Thread: The Weight Of The World

I wasn't sure if I should post this thread in the earth science section (or mathematics) but I guess the topic is general; apologies if otherwise.

I wonder if the center of gravity of the earth, or any object, is in the same position as its gravitation "center of attraction"?

The center of gravity is an imaginary point where all the object's mass can be positioned and from a mechanical perspective the object will "behave" as if it were acted on at this central point (perhaps internal rotations modify this?). The center of gravity is a weighted average of mass * distance integrated linearly over the object in question.

However gravity follows an inverse square law



and this integration is no longer linear. If the integral is performed then the gravitational center of attraction would shift closer to the external attracted object than the center of mass.

If this reasoning is valid, then have we an accurate estimate for the weight of the world? For example we can measure the force of attraction between two objects in a laboratory and given that we know their (inertial) mass the we can estimate the constant G. However when we "weigh" such objects in the earth's gravitational field, can we extrapolate it's weight based on the measured (weight) force and our understanding of its distance to the earth's center?

The center of mass for a symmetrical earth may be at it's center but if the gravitational center is displaced then the weight of the earth would be estimated with a bias (systematic error).

Further, we know the earth is not homogeneous; its density varies with depth. This must further compound our ability to estimate the weight of the earth if we base this just on the weight of objects on its surface relative to inter weight gravitational attraction measurements.

If the attracted object is sufficiently far away compared to the earth's radius the I guess the gravitational center will be adequately positioned at it's center of mass. The moon, for example, has an orbit consistent with such relative dimensions. However can we ascribe its orbit to an estimation of the earth's total mass?

Since the moon acts on the sea and dissipates energy in the form of tides, would this energy expenditure modify its orbit and still add bias to the earth's estimate. Also, the earth is also in orbit with the moon (conjoined pairs) and the mass of the moon influences its motion. Does this mean we still need to know the weight of the moon to know the weight of the world?

Something like this has actually been done. The British physician Henry Cavendish used a torsion balance with lead balls to both determine the gravitational constant and the medium density of the earth ( He said it was 5,48 kg/dm³)
By doing so he could also calculate the mass of the earth. However im not sure where he determined the mass of the earth (I think this is what he did) and calculated the medium density from this, or if he did vice versa.
By the way, what does location: NZ mean?

I'm really confued about how can one wiegh such an unweightable object as our planet...
Perhaps there is a more simplistic explanation for it?

btw, I think NZ stands for New Zealand.. not sure tho... google it.

Would it be possible to function the rate at which weight changes closer to the Earth over a given distance for example? For instance 1 mile deep and 10 miles deep could have a model which varies the closer to the centre of mass? Would it be possible to intergrate a function alongside the mass of the moon as well to find the actually 'weight' of the Earth?

PS. NZ on Wikipedia only comes up with New Zeland as an Earthly location, so I'd assume thats the location he/she is.

Yes, I think that would be a good approach; merely extrapolating from the Cavendish result and assuming the earthly mass is concentrated at a single point seems somewhat inaccurate to me. I guess this could be combined with seismic radar measurements for density - the effective "gravitational center" would still depend, I think, on the distance to the other object, e.g. 1 mile, 10 miles (deep or remote).