You can think of it this way: Consider the earth as a big ball of grains of sand (some of the grains are stuck together into rock, and some are melted, but that doesn’t matter for our purposes). Every individual grain pulls towards you in its direction.
The grains directly underneath you pull you straight down, as expected. Grains underneath and to the right of you pull you down and to the right, but you don’t notice that because for every grain that pulls you down and to the right, there is another grain that pulls you down and to the left with the same force. The right and left components of the force cancel out, and you just get pulled down.
That’s why the overall force seems to come from the Earth’s core. Some parts of the Earth are pulling you in directions other than directly towards the core, but their not-towards-the-core contributions are cancelled out by the pull of other parts of the Earth that are in the opposite direction. The only reason the “towards the core” pull is not cancelled is because there is mass below you that has no corresponding mass above you.
Ironically, if you were actually at the earth’s core, you wouldn’t feel any of the Earth’s gravity at all. The mass of the earth would still be pulling on you, but all of the pulls would cancel out.
Again, this is not true. It’ll work if the object is a sphere, and it’s a good approximation if your distance from the object is much greater than the object’s size, but if you’re close to a nonspherical object, it’s a terrible approximation.
We did what I assume to be the Cavendish Experiment in our college freshman physics lab. The big problem was that, due to the mirror and ‘long throw’ you had to wait a long time for the parts to settle down so you could take a reading. Not one team was able to get a valid reading. Eventually the project was dropped when the head of the Physics Dept slammed a door down the hallway and un-did hours of waiting.
I believe Bill Bryson recounted in “A short history…” of attempts to measure The Gravitational Constant (“G”) using the Cavendish Experiment where one of the two weights was a mountain in Scotland. Unfortunately they did not use the proper value of mass for the mountain.
Sheesh, and your previous post is right ahead of mine. It’s like I didn’t carefully read all the other posts in the thread before I mouthed off or something. :smack:
She turned on the overhead projector. She put two big drops of mercury on the glass. She pushed one drop towards the other drop, and when they got close, we could see their shadows jump together on the screen.
Yes, she really did this.
You can see a similar, yet less dramatic, effect with water drops on a hydrophobic surface, such as a Rain-X treated pane of glass.
One example: when you’re near enough to a uniform disk of matter, the gravitational force stays constant as you move towards or away from the disk - very different than the gravity force from a sphere
Or mercury refluxing, leaving a thin film on the glass. Or updraft from the projector. Or irregularieties in the glass surface, causing the liquid to pool in a depression. Or inertia, making the poked drop continue on its’ path. Or…almost anything?
We’ll start by assuming we have a big room–10 m long and 5 m high (actually, a tad more to leave room for the apparatus). And we’ll use a 1 mm steel wire to hang things from.
Steel has a yield strength of ~250 MPa, and our wire has a cross-sectional area of 7.8e-7 m^2, so it can hold 196 kg. We’ll use 100 kg of mass to leave some margin.
The 100 kg is split between two (we’ll say steel) spheres which have a radius of about 12 cm. And we’ll say that they’re mounted on arms 1 m from the main axis.
The gravitational force equation is:
F = Gm1m2/r^2
If we assume r is 30 cm, and we are testing against another set of 50 kg spheres, then we have F=3.7e-6 N, and a torque of 3.7e-6 N-m.
The angle that a wire is displaced is equal to:
a = TL/GJ
where:
T = torque
L = length
J = torsional constant
G = modulus of rigidity
J is the tricky one. For a circular cross-section, it’s:
J = 0.5pir^4
So for r=0.0005 m, J=9.8e-14 m^-4
Plugging in the rest:
T = 3.7e-6 N-m
L = 5 m (distance to ceiling)
G = 78 GPa (for steel)
We get:
a = 0.0024 rad
Light will reflect off an angled mirror at twice the angle it started at, so over 10 m we have a 48 mm displacement. Quite a lot, all things considered. Of course we could have gone backwards and figured out G from the displacement (since that’s what Cavendish was trying to measure).
Basically all mass attracts all other mass. So you are indeed exerting a pull on, say, a coffee cup on the table in front of you, and it is exerting a pull on you.
However, while you are on the earth, such forces are dwarfed by the force that is pulling you towards it; because g is proportional to mass.
The buoyancy of a cheerio counter acts the gravitational pull of the earth. The gravitational forces of two cheerios make them move together, other forces keep them together.
