Where is the location of centre of gravity of the Earth?

Question

Take the Sun as a point object at the origin.

Take the Earth as a sphere centred at $x=d$ with a radius of $R$

Let $d=150,000,000,000$ m

Let $R=6,400,000$ m

Slice the sphere into vertical disks of mass $\Delta m$

The force on each disk will be $\frac{k\Delta m}{x^2}$ where $k=GM_{Sun}$

The total force will be $F=k\int^{d+R}_{d-R} \frac{\Delta m}{x^2}$

Assume the Earth has a constant density.

Let the mass of the entire Earth be $m$

$$\frac{\Delta m}{volume-of-disk}=\frac{m}{volume-of-earth}$$

$$\frac{\Delta m}{\pi y^2 \Delta x}=\frac{m}{\frac{4}{3}\pi R^3}$$

$$\Delta m = \frac{3my^2}{4R^3} \Delta x$$

$$F=\frac{3km}{4R^3}\int^{d+R}_{d-R} \frac{y^2}{x^2} \delta x$$

Let's deal with the $y^2$ now.

$$R^2=y^2+(d-x)^2$$

$$y^2=R^2-d^2+2dx-x^2$$

$$\frac{y^2}{x^2}=\frac{R^2-d^2}{x^2}+\frac{2d}{x}-1$$

So the force is now: $$F=\frac{3km}{4R^3}\int^{d+R}_{d-R} (\frac{R^2-d^2}{x^2}+\frac{2d}{x}-1) \delta x$$

$$F=\frac{3km}{4R^3} [\frac{d^2-R^2}{x}+2d \ln{|x|}-x]^{d+R}_{d-R}$$

$$F=\frac{3km}{2R^3}(d\ln{|\frac{d+R}{d-R}|}-2R)$$

If the force of gravity is acting on a point at location $x=r$ then the force $F$ is also given by: $$F=\frac{km}{r^2}$$

So:

$$F=\frac{km}{r^2}=\frac{3km}{2R^3}(d\ln{|\frac{d+R}{d-R}|}-2R)$$

$$F=\frac{1}{r^2}=\frac{3}{2R^3}(d\ln{|\frac{d+R}{d-R}|}-2R)$$

Now when I sub in $R=6400000$ and $d=150000000000$ I expect to get a value for $r$ slightly less than $d$ , but I don't get this result. In fact the value of what is in the bracket turns out to be negative: $-0.003868$. Have I done something wrong in my derivation ?

Answer 1

The Sun's gravity is stronger (weaker) at the near (far) side with respect to the center (along the center line, $y=z=0$). If you find the center of gravity of that:

$$ x_{cg} = \frac{\int_{x=-R}^{x=R}{x\frac{G\rho(x)}{(d+x)^2}dx}}{\int_{x=-R}^{x=R}{\frac{G\rho(x)}{(d+x)^2}dx}} < 0 $$

(where I moved the origin to the center of the Earth), that is going to be closer to the Sun than the center of mass:

$$ x_{cm} = \frac{\int_{x=-R}^{+R}{x\rho(x)dx}} {\int_{x=-R}^{x=R}{\rho(x)dx}}=0$$

which is what you were trying to do.

You can estimate the difference in the force at each end from

$$F(x) \approx F(0) + \frac{dF}{dx}(x) = F(0)[1-\frac{2x}{d}] $$

so

$$ F(\pm R) = F(0)[1\mp \frac{2R}d] $$

or

$$ \frac{\Delta F} F = \frac{2R} d $$

for a shift of 1 radius.

The problem is that when you get off the center line, the fields are not parallel, they all point inward towards the center of the sun. Though transverse components cancel by symmetry, the longitudinal component is reduced.

You can estimate by how much:

$$ F(x=0, y=R) \approx F(0)\cos{\frac R d} = F(0)[1-\frac 1 2 (\frac R d)^2] $$

which seems insignificant compared with longitudinal tidal effect because it is quadratic in the small parameter (versus linear).

But: It applies to much more of the mass, since it covers the whole ring and not just the end points (that takes care of 1 power of $R/d$) and the transverse effect has the same magnitude as the longitudinal effect (and thus, the tidal tensor is traceless).

You will have to generalize the 1st formula for $x_{cg}$ to include 3 dimensions (with cylindrical symmetry) to discover the correct answer.

The correct answer is that the force acts as if all the mass is at the center of the Earth--for a spherically symmetric Earth. If there is a quadruple moment (c.f., $J_2$), then it couples to the tensor gradient of the gravitational field to create torques and move centers-of-gravity, but spherically symmetric Earths have no higher order moments.