A Question of gravity!!

Let's see who can answer this!

The gravity at the centre of the Earth is 'zero'.
So will I weigh more in Death Valley or on top of Mt Everest?
And why?

P.S. The answer to this will be posted a little later otherwise it'll spoil the fun!!

more importantly how much more will i weigh the day after a night on the turps. Seriously, death valley for me, only cause i hate the cold!

well if the centre of the earth does equate to 0 gravity...... (just assuming you are right) At first i thought u'd be heavier at mnt everest.... then I was thinking it may be the opposite.... because I've always known that freefall is at a constant speed....... so if gravity is the greatest at centre of earth....then the further you are from the centre the less you weigh.... NOW, this brings up the point that 'size' of object is supposed to have greater gravity...so this assumption of being heavier closer to centre of earth may infact be wrong.... (as its smaller)...

With regards to the freefall comment........could it be you weigh the same in both places? and that gravity of earth is equal anywhere closer or further away from the centre...which means that 0 gravity at centre is not fact?

So..the question is..... is gravity INFACT 0 at the centre of the earth?

Interesting question, one I can't put an opinion on.......

However I have read of a new book about how the matter is in constant expansion/shrinking....so when you jump, its the earth meeting your feet....... weird theory...as it doesn't explain holding something ie thin plastic...and it bends for the earth.....or the fact that a rocket has the power to 'push' the earth away...... so this theory...i don't really hold much faith on.....

Either way, your opinion/fact would greatly satisfy me!

Kind Regards
Merger

Originally posted by Merger

So..the question is..... is gravity INFACT 0 at the centre of the earth?

Interesting question, one I can't put an opinion on.......

Either way, your opinion/fact would greatly satisfy me!

Kind Regards
Merger

Merger, You've brought out some interesting points.
Yes. Gravity at dead centre is Zero!
Could you wait for some more time for the answer? As I said, it'll spoil the fun otherwise!

Mike a tricky question,

The standard form of Gravity, is = (G*m)/d2. [It is supposed to be square]

G = Gravitational Constant, m = mass of object and d = distance

As you would be aware Gravity is different at different places of the earth. There are three things that cause gravity to be different between the poles of the earth and the equator. Gravity increases as you get closer to the center of mass. The spin of the earth is another factor. The farther you are from the spin axis, the lower "gravity" is, because another force (centrifugal, or centripetal) caused by the spin pulls you in the opposite direction. There is more material between you and the center of the earth when you are near the equator because of the flattening. Since there is more mass, gravity increases some. The net effect of these three factors is an increase in gravity near the poles.

To answer to your question, it is important to understand the density of Earth's Mass enclosed within the two radii.

Assuming Death Valley with Sea Level, and assuming that the difference in density of the Earth's Mass is negligible, we can use [assuming m=1], 1/R2 and say that the you would weigh more at Death Valley.

As i have mentioned before, i do not know the density of earths mass and hence cannot pin down the correct answer. I am not 100% sure how far off Death Valley and Mt Everest are from the poles.

k i'm a new member nd this is my first post :-)
i have the answer to this one...

g value at the centre of the earth is zero. the earth being a large solid mass, the gravitational field vectors cancel off symmetrically at the centre, so u weigh nothin at the centre...

g at any point 'h' metres above sea level is given by g'=g(1-2h/R) where g is the value at sea level, R is the radius of the earth

g at a point 'd' metres point below sea level is given by g'=g(1-d/R)

i'm not includin derivations

obviously, in both cases a decrease of g is observed as we move away 4m sea level. mt. everest is 29,000 feet high... i looked up deat valley... its about 400 ft deep...

thus, from the equations its obvious that g (and consequently yr weight) will be lower on mt. everest

as to why g decreases as we go up, its cos gravitaional force is inversely proportional to distance^2 but as we go underground, effective mass surroundin us is less and this decrease outweighs the increase in g due to decrease in distance... (force proportional to mass as well)

just for makin this complete... yr 'weight' is given by w=mg, m is your 'mass' and g is accelerationdue to gravity where u are

sorry, if this is confusin... i learned this in 11th grade physics class :-)

Confusing??? thats an understatement!! LOL

However i believe you!

Kind Regards
Merger

Gravity in a valley and in a mountain top is exactly the same.

Gravity is 0 at the center of the mass of Earth because the gravity "pull" is equal from all sides.

Remember that gravity is a distortion of spacetime, and not a force. A ten kg object and a feather will fall with the same acceleration towards Earth in a tube without air, because mass does not play a role in gravity.

You will weigh more on Everest and less in Death Valley. Actually based on gravity maps, areas at the base of Everest are more negative gravity wise when compared to Death Valley.

General gravity equations assume a flat featureless surface composed of a singular material. Very ideal. Which I suppose is the crux of your trick question. Of course after looking at the maps, altitude is only a partial factor, it looks like there are mineral and structural anomalies in the earth which also influence the strength of gravity.

Enjoy the cheat material...

www.csr.utexas.edu...

Gravity in a valley and in a mountain top is exactly the same.

masterp, i disagree with you... its pretty obvious and trivial that 'gravity' is distance dependent... read my earlier post again

A ten kg object and a feather will fall with the same acceleration towards Earth in a tube without air, because mass does not play a role in gravity.

i think the original question was about 'weight' and not about acceleration de to gravity. True, accn due to gravity is independent of mass, but weight is dependendt on mass w=mg

approximating the earth to be symmetric, when you integrate all r's from the center of the earth, gravity will be zero. gravity is dependent on distance, actually inverse squared, so the further you get from the center the smaller the force wil be. so you will weigh less at the top of the mountain, and more in death valley. your mass will be constant no matter where you are, mass alone does not "weigh" anything.

Hmm I was lead to believe, the further from point zero you got, the less gravitational pull is on you, because it makes sense.

Also, don't forget that mountains (due to their huge mass, above and below the earth) have their own gravitation attraction, so it could possibly have some distortion of the pull on top of the mountain.

Originally posted by Masisoar

Also, don't forget that mountains (due to their huge mass, above and below the earth) have their own gravitation attraction, so it could possibly have some distortion of the pull on top of the mountain.

Not exactly... the mass of a mountain is definitely way less than the earth and thus negligible... it does not make a significant change in the g value... however the distance difference is significant... (this is for points above sea level)

for points below sea level , an entire shell of the earth's mass is being excluded and this does affect g

Mount Everest because the amount of spacetime warping mass in this location is greater then in death valley. Gravity Probe B is trying to figure this out right now as we speak, so any answers arrived at mathematically may have to be revised.

Off2 Infinity,

The mass of the mountain is obviously less than that of earth. How do you make it negligible? For the sake of comparision, it is important to understand the density of earth. Different rocks for example heavier rocks could be formed under Mt Everest and hence refering to the question, mass will increase, which does directly affect the equation.

I am not sure why you are saying gravity is zero at the center of the earth or what you mean by this.

Gravity may be high at Everest because of height and mineral density, but as you move south to the equator into indonesia at sea level, gravity begins to spike again there as well which has a high level of vulcanism.

Originally posted by CAPT PROTON
Gravity may be high at Everest because of height and mineral density, but as you move south to the equator into indonesia at sea level, gravity begins to spike again there as well which has a high level of vulcanism.

Source please?

Originally posted by knowledge23
Off2 Infinity,

The mass of the mountain is obviously less than that of earth. How do you make it negligible? For the sake of comparision, it is important to understand the density of earth. Different rocks for example heavier rocks could be formed under Mt Everest and hence refering to the question, mass will increase, which does directly affect the equation.

what i meant is that due to the mass of the mountain, there is a slight increase in g but this is overcome by the decrease in g due to increase in distance. this is because the mass of the mountain is a "very very " small fraction of that of the earth while the increase in distance from the centre is only a "very" small fraction of the radius (note the difference in the number of verys)... more importatly , force is inversely proportional as the SQUARE of the distance, so the distance effect is more important than the mass effect of the mountain...

sardion2000, just follow the link I posted earlier.

The visual satellite data provided by GRACE shows off the varying differences in global gravity.