Newton's Law of Gravitation — Question 10

Question 10

Central attraction can turn a velocity without changing angular momentum

A particle of mass $m$ at position $\vec r$ relative to a fixed point source $M$ feels $\vec F=-GMm\vec r/r^3$. Its velocity need not be radial.

1. Find the torque about the source and deduce the behavior of angular momentum.

2. If the velocity is initially perpendicular to $\vec r$, determine the instantaneous power and explain why the acceleration is not zero.

Question 10 - Solution

Central attraction can turn a velocity without changing angular momentum

(a) Derivation and calculation

For a fixed source, $$\vec\tau=\vec r\times\vec F=-\frac{GMm}{r^3}(\vec r\times\vec r)=\vec0.$$ Since $d\vec L/dt=\vec\tau$, $$\boxed{\vec L=\vec r\times m\vec v=\text{constant}}.$$ The force can change velocity without changing angular momentum about the source.

(b) Result and interpretation

At the specified instant $\vec r\cdot\vec v=0$, so $$\boxed{P=\vec F\cdot\vec v=-\frac{GMm}{r^3}\vec r\cdot\vec v=0}.$$ Nevertheless, $$\boxed{\vec a=-\frac{GM}{r^3}\vec r\ne\vec0},\qquad \frac{d(v^2)}{dt}=2\vec v\cdot\vec a=0.$$ Acceleration changes direction instantaneously; transverse velocity alone does not establish a circular orbit.

Physical check

The orbit is circular only if the transverse speed also satisfies $v^2=GM/r$.