Exercises — Newton's law of gravitation — universal, action at distance

WHAT: direct substitution into $F=G{m}_1{m}_2/r^2$. WHY: both masses and the separation are given — nothing to rearrange. $F=\frac{(6.674\times 10^{-11})(1)(1)}{(0.5{)}^2}=\frac{6.674\times 10^{-11}}{0.25}$ $F=2.67\times 10^{-10}\text{N}$ This is the magnitude; each bag is pulled toward the other along the line joining them. Lesson: absurdly tiny — a fraction of a bacterium's weight. This is why we never feel gravity between everyday objects; only the huge Earth is heavy enough to matter.

No. $G$ is the universal constant — identical on Earth, the Moon, and in the farthest galaxy. What changed is $g=GM/r^2$, which depends on the local body's mass $M$ and radius $r$. The Moon has much smaller $M$, so its $g$ is smaller. See Gravitational Field & Potential.

WHAT: set the gravity force equal to weight, $\frac{G{M}_{\oplus }m}{R_{\oplus }^2}=mg$, then cancel $m$. WHY cancel $m$? The test mass $m$ appears on both sides. Killing it proves every object gets the same $g$ — Galileo's leaning-tower result falls straight out of the algebra. $g=\frac{G{M}_{\oplus }}{R_{\oplus }^2}=\frac{(6.674\times 10^{-11})(5.97\times 10^{24})}{(6.37\times 10^6{)}^2}$ $g=\frac{3.984\times 10^{14}}{4.057\times 10^{13}}\approx 9.82{\text{m/s}}^2✓$

WHAT: compute $g$ at the new $r$, then $W=mg$. WHY not just use 9.8? Because $g$ shrinks as $1/r^2$; at higher altitude the field is weaker. $g^{\prime }=\frac{G{M}_{\oplus }}{(1.02R_{\oplus }{)}^2}=\frac{g}{1.02^2}=\frac{9.82}{1.0404}=9.44{\text{m/s}}^2$ $W=m{g}^{\prime }=70\times 9.44=661\text{N}$ Her mass is still $70\text{kg}$; only the pull (weight) dropped by about $4\%$.

WHAT: straight substitution with $m_1=M_{\oplus }$, $m_2=M_{\text{Moon}}$. $F=\frac{(6.674\times 10^{-11})(5.97\times 10^{24})(7.35\times 10^{22})}{(3.84\times 10^8{)}^2}$ Numerator $=2.929\times 10^{37}$. Denominator $=1.475\times 10^{17}$. $F\approx 1.99\times 10^{20}\text{N}$ Huge — because both masses are astronomical, even though they're far apart.

(a) WHAT: the same law, scaled by $1/r^2$. At $60$ Earth-radii the field is $60^2=3600$ times weaker. $a_{\text{grav}}=\frac{g}{60^2}=\frac{9.8}{3600}=2.72\times 10^{-3}{\text{m/s}}^2$ (b) WHY centripetal? The Moon travels a near-circle; a circular path requires an inward acceleration $a=4{\pi }^{2}r/T^2$, where $T$ (the period) is how long one orbit takes (Circular Motion & Centripetal Force). Convert $T$ into seconds: $27.3\times 86400=2.359\times 10^6\text{s}$. $a_{\text{circ}}=\frac{4{\pi }^2(3.84\times 10^8)}{(2.359\times 10^6{)}^2}=\frac{1.516\times 10^{10}}{5.564\times 10^{12}}=2.72\times 10^{-3}{\text{m/s}}^2$ They match. The force that pulls the apple down is the same force steering the Moon — that agreement is the empirical proof of universality.

WHAT: track each proportionality separately, since $F\propto m_1{m}_2/r^2$.

• Triple $m_1$: $\times 3$.
• Halve $m_2$: $\times \frac{1}{2}$.
• Double $r$: divide by $2^2=4$, so $\times \frac{1}{4}$. $\text{factor}=3\times \frac{1}{2}\times \frac{1}{4}=\frac{3}{8}=0.375$ New force is $0.375F_0$ — the distance-squared term does the heaviest lifting.

WHAT: let the probe sit distance $x$ from Earth and $d-x$ from the Moon. Set the two pulls equal (a red probe in the figure feels a black arrow each way). WHY equal, not sum? Using our sign convention, the pull toward Earth is $-$ and the pull toward the Moon is $+$ (opposite directions along the axis). "Net zero" means $(+)F_{\text{Moon}}+(-)F_{\text{Earth}}=0$, i.e. the two magnitudes are equal. $\frac{G{M}_{\oplus }m}{x^2}=\frac{G{M}_{\text{Moon}}m}{(d-x{)}^2}$ Cancel $G$ and the probe mass $m$. WHY cancel $m$ here? The probe's own mass $m$ appears on both sides — it is a common factor, so it divides out. Physically this means the balance point is the same for a feather or a boulder; the probe's mass never chooses where equilibrium sits. $\frac{M_{\oplus }}{x^2}=\frac{M_{\text{Moon}}}{(d-x{)}^2}\Rightarrow \frac{(d-x{)}^2}{x^2}=\frac{M_{\text{Moon}}}{M_{\oplus }}=\frac{1}{81.3}$ Take the square root (positive, since both lengths are positive): $\frac{d-x}{x}=\frac{1}{\sqrt{81.3}}=\frac{1}{9.017}$ $d-x=\frac{x}{9.017}\Rightarrow d=x\left(1+\frac{1}{9.017}\right)=1.1109x$ $x=\frac{d}{1.1109}=\frac{3.84\times 10^8}{1.1109}=3.457\times 10^8\text{m}$ So the null point sits about $3.46\times 10^8\text{m}$ from Earth's centre — roughly $90\%$ of the way to the Moon, because Earth is far heavier and dominates most of the gap. Why the negative root is rejected: $\sqrt{\cdot }$ also allows $(d-x)/x=-1/9.017$, giving $x>d$ (probe beyond the Moon). There the pulls point the same way, so they add (both same sign), never cancel — physically impossible, so we drop it.

WHAT: set gravitational pull equal to the required centripetal force $m{v}^2/r$. WHY: in a circle the only inward force available is gravity, so it must supply exactly $m{v}^2/r$ (Circular Motion & Centripetal Force). $\frac{G{M}_{\oplus }m}{r^2}=\frac{m{v}^2}{r}\Rightarrow v^2=\frac{G{M}_{\oplus }}{r}\Rightarrow v=\sqrt{\frac{G{M}_{\oplus }}{r}}$ The satellite mass $m$ cancels — orbital speed depends only on how far out you are. Now $r=6.37\times 10^6+4.0\times 10^5=6.77\times 10^6\text{m}$: $v=\sqrt{\frac{(6.674\times 10^{-11})(5.97\times 10^{24})}{6.77\times 10^6}}=\sqrt{5.885\times 10^7}$ $v\approx 7.67\times 10^3\text{m/s}\approx 7.67\text{km/s}$ That's about $27,600\text{km/h}$ — matching the real ISS. This same $v^2=GM/r$ is the seed of Kepler's Laws.

WHAT: combine "gravity = centripetal" with $v=2\pi r/T$ (speed = circumference over period). WHY this route? We can't measure $M$ directly, but we can time an orbit. The formula lets the orbit report the mass back to us — this is literally how astronomers weigh planets and stars. Start from $\frac{GMm}{r^2}=\frac{m{v}^2}{r}$ and substitute $v=\frac{2\pi r}{T}$: WHY cancel the small mass $m$? The orbiting moon's mass $m$ sits on both sides as a common factor, so it divides out — the orbit's size and timing depend only on the central mass $M$, not on how heavy the orbiting body is. That is why we can weigh the planet without knowing its moon's mass. $\frac{GM}{r}=v^2=\frac{4{\pi }^2{r}^2}{T^2}\Rightarrow M=\frac{4{\pi }^2{r}^3}{G{T}^2}$ Convert $T=5.0\times 86400=4.32\times 10^5\text{s}$. $M=\frac{4{\pi }^2(2.0\times 10^8{)}^3}{(6.674\times 10^{-11})(4.32\times 10^5{)}^2}$ Numerator $=4{\pi }^2(8.0\times 10^{24})=3.158\times 10^{26}$. Denominator $=(6.674\times 10^{-11})(1.866\times 10^{11})=1.246\times 10^1$. $M=\frac{3.158\times 10^{26}}{12.46}\approx 2.53\times 10^{25}\text{kg}$ Notice $M\propto r^3/T^2$ — that ratio is exactly Kepler's Third Law.

Large $r$ (WHAT): $F=G{m}_1{m}_2/r^2\to 0$ as $r\to \infty$, but it never becomes exactly zero for any finite distance. Gravity has infinite range — the most distant galaxy still tugs on you, just immeasurably faintly. This is why "universal" is literal. Small $r$ for point masses (WHAT): as $r\to 0$, $F\to \infty$ — the formula diverges. That looks alarming. WHY it's not a real crisis: real objects are not points. Once you enter a sphere, the Shell Theorem says only the mass inside radius $r$ pulls you, and that mass shrinks as $r^3$. So the pull inside a uniform sphere actually goes as $F\propto r$ and drops smoothly to zero at the centre — no infinity. The blow-up is only for idealised points sitting exactly on top of each other, which never happens. (True point singularities live in General Relativity.)

WHAT: set total mechanical energy to zero — "just barely escapes" means it arrives at infinity with zero speed left over. WHY zero? At infinity $U\to 0$; the minimum launch has no leftover kinetic energy either, so the total (which is conserved) must be $0$ from the start. $\frac{1}{2}m{v}_{\text{esc}}^2+\left(-\frac{GMm}{R}\right)=0\Rightarrow \frac{1}{2}m{v}_{\text{esc}}^2=\frac{GMm}{R}$ WHY cancel the launch mass $m$? It appears on both sides as a common factor, so it divides out — a pebble and a rocket need the same escape speed. $\frac{1}{2}{v}_{\text{esc}}^2=\frac{GM}{R}\Rightarrow v_{\text{esc}}^2=\frac{2GM}{R}\Rightarrow \boxed{v_{\text{esc}}=\sqrt{\frac{2GM}{R}}}$ Now evaluate for Earth ($M=5.97\times 10^{24}\text{kg}$, $R=6.37\times 10^6\text{m}$): $v_{\text{esc}}=\sqrt{\frac{2(6.674\times 10^{-11})(5.97\times 10^{24})}{6.37\times 10^6}}=\sqrt{1.251\times 10^8}$ $v_{\text{esc}}\approx 1.12\times 10^4\text{m/s}\approx 11.2\text{km/s}$ Notice it's exactly $\sqrt{2}$ times the low-orbit speed of L4.2 — a neat sanity check.

• Force between two 1 kg bags at 0.5 m? ::: $2.67\times 10^{-10}\text{N}$
• Which is universal, $g$ or $G$? ::: $G$; $g=GM/r^2$ is local.
• Orbital speed formula? ::: $v=\sqrt{GM/r}$ (satellite mass cancels).
• Weigh a planet from an orbit? ::: $M=4{\pi }^2{r}^3/(G{T}^2)$ — Kepler's third law.
• Escape speed? ::: $v_{\text{esc}}=\sqrt{2GM/R}$, about $11.2\text{km/s}$ for Earth.
• Does gravity's range end? ::: No — it fades as $1/r^2$ but never hits zero at finite $r$.