Newton's Law of Gravitation — Question 2

Section 13-1 · Physics pages 1894–1895

Question 2 Equal gravitational forces can produce very unequal accelerations Two isolated point masses M = 9m and m are separated by r. Work in an inertial frame and neglect all other interactions. M m r Point masses or spherical bodies; separation is measured between centers. (a) Compare force magnitudes and acceleration magnitudes. (b) Find the acceleration of their center of mass and explain the distinction.
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Question 2 - Solution Equal gravitational forces can produce very unequal accelerations M m r (a) Derivation and calculation Take +x from M toward m. Newton’s third law gives⃗ FM = +9Gm2 r2 ˆx,⃗ Fm = −9Gm2 r2 ˆx. Dividing each force by its own mass, aM = Gm r2 , am = 9Gm r2 , am/aM = 9 . (b) Result and interpretation For the isolated pair,⃗ aCM = M⃗aM + m⃗am M + m =⃗FM +⃗Fm 10m =⃗0 . Equal interaction forces give unequal individual accelerations while preserving the center-of-mass velocity. Physical check The larger mass accelerates less; the force pair cannot accelerate the isolated mass center.

Question 2

Equal gravitational forces can produce very unequal accelerations

Two isolated point masses M=9mM=9m and mm are separated by rr. Work in an inertial frame and neglect all other interactions.

Diagram shown in the original SVG worksheet.

  1. Compare force magnitudes and acceleration magnitudes.

  2. Find the acceleration of their center of mass and explain the distinction.

Question 2 - Solution

Equal gravitational forces can produce very unequal accelerations

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Take +x+x from MM toward mm. Newton’s third law gives F→M=+9Gm2r2x̂,F→m=−9Gm2r2x̂.\vec F_M=+\frac{9Gm^2}{r^2}\hat x,\qquad \vec F_m=-\frac{9Gm^2}{r^2}\hat x. Dividing each force by its own mass, aM=Gmr2,am=9Gmr2,am/aM=9.a_M=\frac{Gm}{r^2},\quad a_m=\frac{9Gm}{r^2},\quad \boxed{a_m/a_M=9}.

(b) Result and interpretation

For the isolated pair, a→CM=Ma→M+ma→mM+m=F→M+F→m10m=0→.\vec a_{\mathrm{CM}}=\frac{M\vec a_M+m\vec a_m}{M+m} =\frac{\vec F_M+\vec F_m}{10m}=\boxed{\vec 0}. Equal interaction forces give unequal individual accelerations while preserving the center-of-mass velocity.

Physical check

The larger mass accelerates less; the force pair cannot accelerate the isolated mass center.