the farther you are from the center of the planet, the weaker the pull between the planet and your body. The force gets weaker quite rapidly. If you double your distance from the planet, the force is one-fourth. If you triple your separation, the force drops by one-ninth. Ten times the distance, one-hundredth the force. See the pattern? The force drops off with the square of the distance.
Does this mean I would weigh less if I were standing on a mountain as opposed to a location that is below sea level? +++++++++++++++++++++++++++++++++++++++++++++++++++ 07-18-02, 02:50 PM Bibc14 you are correct.
i think that it is reffered to as the law of inverse squares, the same thing applies to the intensity of light radiating from a point.
you DO weigh less on top of a mountain than you
though compared to the radius of the earth, the distance from the top of the tallest mountain to the bottom of the lowest crevace on land, is not that far,(comparatively) so it is nowhere near the 2X as far away you would need to be to weigh 1/4 as much, so the change is only minimal.
07-18-02, 08:24 PM frankvan about one quarter as much if you were about 4,000 miles away from the surface of the earth. Your weight is equal to G*M*m divided by the distance squared. Where M is the mass of the planet, m is your mass, and G is the gravitational constant. Earth's mass is 5.98*10^24, it's radius is 6.38*10^6 meters, and G is 6.67*10^-11 n.m^2/kg^2. plug in your mass in kg and it should result in your weight.
07-18-02, 09:28 PM Bibc14 Frank is right, with the earths radius being about 8,000 miles, you would have to be 4,000 miles away to weigh 1/4 your weight
but Everest(though not the highest point on earth) is 5.498 Miles high, so 8005.498/80000 = 1.00068725 times as far, so 1/ (1.00068725)^2= 0.998626915640414742341036727146232 Times as much.
so if you weigh 100 Pounds at sea level, you weigh 99.8629 Lbs on top of everest.
07-19-02, 01:24 PM frankvan You substituted the earth's diameter for the radius. As you know the earth is about 8,000 miles in diameter.
07-20-02, 11:13 PM coldfuse The summit of Mt. Chimborazo, because of the variable radius of the earth, is the farthest surface point from the center of the earth. Its elevation is 6310 meters, and its peak's distance from the center of the earth is 6,384,404 meters. The difference between the two is 6,378,094 meters (at sea level).
The formula for your weight places these distances squared in the denominator. The ratio of the distances from the center of the earth squared at the top of Mt. Chimborazo and at sea level is:
6,384,40422 : 6,378,09422
which is approximately equal to 1.002. Therefore, if you weigh 100 pounds atop the mountain, you will weigh 100.2 pounds at sea level. Not a heck of a lot of difference, on my scales!
07-21-02, 09:29 AM maiku There is a false assumption being made in some of the previous replies which, when allowed for, makes the difference in weight at sea level and atop Mt. Everest even samller than so far calculated--though exactly how much smaller I'm not prepared to say.
The false assumption is that it is distance from the geographical center of a spherical earth that counts, and that we may just add the height of a mountain to the radius at sea level to get the distance we want. This is erroneous. It is the distance from the center of mass of the earth that counts. You have to take into account the mass of the mountain you're standing on itself. The earth's center of mass is ever so slightly closer to the peak of Mt. Everest than you might think, because of the slightly greater mass concentrated directly below it.
I don't think the difference can be calculated with any reliablility closer than frankvan and coldfuse have already done. We don't, in fact, know where the center of mass of the earth is, precisely. We can only estimate it by experimental observation.
07-21-02, 10:09 AM gatman jeees....some people will do anything for a diet !!
07-21-02, 11:16 PM coldfuse Is it close enough to only three decimal places? Actually, this is a good point and I didn't consider it at all. However, considering it would have driven me quite mad!
10-13-02, 11:37 AM Squirre| Is it true that the inverse square rule only exactly applies to a particle?
If it does then it means that, when you tunnel down to below sea level (say, 4000 miles) then the gravity force pulling you further "down" won't be exactly 4 times (1 / ½²) normal, i.e. it will not be 39.2N (9.8N x 4) towards the centre of the Earth. It would be 4x if you halved your distance from a particle with mass.
I wish I could draw a diagram at this stage, but what I'm trying to say is that when you tunnel halfway towards the centre of the Earth (¼ diameter), this only leaves ¾ diameter directly "below" you and ¼ "above you" which will cause you to weigh less, not more. Likewise if you only go as deep under the ground as Everest is high, you'll probably not weigh a little more, but a little less. I would say that the cause of this is that all the particles in the Earth have a pull of gravity, not just the ones in the middle.
If you kept travelling down towards the centre, you might start to weigh less downwards, more upwards, and more along the x- and z-axes until you reached the middle and you were being pulled evenly outwards (ouch).
I hope amongst all my babbling I've helped answer your good question in some way, and not totally contradicted all the people who know a damn sight more than I do. --- Squirrel
10-13-02, 12:54 PM Minnesota As you approached the center of the Earth you would weigh less and less. Once at the center the gravitational attraction would be equal in all directions, or in effect, zero. It would be like floating in space.
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