# Massless in free fall

1. Mar 29, 2012

### doey

we are in weightless condition because there are no resultant force between earth and me,at this instant, earth and me are in the opposite direction moving toward each other,will earth oso can consider weightless??? when i free falling

p/s : beg pardon at 1st typing massless ,typing error ==

Last edited: Mar 29, 2012
2. Mar 29, 2012

### xAxis

Your question hardly makes any sense. Is this a question at all?

3. Mar 29, 2012

### Michael C

The word "weightless" can be problematic since "weight" can have different meanings. The two most-used definitions of "weight" are:

1. The force exerted by gravity on an object. If we use this definition, we cannot say that an object near the Earth in free fall is weightless: gravity is still exerting a force on the object. in this case you might say that the apparent weight of the object in free fall is zero.

2. ("operational definition") The force measured by weighing the object with a force-sensitive scale. When this definition is used, an object in free fall is indeed "weightless".

How do you define the weight of the Earth? If you decide that it's the force exerted by the Earth's gravity on the Earth itself, then the answer is that the Earth's weight is always zero. Since the Earth is orbiting the Sun, you might prefer to define the Earth's weight as "the force exerted on it by the Sun", which you can calculate using Newton's law of gravitation.

4. Mar 29, 2012

### Staff: Mentor

Your post is confusing and hard to understand, as it contains a great many typing and grammar errors. It would help if you could spend a little more time making sure your post is clear, understandable, and concise.

5. Mar 29, 2012

### doey

sry for all that my posted consisting unclear and make all confuse.I knew I was mistaken the concept ==

Last edited: Mar 29, 2012
6. Mar 29, 2012

### doey

is it means we are just weightless because of we cannot weight it in scale?I was confounded.it just stated when an object is in free fall,it is pulling by gravity ,it cannot be weightless.then how come it become weightless indeed as you told in number 2(above)

as i know ,it might be apparently weightless because there is no FORCE pushing or pulling it except gravity,just like when I 'm in universe(space) ,i will having the same feeling as i free falling.but different is in universe,i 'm at rest.Gravity is a field or a force actually?,i seem to confused now.....or what i was thinking are totally wrong?

Last edited: Mar 29, 2012
7. Mar 29, 2012

### tiny-tim

weight is really only relevant if we are in contact with something (eg the floor) …

we cannot feel our weight, we can only feel reaction forces between us and other objects

we only "feel" our weight when eg the floor pushes on us with a force whose magnitude (but not direction) we call our weight

if the floor is accelerating up (eg the floor of a rocket), then the reaction force is greater, so we feel heavier

but in free fall, there is no contact, so nothing to feel

8. Mar 30, 2012

### doey

ya,i was thinking this a whole night time==
here i come out an answer:

just like a person stand still on the floor,he has weight because the floor acting an upward force which respond toward the gravity.
F=w(W is the upward force ,we call it Weight)

when you having another force that pulling or pushing like in the elevator
F+w=ma or w-F=ma

when free falling,there are only the gravity force acting .
F=ma
in this case there have no W in the equation ..... it seem to be weightless

here i have a question,is the free falling acceleration not relate to our weight?if it is relate to our weight ,then the acceleration will only applicable when Free fall is happening on the surface of earth.

f=GMm assume an mass (m) is 1kg
r^2

the r is radius of earth,we can get the f is 9,81 then using F=ma to calculate the free fall acceleration.that means all our calculation relate to weight are only applicable when the object is on the surface of earth,right? W=m(9.81) ,if it is not on surface of earth,the 9.81 shouldn't be the gravitational acceleration because different radius is using now.

am i correct?????????

and as my inference,you can really feel if you are free falling,what about the tidal stress?despite i knew that there will be vry vry small and need not to be noticed

Last edited: Mar 30, 2012
9. Mar 30, 2012

### K^2

It's important to keep in mind that first definition only makes sense in Newtonian Gravity. So while it seems like it's a good definition for "true weight", ti becomes completely meaningless when you go to General Relativity. When you treat gravity as an artifact of space-time curvature, you find that second definition is the only one that makes sense. And in that context, a free-falling object is truly weightless.

10. Mar 30, 2012

### tiny-tim

hi doey!
yes, if we're in space, the acceleration is less than 9.81 …

at radius R, it's 9.81 r2/R2, where r is the Earth's radius

(actually, even on Earth, at sea level, g varies slightly, eg because of the varying thickness of the continental crust)
yes, that's right

it will be miniscule, but yes there's a slight stretching!

now you've got the hang of it, i'll say something that i didn't want to say earlier, in case it was confusing …

by "weight", we usually mean the gravitational force (from the Earth), mg, and we even call it "W"

but that isn't what is meant by "weight" or "weightless" in ordinary English …

in ordinary English (and in some exam questions ), it refers to the reaction force, N, that we feel from the floor etc, not the gravitational force!​

11. Mar 30, 2012

### sophiecentaur

On any diagram of forces you would usuall find a downwards arrow, labelled 'W', whether on a table or in free fall. On a table, the force 'N' would be equal and opposite to W and it would be only then that you would be able to measure the Weight force.
I don't think that your 'weight' is ever anything other than your Effective Weight, including all the extra small forces on top of Earth's gravity. It's not a quantity that's of great interest in Science, compared with Mass.