Gravitation and the Principle of Superposition — Question 3

Question 3

Perpendicular attractions combine into a diagonal field

Two point sources of mass $M$ are at $(a,0)$ and $(0,a)$. A test mass $m$ is at the origin.

1. Find the force vector and magnitude on the test mass.

2. Find the mass that must be placed at $(-a,-a)$ to cancel the force.

Question 3 - Solution

Perpendicular attractions combine into a diagonal field

(a) Derivation and calculation

The individual forces are $$\vec F_1=\frac{GMm}{a^2}\hat i,\qquad \vec F_2=\frac{GMm}{a^2}\hat j.$$ Writing $F_0=GMm/a^2$, $$\boxed{\vec F=F_0(\hat i+\hat j)},\qquad \boxed{|\vec F|=\sqrt{F_0^2+F_0^2}=\sqrt2F_0}.$$

(b) Result and interpretation

The third source is at distance $\sqrt2a$ with direction $-(\hat i+\hat j)/\sqrt2$: $$\vec F_c=-\frac{GM_cm}{2\sqrt2a^2}(\hat i+\hat j).$$ Component cancellation requires $$\frac{GM_cm}{2\sqrt2a^2}=\frac{GMm}{a^2},\qquad \boxed{M_c=2\sqrt2M}.$$

Physical check

The cancelling mass is farther away than either original source and must supply a larger force magnitude.