Gravitation and the Principle of Superposition — Question 4

Question 4

Removing one corner mass can be solved by subtracting its field

Four equal point masses $M$ originally occupy the corners $(\pm a,\pm a)$ of a square. The mass at $(a,a)$ is removed.

1. Find the field at the center without summing three separate vectors.

2. Give its magnitude and direction.

Question 4 - Solution

Removing one corner mass can be solved by subtracting its field

(a) Derivation and calculation

By symmetry, the full square has $\vec g_{\mathrm{full}}=0$. The removed source contributes $$\vec g_{\mathrm{removed}}=GM\frac{a\hat i+a\hat j}{(\sqrt2a)^3} =\frac{GM}{2\sqrt2a^2}(\hat i+\hat j).$$ Therefore $$\boxed{\vec g_{\mathrm{remaining}}=\vec g_{\mathrm{full}}-\vec g_{\mathrm{removed}} =-\frac{GM}{2\sqrt2a^2}(\hat i+\hat j)}.$$

(b) Result and interpretation

Its norm and unit direction are $$\boxed{|\vec g|=\frac{GM}{2a^2}},\qquad \hat g=-\frac{\hat i+\hat j}{\sqrt2}.$$ The field points toward $(-a,-a)$, opposite the empty corner. Subtraction is a superposition calculation, without introducing physical negative mass.

Physical check

The remaining field has exactly the magnitude of the removed contribution and the opposite direction.