Q. How does weight differ between objects on Earth and on the Moon?

[Cliff Hanley]

Are the following definitions of weight valid;

The force of gravity pulling on a mass?
The force between two bodies due to the force of gravity?

Q. Does weight only apply to the force due to gravity on a body in relation to a planet (or satellite, asteroid etc), eg, a man standing on Earth or on The Moon?
Or can we talk about the weight of, say, a very small object (eg, a pebble) in relation to a relatively larger object (eg, a person)?

 

[Cliff Hanley]

Is it sensible to talk about "my weight" or your "weight", or even "my weight on The Moon" etc, given that the weight is not mine as such?

 

[Cliff Hanley]

Given that weight is a vector quantity / measurement should it be represented by an arrow (pointing downwards) as well as a magnitude in N (Newtons)?

 

[Cliff Hanley]

Does weight only apply to a body in contact with another body, eg, a man standing on Earth? Or does it apply also to the man falling to earth? I'm guessing the latter given that the falling man would be accelerating at 10/m/s^2 ; the reason I'm asking is that I'm wondering if Earth has weight in relation to the Sun. I know that there is a gravitational pull on Earth towards the Sun, or there's gravitational attraction between Earth and the Sun (keeping it from 'flying off' at a right angle, in other words, keeping it in orbit) but I don't know if there is (technically) weight involved here.

 

[Qwertywerty]

Given that weight is a vector quantity / measurement should it be represented by an arrow (pointing downwards) as well as a magnitude in N (Newtons)?

Yes , it is a vector , and is represented by both magnitude and direction . The arrow you refer to only tells you it is a vector , and doesn't have to point downwards .

Does weight only apply to a body in contact with another body, eg, a man standing on Earth?

No , it essentially counts as a non - contact force .

Another point - All objects exert a force of gravity on another , but in most cases , this force is negligible .
For example , force of gravity on a rock by me might be in the range of 10^-9 N , and thus plays a more significant role when heavenly / larger bodies are involved .

Hope this helps .

 

[Cliff Hanley]

Thanks. If weight is represented by direction as well as magnitude why do I only ever see the N (for Newtons) and no arrow in answers to weight questions online? And if the arrow doesn't have to point downwards what would be an example of it pointing otherwise (bodies fall to Earth, or fall to the Moon, ie, they fall 'down', no?)?

Also; so I pull the Earth towards me as well as it pulling me towards it? If so, is that force, although extremely negligible, technically calculable?

 

[Qwertywerty]

To your first point - I meant you would write the force as :

( Vector form ) .To your second point - Yes , according to Newton's third law .
Please note , I said it is negligible in many cases , but not when larger bodies like the Earth are involved .

Thus , force between you and the Earth is not negligible - look at the formula , F doesn't remain negligible ( Weight of the Earth is 5.972 × 10^24 kg ) .

The force on the Earth by man is equal to force on man by the Earth = mg , where ' m ' is mass of the man .

 

[Chestermiller]

The word "weight" is usually reserved for the gravitational pull on an object caused by a heavenly body.

Chet

 

[Cliff Hanley]

"The force on the Earth by man is equal to force on man by the Earth = mg , where ' m ' is mass of the man ."

But when talking about "my" weight on Earth the 'm' in the equation is my mass; 900N = 90kg (my mass) x 10kg/N.
Wouldn't talking about Earth's weight in relation to me use, w = m (of Earth) x g (my 'gravitational field strength', ie, my equivalent of Earth's 10N/kg, and therefore make Earth's weight in relation to me different from my weight in relation to it?

 

[Cliff Hanley]

"The word "weight" is usually reserved for the gravitational pull on an object caused by a heavenly body."

Thanks. But can it (technically) be used in terms of the relationship between the pull on, say a pebble, by a man?

 

[Qwertywerty]

Wouldn't talking about Earth's weight in relation to me use, w = m (of Earth) x g (my 'gravitational field strength', ie, my equivalent of Earth's 10N/kg, and therefore make Earth's weight in relation to me different from my weight in relation to it?

No , the resultant force would be the same in both the cases .

If you do use -

w = m₁ (of Earth) x g (my 'gravitational field strength'

Then your gravitational field strength would be Gm/r² , where m is your mass .

This would be the same as w = m × Gm₁/r² .

Hope this helps .

 

[Chestermiller]

"The word "weight" is usually reserved for the gravitational pull on an object caused by a heavenly body."

Thanks. But can it (technically) be used in terms of the relationship between the pull on, say a pebble, by a man?

Why is the answer to this question so important to you? Do you think you will ever encounter it in practice? As far as I'm concerned, you can call it whatever you want.

Chet

 

[Qwertywerty]

Thanks. But can it (technically) be used in terms of the relationship between the pull on, say a pebble, by a man?

https://www.google.co.in/url?sa=t&s...Wq9AL4&usg=AFQjCNFtM3gUP3VTWlUh2_S7tEMifLUt8Q

 

[maline]

The word "weight" is usually reserved for the gravitational pull on an object caused by a heavenly body.

Chet, do you have a source for that convention?

Wouldn't talking about Earth's weight in relation to me use, w = m (of Earth) x g (my 'gravitational field strength', ie, my equivalent of Earth's 10N/kg, and therefore make Earth's weight in relation to me different from my weight in relation to it?

Your "g" is so tiny that the two forces come out equal & opposite.

But can it (technically) be used in terms of the relationship between the pull on, say a pebble, by a man?

Sure, if you like. But it sounds more "normal" if you just talk about force. The name "weight" doesn't add much information.

 

[Cliff Hanley]

Why is the answer to this question so important to you? Do you think you will ever encounter it in practice? As far as I'm concerned, you can call it whatever you want.

Chet

Thanks. It's not that I'm concerned with encountering it in practice, it's more an attempt to understand what weight really means.

 

[Cliff Hanley]

The Earth's mass is 6 x 10^24 kg; is this 6 billion quadrillion kg? Or would we say 6 trillion trillion kg?

 

[Qwertywerty]

The Earth's mass is 6 x 10^24 kg; is this 6 billion quadrillion kg? Or would we say 6 trillion trillion kg?

https://www.google.co.in/url?sa=t&s...YjtBRY&usg=AFQjCNHPFOf-CqS9_BCYKHVVn8eYyDKORQ

 

[Chestermiller]

Chet, do you have a source for that convention?

No. Just personal experience. If I were writing about the gravitational force that the man exerts on the pebble, I would always refer to it "the gravitational force that the man exerts on the pebble," and not "the weight of the pebble relative to the man."

Chet

 

[Cliff Hanley]

Thanks. Septillion. Nice. I like that. The mass of Earth is a septillion kg.

 

[maline]

Five septillion, nine hundred and seventy-two sextillion, one hundred and ninety-something quintillion...

 

[tony873004]

w = mg
Weight is the force exerted on a mass in a gravity field.
But in everyday life, we usually use the word weight to refer to contact force.

Like all forces, weight comes in pairs. That's Newton's 3rd Law. How much does the Earth weigh? It weighs 160 pounds. It weighs the same as me. My body produces a gravity field. It is not very strong. An object placed on me would need to be 6 trillion trillion kilograms before it weighed 160 pounds.

Do a handstand on a bathroom scale and have a friend take a picture. Turn the picture upside down. In this orientation, the scale is placed on you, and the Earth is placed on the scale. You are "weighing" the Earth. If I did this, it would read that the Earth weighs 160 pounds.

Earth is very massive. It takes only about 73 kg of mass placed on Earth to produce a force of 160 pounds. The above-mentioned picture viewed right-side up would confirm this.

In everyday language, weight is a contact force. When in orbit, the "weightless" astronauts still have mass, and are still in Earth's gravity field. Therefore, according to w = mg, they have weight. But they are falling with their spacecraft and have no contact force against the floor. We refer to this condition as weightlessness.

Your bathroom scale actually does not tell you your weight. It tells you your contact force (or Normal force). If you place a hand on the sink and push up slightly, the reading on the scale goes down. w = mg. What changed, your mass or the acceleration due to Earth's gravity? Neither! Your weight didn't change despite the different scale reading. But your Normal force did. Place the scale on a 30 degree incline. It will read 139 pounds if I stand on it. My m didn't change. Earth's g didn't change. So my weight didn't change. But the normal force between me and the scale did change. When the scale is on level ground, and the person on the scale is not touching anything, normal force and weight are equal in magnitude which is why we use such a scale to determine weight.

In an elevator accelerating up, you feel heavier. If it accelerates down, you feel lighter. If you ride the "Drop Zone" you are briefly "weightless". Not really. M never changed, g never change, but that's just what we call it.

 

[Cliff Hanley]

Five septillion, nine hundred and seventy-two sextillion, one hundred and ninety-something quintillion...

Thanks. So we could say the mass of the Earth is around 6 septillion kg. I do enjoy learning these new numbers.

 

[Cliff Hanley]

Thanks. Great post. So much for me to think about. I'll get back to you when I've given it lots more thought.

 

[Cliff Hanley]

If one astronaut attempted to lift another astronaut on the Moon would it be around 6 times easier to do so than on Earth? Or would the first astronaut's (the 'lifter') ability to lift also be affected by his own change in weight?

 

[Qwertywerty]

If one astronaut attempted to lift another astronaut on the Moon would it be around 6 times easier to do so than on Earth? Or would the first astronaut's (the 'lifter') ability to lift also be affected by his own change in weight?

Yes , it would be easier .