational field is calculated by including all of the sources of gravity, not just one. You cannot deliberately ignore significant gravitating masses. That is just silly.

 

[rajen0201]

How is the gravitational force on the person "reduced" here by the moon?
View attachment 264001
Zz.

Moon reduces whole Earth's gravity it is not like near side & far side.

 

[rajen0201]

Are you asking about how gravity from different sources is combined? In Newtonian Gravity it's just vector addition. In GR it's complicated.

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

 

[rajen0201]

First, the claim isn’t true.

How?

 

[rajen0201]

1. Have you read all the replies on the thread?
2. Do you have any citations for your idea?

I am giving relies on all members. please see the thread discussion. also, give sufficient time to reply all.
What is the meaning if we require citations to discuss well-established science. This is a general discussion and nothing more.
Earlier my two threads were blocked in the name of citations.

 

[ZapperZ]

Moon reduces whole Earth's gravity it is not like near side & far side.

Please show me the physics you are using the arrive at this error. In particular, draw a free-body diagram for that person.

Zz.

 

[Dale]

What is the meaning if we require citations to discuss well-established science.

It means that we don’t think what you are describing actually is well-established science. Requiring you to post what you believe supports your concept will allow us to determine if you are misunderstanding a correct source or simply using a bad source.

All posts on PF should be consistent with the professional scientific literature. Providing such references upon request is a key part of how we accomplish this. Such requests should always be respected

Earlier my two threads were blocked in the name of citations.

Yes, if references were requested and not provided they will get blocked