Which of the following places would an object have the greatest weight?

Question

- on the surface of the earth
- at the bottom of a mine
- out in space
- on the moon
- on top of a mountain

Answer 1

We can easily eliminate both the moon and space as possible answers. Both of these answers are fairly ridiculous. Let's examine the other three in more detail. Weight is basically a force acting on a massive object as a result of a graviational attraction. Newton's Law of Gravitation which says that weight is given by the formula:

W=GmM/r^2

where m is your mass, M is the planet mass, G is a constant and r is the distance between your center of mass and the planet's center of mass.

This would tend to suggest that on top of a mountain you would have slightly smaller weight than at sea level.

When we start to mine into the Earth, it has been said by some that the rock above us will pull us up, which is in fact correct, meaning we have to change Newton's Law of Gravitation. I will use something very similar to Gauss's Law on a charged sphere to explain this. Only mass that is enclosed by the spherical surface with radius r, how far you are from the center of Earth, contributes to your weight. Basically if you go through a bunch of wonderful calculus, you will see that your weight inside a mine will be slightly less than that on the surface of the Earth.

How big are these differences is the next question. The height of the highest mountain (Everest) is about 9000 m. The radius of the Earth is on the order of 10^6 m. You change your distance to the center of the Earth by 0.9% which would change your weight a whopping 0.0081% (as weight is inversely proportional to r^2). If we did a similar calculation for the lowest mine (maybe 4000 m), we would find a very similar result. Basically your weight on the surface of the Earth, in a mine, and on top of a mountain would be the same for all practical purposes.

Answer 2

The further you are from the center of mass (gravity) of an object, the less its force on you. So - moon and space are wrong. The top of a mountain is wrong for the same reason. (Also at the top of a mountain your momentum which sends you out into space isn't as well counteracted by gravity - i. e. you weigh less.) At the bottom of a mine, you have some mass above you which is counteracting the rest of the mass of the Earth. So your weight would be greatest at the surface of the Earth.
You can lose weight by jumping off a cliff or a tall building. The problem is that you gain a very large amount of weight when you hit the surface of the Earth. Ouch!!

Answer 3

I think the object would weigh essentially the same on the surface of the earth, or in a mine, but I believe it would weigh the most on the surface. Even though you'd be closer to the center of the earth in a mine (and therefore more gravity), the gravity of the rock above would pull upward. From what I've heard, you'd essentially have no weight at the center of the earth. I've never answered a question on yahoo before, I hope Im doing this right.

Wow, I didn't know I could edit a question. Well like that one guy at the bottom said, gravity is equal to g*m/r^2.
g is the gravitational constant 6.67*10^-11. m is the mass of earth- 5.9742*10^24. And r is the radius of the earth, which is 6373000 meters.

Put that into the above equation and you get 9.811 meters per second per second. If you climb to the top of mount everest- some 9000 meters up I believe, you'd get 9.78 m/s^2.

So... essentially, if you increase your height above the earth, you'll experience less gravity- less weight. However, if you go into a mine, you'll have mass above you pulling you up and that completely cancels out the mass above you. I'm really bored, so I'm actually going to do this calculation... and I got 9.806 m/s^s if you go down 3 kilometers. should I explain how I did this... only if you want me to. So yeah, I still think your weight is greatest at the surface.

Answer 4

Let's get rid of the easy one first,

out in space - effectively, no weight
moon - 1/6 weight approx.

on top of mtn. - lets say 1 mile high versus 4000 mile radius of earth. (r1/r2)^2 would be the ratio of the gravity from surface to top of mtn. Therefore, gravity would be ~ 99.95 % at top of mtn versus surface of earth. I think if anyone felt heavier up there it would be due to lack of oxygen not increased weight.

On surface of earth - normal wieght.

In bottom of mine - I had to think of this in terms I could visualize. Suppose you were on a planet 20' tall and square. If you stood in the center of a side, the distance from the center of the planet would be 10'. If you were to go down 1', 95 % of the mass of the planet would be pulling you towards the center, while a portion of the 5% would pull you up to the surface. So, in that regards alone, you would be slightly lighter. However, if you take into account the square of the distances from the center of mass, (you were 10' from center, now 9' from center), the change would be 10^2/9^2 or 123% more force. So, the change in distance from the center would produce a greater change than moving a small percentage of the depth. Therefore, I believe you would increase in weight the further you went until some point where the weight would then begin to reduce until it got to zero at the center. I am sure someone could graph the forces out and could probably use a lot more accurate model than I did.

So, I would say bottom of mine.

Answer 5

Gravitational force is what creates Weight. W = M x Acceleration due to Gravity. In this case, M is constant and the acceleration due to gravity is varying.

Since A depends on the distance from the center of the object, it means that as i reach the core of the earth, A is already zero, therefore a body will experience weightlessness.

Thus, we can conclude that while we are still on earth, the weight of an object is highest on the surface of the earth.

As we climb up a mountain, the distance from the center increases, and thus we feel heavier.

When we escape the force of gravity, which happens at escape velocity, A will decrease exponentially, and when we leave the earths gravitational pull, the body again becomes weightless. This is the principle of the rocket, and this escape velocity calculation can be done very accurately.

That, however, is beyond the scope of this question.

As far as the moon is concerned, A is again dependent on the weight of the object attracting us. Since the moon is much lighter than earth, its gravitational pull is much lesser (1/6th) than earth.

Answer 6

On the surface of the moon the weight is 1/6 th of the weight in earth.

Out in space and and on the top of a mountain the acceleration due to gravity is less than at the surface of earth ,as the distance increases from the center of the earth.

On the surface of earth we find g as if the whole mass is concentratred on the center.

But when finding the g inside earth's serface, a portion of earth above us will pulling us up while another portion will be pulling us down

The formula used to find g = GM/ R^2 is not valid inside earth.

It is valid on the surface and above the surface.

Calculation and experiment show that g inside mine will be less.

Hence g on the surface of earth is maximum.

At the center of earth g = 0.

Answer 7

Weight is directly proportional to the the gravitational force. Gravitational force on the other hand is inversely proportional to the square of the distances between the two objects. Therefore we can say that weight is also inversely proportional to the square of the distances between the two bodies.

That said and assuming that the mine is lower than the surface of the earth, we can conclude that an object will have the greatest weight at the bottom of a mine.

Answer 8

Definitely not on the moon and not out in space. For the rest of the options, it has to depend on where is it located. The Earth is not a perfect sphere hence at different locations, the gravitional field strength would not be the same.

The most possible answer is at the bottom of the mine due to the equation F= GMm/r^2

Answer 9

weight = mass * g

mass remains same everywhere

now

1)
value of g decreases as we go upwards from earth surface

or go inwards towards the earth's core.

so top of mountain and bottom of mine cases have lesser weight than when the object is at surface

for variation with height

g(at height h)= g(surface)[1-2h/R]

for variation with depth

g(at depth d) = g(surface)[1-d/R]

where R is the radius of earth


2)
moon has value of g less than that of earth

(1/6 that of earth)

3)
in space g = 0

so the maximum value of weight is on the surface of earth

Answer 10

it is not some rule-of-thumb assertion. It replaced into got here across through Archimedes and is definitely what created the tale of him working during the streets bare to make certain King Midas. a effective google seek once you have time. anyhow, the entire concept of a buoyant tension is predicated on the actuality that, as you get deeper in a fluid, the strain will boost. (why your ears harm once you dive to the backside of a pool.) So, in case you immerse something, the upward tension on the backside is larger than the downward tension on the precise, it particularly is what makes stuff look lighter whilst below water. (think of returned to lifting the anchor and how lots heavier it by surprise will become whilst it leaves the water. The anchor does not glide, even regardless of the indisputable fact that it nevertheless has an upward tension on it which makes it lighter whilst immersed in water.) So, products which glide have an upward buoyant tension extra beneficial than their very own weight. products which sink have a buoyant tension below their very own weight. They nevertheless sink, yet seem lighter. i ought to do a information for you at this factor, yet i'm in basic terms going to bypass to the generally happening and generally happening equation for buoyant tension: Buoyant tension = density of the fluid X quantity of the immersed element of the object X acceleration with the aid of gravity. Density additionally equals mass / quantity, so in case you multiply density X's quantity, it grants mass. So the above equation for buoyant tension simplifies to: mass of the displaced fluid X acceleration with the aid of gravity, and because Weight = mass X gravity, Buoyant tension equals weight of the fluid displaced. F = dVg = mg = weight