Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / Section E / E15

For point masses, Newton's force is attractive along their separation with magnitude

G m_{1} m_{2} / r^{2}

. Treat the narrow rod as uniform, with linear mass density

μ = M / (2 a)

; a mass profile would otherwise be needed beyond total mass and length.

b > a

, integrating the element force gives

F_{1} = G m μ \int_{- a}^{a} \frac{d x}{( b - x ) ^{2}} = \frac{GM m}{b ^{2} - a ^{2}},

directed toward the rod. On the perpendicular bisector, axial components cancel and transverse components add:

F_{2} = G m μ r \int_{- a}^{a} \frac{d x}{( x ^{2} + r ^{2} ) ^{3/2}} = \frac{GM m}{r r ^{2} + a ^{2}} .

r = b ≫ a

F_{1} = \frac{GM m}{b ^{2}} [1 + \frac{a ^{2}}{b ^{2}} + O (a^{4} / b^{4})], F_{2} = \frac{GM m}{b ^{2}} [1 - \frac{a ^{2}}{2 b ^{2}} + O (a^{4} / b^{4})] .

Both approach the point-mass force, but

F_{1} / F_{2} = 1 + 3 a^{2} / (2 b^{2}) + O (a^{4} / b^{4})

, so the axial attraction is slightly larger.

m ≪ M

, centrifugal balance gives

v (r) = \frac{GM}{r ^{2} + a ^{2}} = \frac{GM}{a} [1 + (r / a)^{2}]^{- 1/4} .

For

r ≫ a

, it is Keplerian,

v \sim GM / r

. For rod thickness

≪ r ≪ a

, the force approximates that of an infinite line and $v \to GM / a$ , a flat speed limit rather than a divergence.

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