Does the weight of an object change in the slightest relative to the position of the Moon, Sun, Other objects?

[Silvestru]

If I have a very accurate scale and place a weight of exactly 1 Kg at a certain time, leaving it there and observing it over a prolonged period of time.

Will there be any variations caused by the movement of the objects in our Solar system?

[Mordred]

Yes there will be variations in weight, assuming an accurate enough scale.

Edited June 26, 2017 by Mordred

[Silvestru]

Yes but caused by what exactly? I mean I know an object weights more on the poles for example than at the equator and even the density of the ground can have a very slight effect but I really don't know how and what affects the object the most regarding factors outside of these.

[Mordred]

the moon would exert an opposing force on the weight depending on its position. Much like it affects the tides.

[DrKrettin]

Yes but caused by what exactly? I mean I know an object weights more on the poles for example than at the equator and even the density of the ground can have a very slight effect but I really don't know how and what affects the object the most regarding factors outside of these.

 

You are nearer the centre of the Earth at the poles than on the equator, so with the inverse square law, the attraction is greater.

 

In a Newtonian solar system, every object exerts a force on every other according to this inverse square law, so in principle, say, Jupiter directly overhead exerts and upward force which reduces the effect of the downward gravitational force. This is often used by astrologers to argue that the planets exert some kind of influence at birth, which is total nonsense because these forces exerted by the planets are far less than the forces exerted by the building in which you are being born. The only significant forces are those of the sun and the moon.

Edited June 26, 2017 by DrKrettin

[Silvestru]

 

You are nearer the centre of the Earth at the poles than on the equator, so with the inverse square law, the attraction is greater.

Thank you and I know this, I was referring to factors outside of these two that I gave as example. Like the position of the earth relative to the sun. (considering the elliptical orbit)

[Handy andy]

Ah so. It is better to get weighed on a spring high tide, with the sun and the moon in alignment.

[DrKrettin]

Ah so. It is better to get weighed on a spring high tide, with the sun and the moon in alignment.

 

I guess that if you walk to the equator and weigh yourself, you will weigh less than when you started out.

[swansont]

in the 1930s Alfred Loomis measured the variation of pendulum clocks compared to quartz clocks, and showed the effect of the moon on the clocks as g varied, along with local effects from mass shifting from the tides (which was of similar size as the direct variation from g changing)

[Mordred]

in the 1930s Alfred Loomis measured the variation of pendulum clocks compared to quartz clocks, and showed the effect of the moon on the clocks as g varied, along with local effects from mass shifting from the tides (which was of similar size as the direct variation from g changing)

I had forgotten who did that experiment thanks for that

[dimreepr]

Thank you and I know this, I was referring to factors outside of these two that I gave as example.

 

The density of the ground beneath you.

[stefan r]

If the "very accurate scale" is a balance scale then it will read the same anywhere on earth, on the moon, or on Jupiter. Changes in gravity effect both sides of the balance. A spring scale changes measurement when gravitational force changes.

 

All objects with mass have a gravitational effect. They do not need to be in our solar system. It gets smaller with distance.

 

Distance from earth to moon: 384,399 km (average)

Mass of moon: 7.342×10²² kg

G = 6.67408(31)×10⁻¹¹ m³⋅kg⁻¹⋅s⁻²

 

F = G m₁m₂r^-2

Force is equal to the gravitational constant multiplied by mass #1 mass #2 and divided by radius squared.

 

F = 6.674×10⁻¹¹ m³⋅kg⁻¹⋅s⁻² x 7.342×10²² kg x 1 kg / (384,399,000m)²

 

F = 3.316 x 10^-5 kg⋅m⋅s^-2

 

So near the equator the scale will increase and decrease around +/- 3.3 x 10^-5 newtons with the lunar cycle.

 

Sun

1.496x 10¹¹ m (average)

1.989x 10³⁰ kg

F = 6.674×10⁻¹¹ m³⋅kg⁻¹⋅s⁻² x 1.989×10³⁰ kg x 1 kg / (1.496x 10¹¹m)²

F = 0.00593 kg⋅m⋅s^-2

 

F is in units of Newtons. So the "weight" of one kilogram is 9.8 newtons, 9.8 kg⋅m⋅s^-2.

 

 

Yes but caused by what exactly? I mean I know an object weights more on the poles for example than at the equator and even the density of the ground can have a very slight effect but I really don't know how and what affects the object the most regarding factors outside of these.

 

Rotation of earth has a centripetal force so a kg exerts less force on the spring. Near the poles the gravity from the moon and sun are close to perpendicular.

[John Cuthber]

On a related note...
Here on Earth, one of the things we do to keep track of our health is to weigh ourselves.

NASA and the like are interested in the health of crews on space ships.

How do they weigh them?

[swansont]

On a related note...

Here on Earth, one of the things we do to keep track of our health is to weigh ourselves.

NASA and the like are interested in the health of crews on space ships.

How do they weigh them?

 

 

They get SLAMMD or ride the BMMD. Calibrated springs are involved.

 

http://www.airspacemag.com/daily-planet/how-do-astronauts-weigh-themselves-space-180953884/