Flat Earth Theory (Part 2)

Classical Mechanics Level 3

Suppose there is a solid, uniform circular disk of mass $M$ and radius $R$ . This disk represents a "flat Earth".

The center of the disk is at $(x,y,z) = (0,0,0)$ , and the disk is parallel to the $xy$ plane. A massive point particle is located at $(x,y,z) = (1.2185 R,0,0)$ . What gravitational acceleration does it experience (in $\text{m/s}^2$ )?

Details and Assumptions:
1) $M = \text{Earth mass} = 5.972 \times 10^{24} \, \text{kg}$
2) $R = \text{Earth radius} = 6356 \times 10^{3} \, \text{m}$
3) $\text{Gravity constant} = 6.674 \times 10^{-11} \text{N} \, \text{m}^2 \, \text{kg}^{-2}$
4) Give your answer as a positive number

The answer is 9.81.

1 solution

Mark Hennings
Jun 2, 2019

With $\lambda = 1.2185$ , and with $\rho = \tfrac{M}{\pi R^2}$ being the mass density of the disk, the force on a test particle of mass $m$ is $\begin{aligned} Gm\iint_{x^2+y^2 \le R^2} \frac{\rho\,dx\,dy}{(\lambda R - x)^2 + y^2}\, \frac{\lambda R - x}{\sqrt{(\lambda R - x)^2 + y^2}} \; = \; Gm\rho \iint_{X^2+Y^2 \le 1}\frac{(\lambda - X)\,dX\,dY}{((\lambda-X)^2 + Y^2)^{\frac32}} \; = \; \frac{GMm}{\pi R^2}\iint_{X^2+Y^2 \le 1}\frac{(\lambda - X)\,dX\,dY}{((\lambda-X)^2 + Y^2)^{\frac32}} \end{aligned}$ acting along the $x$ -axis. Thus the particle experiences an acceleration of magnitude $\frac{GMm}{\pi R^2}\iint_{X^2+Y^2 \le 1}\frac{(\lambda - X)\,dX\,dY}{((\lambda-X)^2 + Y^2)^{\frac32}} \; = \; \boxed{9.809\;}\mathrm{m\,s}^{-2}$

Flat Earth Theory (Part 2)

The answer is 9.81.

1 solution

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