请求证明线性运动下的开普勒第三定律（不掌握二阶微分方程解法）

Got it, let's walk through this step by step—you don't need deep expertise in second-order ODEs to make this work. We'll use a simple trick to reduce the equation's order and solve it with integrals instead.

First, let's define our linear motion scenario: imagine an object starting at rest from a distance ( r_0 ) from a massive body (mass ( M )), moving straight toward it, then symmetrically bouncing back to ( r_0 ). The total round-trip time is our period ( T ).

Step 1: Reduce the second-order ODE to first-order #

The given differential equation is:
$$\frac{d^2r}{dt2}= -\frac{GM}{r^2}$$
We can rewrite the second derivative using velocity ( v = \frac{dr}{dt} ). Since ( \frac{d^2r}{dt2} = \frac{dv}{dt} = \frac{dv}{dr} \cdot \frac{dr}{dt} = v \frac{dv}{dr} ), substitute this into the original equation:
$$v \frac{dv}{dr} = -\frac{GM}{r^2}$$

Step 2: Integrate to find velocity as a function of ( r ) #

We use initial conditions: when ( r = r_0 ), ( v = 0 ) (the object starts at rest). Integrate both sides:

• Left side: ( \int_0^v v , dv = \frac{1}{2}v^2 )
• Right side: ( -GM \int_{r_0}^r \frac{1}{r^2} dr = -GM \left( -\frac{1}{r} + \frac{1}{r_0} \right) = GM \left( \frac{1}{r} - \frac{1}{r_0} \right) )

Equate the results:
$$\frac{1}{2}v^2 = GM \left( \frac{1}{r} - \frac{1}{r_0} \right)$$
Since the object moves toward the massive body, ( \frac{dr}{dt} ) is negative. Take the negative root for velocity:
$$v = \frac{dr}{dt} = -\sqrt{2GM \left( \frac{1}{r} - \frac{1}{r_0} \right)} = -\sqrt{\frac{2GM(r_0 - r)}{r r_0}}$$

Step 3: Integrate to find one-way travel time #

We want the time ( t_1 ) to go from ( r = r_0 ) to ( r = 0 ). Rearrange the velocity equation for ( dt ):
$$dt = -\sqrt{\frac{r r_0}{2GM(r_0 - r)}} dr$$
Flip the integral limits to remove the negative sign:
$$t_1 = \int_{r_0}^0 dt = \int_0^{r_0} \sqrt{\frac{r r_0}{2GM(r_0 - r)}} dr$$

Simplify the integral with substitution #

Let ( r = r_0 \sin^2\theta ), so ( dr = 2r_0 \sin\theta \cos\theta d\theta ). When ( r=0 ), ( \theta=0 ); when ( r=r_0 ), ( \theta=\frac{\pi}{2} ). Substitute into the integral:
$$t_1 = \int_0^{\frac{\pi}{2}} \sqrt{\frac{r_0 \sin^2\theta \cdot r_0}{2GM(r_0 - r_0 \sin^2\theta)}} \cdot 2r_0 \sin\theta \cos\theta d\theta$$
Simplify the square root term (the ( \cos\theta ) cancels out):
$$\sqrt{\frac{r_0^2 \sin^2\theta}{2GM r_0 \cos^2\theta}} = \frac{\sin\theta}{\cos\theta} \sqrt{\frac{r_0}{2GM}}$$
This reduces the integral to:
$$t_1 = 2r_0 \sqrt{\frac{r_0}{2GM}} \int_0^{\frac{\pi}{2}} \sin^2\theta d\theta$$
The integral ( \int_0^{\frac{\pi}{2}} \sin^2\theta d\theta = \frac{\pi}{4} ), so:
$$t_1 = \frac{\pi r_0^{3/2}}{2\sqrt{2GM}}$$

Step 4: Derive the period proportionality #

The full period ( T ) is twice the one-way time (symmetric return trip):
$$T = 2t_1 = \frac{\pi r_0^{3/2}}{\sqrt{2GM}}$$
Square both sides:
$$T^2 = \frac{\pi^2 r_0^3}{2GM}$$
This clearly shows ( T^2 \propto r_0^3 )—exactly Kepler's third law, adapted for linear motion.

Why Wolfram Alpha didn't give an r-dependent result #

The original ODE doesn't have an explicit solution for ( r(t) ) using elementary functions. The solution is an implicit function, so we have to use integral methods like this to calculate the period directly, instead of solving for ( r(t) ) first.

内容的提问来源于stack exchange，提问作者Archer