Exercises — Two-body problem — equations of motion, reduction to one-body

$\mathbf{r}={\mathbf{r}}_2-{\mathbf{r}}_1.$ Subtracting ${\mathbf{r}}_1$ from ${\mathbf{r}}_2$ gives the arrow that starts at body 1 and ends at body 2 — think of it as the stretchy rope drawn from kid 1 to kid 2. Its length $r=|\mathbf{r}|$ (a single positive number) is the separation between the bodies.

Exact: $\mu =G(M_{\oplus }+m_{\text{sat}})=G(5.97\times 10^{24}+1200)$. The $1200$ is about $10^{21}$ times smaller than Earth's mass — invisible. So $\mu \approx G{M}_{\oplus }=6.674\times 10^{-11}\times 5.97\times 10^{24}\approx 3.985\times 10^{14}{\text{m}}^3{\text{s}}^{-2}.$ This is the famous $G{M}_{\oplus }$. The satellite mass drops out only because it is tiny, not because the formula ignores it.

${\mu }_{\text{red}}=\frac{m\cdot m}{m+m}=\frac{m^2}{2m}=\frac{m}{2}.$ Product over sum: the reduced mass of two equal masses is exactly half one of them.

$\mu =6.674\times 10^{-11}(1.989\times 10^{30}+1.898\times 10^{27}).$ Sum of masses $=1.990898\times 10^{30}\text{kg}$, so $\mu =1.3287\times 10^{20}{\text{m}}^3{\text{s}}^{-2}.$ Fractional excess over $G{M}_{\odot }$: $\frac{m_J}{M_{\odot }}=\frac{1.898\times 10^{27}}{1.989\times 10^{30}}=9.54\times 10^{-4}$, i.e. about $0.095\%$. Small — but for Jupiter it is measurable, unlike the satellite case.

First, derive the distance on our own. Put Earth (mass $M_{\oplus }$) at ${\mathbf{r}}_1$ and the Moon (mass $m_M$) at ${\mathbf{r}}_2$. The COM is the mass-weighted average $\mathbf{R}=\frac{M_{\oplus }{\mathbf{r}}_1+m_M{\mathbf{r}}_2}{M_{\oplus }+m_M}.$ Earth's distance from the COM is $|{\mathbf{r}}_1-\mathbf{R}|$. Subtract: ${\mathbf{r}}_1-\mathbf{R}={\mathbf{r}}_1-\frac{M_{\oplus }{\mathbf{r}}_1+m_M{\mathbf{r}}_2}{M_{\oplus }+m_M}=\frac{(M_{\oplus }+m_M){\mathbf{r}}_1-M_{\oplus }{\mathbf{r}}_1-m_M{\mathbf{r}}_2}{M_{\oplus }+m_M}=\frac{m_M({\mathbf{r}}_1-{\mathbf{r}}_2)}{M_{\oplus }+m_M}=-\frac{m_M}{M_{\oplus }+m_M}\mathbf{r}.$ Taking lengths (and $r=|\mathbf{r}|$ is the separation): $d_{\oplus }=|{\mathbf{r}}_1-\mathbf{R}|=\frac{m_M}{M_{\oplus }+m_M}r.$ Intuitive read: Earth's share of the "lever" is set by the other body's mass fraction. Since the Moon is light, that fraction is tiny, so Earth barely moves from the COM. $d_{\oplus }=\frac{7.35\times 10^{22}}{5.97\times 10^{24}+7.35\times 10^{22}}\times 3.84\times 10^8.$ The fraction is $0.01216$, so $d_{\oplus }=4.67\times 10^6\text{m}$. Since $4.67\times 10^6<6.37\times 10^6$ (Earth's radius), the COM lies inside Earth — about $73\%$ of the way to the surface. Earth wobbles about this internal point.

$M=m_1+m_2=8\times 10^{30}$ kg. $d_1=\frac{m_2}{M}r=\frac{6}{8}\times 10^{11}=7.5\times 10^{10}\text{m},$ $d_2=\frac{m_1}{M}r=\frac{2}{8}\times 10^{11}=2.5\times 10^{10}\text{m}.$ Check: $d_1+d_2=10^{11}=r$ ✓. The heavier star ($m_2$) sits closer ($2.5\times 10^{10}$) to the COM — its coefficient $m_1/M$ is the smaller mass fraction.

Add the two equations. The right sides are equal and opposite, so they cancel: $m_1{\ddot{\mathbf{r}}}_1+m_2{\ddot{\mathbf{r}}}_2=0.$ Because $M\mathbf{R}=m_1{\mathbf{r}}_1+m_2{\mathbf{r}}_2$ (definition of COM times total mass) and $M$ is constant, differentiating twice gives $M\ddot{\mathbf{R}}=m_1{\ddot{\mathbf{r}}}_1+m_2{\ddot{\mathbf{r}}}_2=0\Rightarrow \ddot{\mathbf{R}}=0.$ Physical meaning: the only forces present are the two bodies pulling on each other (internal). Newton's third law makes them equal and opposite, so they cancel in the sum. With no force from outside the system, the COM cannot accelerate — it glides at constant velocity forever. This is why we can ride along with it and ignore it.

Factor out $m_1$: $\mu =G{m}_1\left(1+\frac{m_2}{m_1}\right)$. As the mass ratio $m_2/m_1\to 0$, the bracket $\to 1$, so $\mu \to G{m}_1=G{M}_{\oplus }$. For the satellite, the fractional error is exactly $\frac{m_2}{m_1}=\frac{1000}{5.97\times 10^{24}}=1.68\times 10^{-22}$ (a pure ratio, no units). To turn a fraction into a percent you multiply by $100$: $1.68\times 10^{-22}\times 100=1.68\times 10^{-20}\%$. Either way it is smaller than any instrument can measure. So the "fixed Earth" approximation is essentially exact here, but the equation itself never drops $m_2$.

The relative equation is $\ddot{\mathbf{r}}=-\frac{G(m_1+m_2)}{r^2}\hat{\mathbf{r}}$ (with $\hat{\mathbf{r}}=\mathbf{r}/r$). Multiply both sides by ${\mu }_{\text{red}}=\frac{m_1{m}_2}{m_1+m_2}$: ${\mu }_{\text{red}}\ddot{\mathbf{r}}=\frac{m_1{m}_2}{m_1+m_2}\cdot \left(-\frac{G(m_1+m_2)}{r^2}\right)\hat{\mathbf{r}}=-\frac{G{m}_1{m}_2}{r^2}\hat{\mathbf{r}}.$ The $(m_1+m_2)$ factors cancel exactly, leaving Newton's gravity magnitude $G{m}_1{m}_2/r^2$. So "mass ${\mu }_{\text{red}}$ pushed by the real force $\mathbf{F}$" reproduces the relative motion — that is why energy uses ${\mu }_{\text{red}}$ while the orbit shape uses $\mu$.

(a) $\mu =G(m_1+m_2)=6.674\times 10^{-11}\times 6\times 10^{30}=4.0044\times 10^{20}{\text{m}}^3{\text{s}}^{-2}.$ (b) Plug into the relative Kepler law (it uses the separation $r$, not each orbit radius): $T^2=\frac{4{\pi }^2}{\mu }{r}^3=\frac{4{\pi }^2}{4.0044\times 10^{20}}(2\times 10^{11}{)}^3.$ Numerator: $4{\pi }^2\times 8\times 10^{33}=3.1583\times 10^{35}$. Divide: $T^2=7.887\times 10^{14}{\text{s}}^2$, so $T=2.808\times 10^7\text{s}\approx 0.89\text{years}.$ This is the third law written for the reduced one-body problem — note $\mu$ (the sum-based one) appears, exactly as the parent note built.

$M=4\times 10^{24}$ kg. Using ${\mathbf{r}}_1=\mathbf{R}-\frac{m_2}{M}\mathbf{r}$ and ${\mathbf{r}}_2=\mathbf{R}+\frac{m_1}{M}\mathbf{r}$: ${\mathbf{r}}_1=-\frac{3}{4}(3\times 10^8,0)=(-2.25\times 10^8,0)\text{m},$ ${\mathbf{r}}_2=+\frac{1}{4}(3\times 10^8,0)=(7.5\times 10^7,0)\text{m}.$ Check: ${\mathbf{r}}_2-{\mathbf{r}}_1=(7.5\times 10^7-(-2.25\times 10^8),0)=(3\times 10^8,0)=\mathbf{r}$ ✓. The heavier body 2 sits closer to the COM ($7.5\times 10^7$ vs $2.25\times 10^8$), matching L2.3's rule.

Differentiate the transforms: ${\dot{\mathbf{r}}}_1=\dot{\mathbf{R}}-\frac{m_2}{M}\dot{\mathbf{r}}$, ${\dot{\mathbf{r}}}_2=\dot{\mathbf{R}}+\frac{m_1}{M}\dot{\mathbf{r}}$. Square each (using ${\mathbf{a}}^2=\mathbf{a}\cdot \mathbf{a}$): ${\dot{\mathbf{r}}}_1^2={\dot{\mathbf{R}}}^2-2\frac{m_2}{M}\dot{\mathbf{R}}\cdot \dot{\mathbf{r}}+\frac{m_2^2}{M^2}{\dot{\mathbf{r}}}^2,$ ${\dot{\mathbf{r}}}_2^2={\dot{\mathbf{R}}}^2+2\frac{m_1}{M}\dot{\mathbf{R}}\cdot \dot{\mathbf{r}}+\frac{m_1^2}{M^2}{\dot{\mathbf{r}}}^2.$ Now build $T=\frac{1}{2}{m}_1{\dot{\mathbf{r}}}_1^2+\frac{1}{2}{m}_2{\dot{\mathbf{r}}}_2^2$:

• ${\dot{\mathbf{R}}}^2$ term: $\frac{1}{2}(m_1+m_2){\dot{\mathbf{R}}}^2=\frac{1}{2}M{\dot{\mathbf{R}}}^2$. ✓
• Cross term: $\frac{1}{2}(-2m_1\frac{m_2}{M}+2m_2\frac{m_1}{M})\dot{\mathbf{R}}\cdot \dot{\mathbf{r}}=0$. The two pieces cancel — this is why the coordinates decouple.
• ${\dot{\mathbf{r}}}^2$ term: $\frac{1}{2}\left(m_1\frac{m_2^2}{M^2}+m_2\frac{m_1^2}{M^2}\right){\dot{\mathbf{r}}}^2=\frac{1}{2}\frac{m_1{m}_2(m_2+m_1)}{M^2}{\dot{\mathbf{r}}}^2=\frac{1}{2}\frac{m_1{m}_2}{M}{\dot{\mathbf{r}}}^2=\frac{1}{2}{\mu }_{\text{red}}{\dot{\mathbf{r}}}^2.$ Summing: $T=\frac{1}{2}M{\dot{\mathbf{R}}}^2+\frac{1}{2}{\mu }_{\text{red}}{\dot{\mathbf{r}}}^2$. ∎ The vanishing cross term is what lets the orbit (energy) problem be solved independently of the COM drift.

Direct: $T=\frac{1}{2}(2)(1^2)+\frac{1}{2}(6)(2^2)=1+12=13\text{J}.$ Split: $M=8$. $\dot{\mathbf{R}}=\frac{m_1{\dot{\mathbf{r}}}_1+m_2{\dot{\mathbf{r}}}_2}{M}=\frac{(2,0)+(0,12)}{8}=(0.25,1.5)$. $\dot{\mathbf{r}}={\dot{\mathbf{r}}}_2-{\dot{\mathbf{r}}}_1=(-1,2)$. ${\mu }_{\text{red}}=\frac{2\cdot 6}{8}=1.5$. $\frac{1}{2}M{\dot{\mathbf{R}}}^2=\frac{1}{2}(8)(0.25^2+1.5^2)=4(0.0625+2.25)=9.25$. $\frac{1}{2}{\mu }_{\text{red}}{\dot{\mathbf{r}}}^2=\frac{1}{2}(1.5)(1+4)=0.75\times 5=3.75$. Sum $=9.25+3.75=13\text{J}$ ✓. The two decoupled energies rebuild the total exactly.

Each ladder rung is one idea: (1) name the rope $\mathbf{r}$, (2) plug masses into the two $\mu$'s, (3) prove the middle point drifts because internal forces cancel, (4) combine everything into a real orbit and recover both bodies, (5) prove the energy splits cleanly. If you can do L5 unaided, you own the reduction from two bodies to one.