Variable forces and inverse-square gravity

What you’ll learn

• Form differential equations from variable-force models.
• Use resistance and inverse-square gravitational force laws.

Understand the idea

Newton's second law remains valid when a force depends on time, speed or position. Select $m\,dv/dt$ or $mv\,dv/dx$ according to the independent variable that makes the equation separable. A resistance force opposes the velocity, so its sign must reflect the chosen direction and the motion being modelled.

Under constant driving force and resistance proportional to speed, the acceleration falls towards zero as speed approaches its terminal value. Gravitational attraction outside a spherical body varies inversely with squared distance from its centre; using surface height instead of centre distance gives an incorrect force. Integrating the force can connect speed to distance.

Key formulae

$$m\frac{dv}{dt}=F(v,t,x)$$

$$m\frac{dv}{dt}=P-kv\quad\text{(constant driving force }P\text{)}$$

$$v_{\mathrm{terminal}}=\frac Pk$$

$$F=\frac{GMm}{r^2},\qquad g(r)=g(R)\frac{R^2}{r^2}$$

Worked example

A $2\,\mathrm{kg}$ particle starts from rest, driven by $12\,\mathrm N$ against resistance $4v\,\mathrm N$, where $v$ is in $\mathrm{m\,s^{-1}}$. Find $v(t)$.

1. Newton's law gives $2dv/dt=12-4v$, or $dv/dt+2v=6$.
2. Solving gives $v=3+Ce^{-2t}$.
3. Use $v(0)=0$ to get $C=-3$.

Answer: $v=3(1-e^{-2t})\,\mathrm{m\,s^{-1}}$.