Exercises — Gravitational potential energy — U = −GMm - r (not mgh)

WHAT we do: compare $-GMm/r$ at two radii — no numbers needed, just the shape of $1/r$. WHY: dividing by a bigger $r$ makes $GMm/r$ smaller in size, and the minus in front means a smaller-size number is closer to zero, i.e. larger on the number line.

• At $r_1=8000$ km: small $r$ → big $GMm/r$ → very negative $U$.
• At $r_2=20000$ km: bigger $r$ → smaller $GMm/r$ → less negative $U$.

So moving out, $U$ increases (moves up toward $0$). The value at $r_1$ (8000 km) is the more negative one. Picture: on a number line, $r_1$ sits deeper below zero; $r_2$ sits nearer to zero. You had to lift the satellite, so its energy went up. ✔

The deciding question is always: does $g$ stay effectively constant over the motion? $mgh$ assumes a uniform field; it is only the small-$h$ approximation of $-GMm/r$.

• (a) $h=2\text{ m}\ll R_{\oplus }$ → $g$ constant → $mgh$ is fine.
• (b) height comparable to $R_{\oplus }$ and far beyond → $g$ drops enormously → must use $-GMm/r$.
• (c) $h=10\text{ m}\ll R_{\oplus }$ → $mgh$ is fine.

Rule of thumb: if $h$ is a few kilometres or less, $mgh$; if it's a large fraction of a planet radius or more, $-GMm/r$. See Acceleration due to gravity g.

WHY this formula: escape means "just barely reach infinity with zero speed", i.e. total energy $=0$: $\frac{1}{2}m{v}^2-GMm/R=0$. The $m$ cancels, leaving $v_{\text{esc}}=\sqrt{2GM/R}$ (see Escape Velocity). $v_{\text{esc}}=\sqrt{\frac{2(6.67\times 10^{-11})(7.35\times 10^{22})}{1.74\times 10^6}}.$ Numerator inside: $2\cdot 6.67\times 10^{-11}\cdot 7.35\times 10^{22}=9.805\times 10^{12}$. Divide by $1.74\times 10^6$: $5.635\times 10^6$. Square root: $\approx 2.37\times 10^3\text{m/s}$. $\boxed{v_{\text{esc}}\approx 2.37\text{km/s}}$ That's about a fifth of Earth's $11.2$ km/s — the Moon's shallower gravity well is easier to climb out of.

WHAT: use $U=-GMm/r$ with $r=R_{\oplus }+h$. $r=6.37\times 10^6+0.41\times 10^6=6.78\times 10^6\text{m}.$ $U=-\frac{(6.67\times 10^{-11})(5.97\times 10^{24})(4.2\times 10^5)}{6.78\times 10^6}.$ Numerator: $6.67\times 10^{-11}\cdot 5.97\times 10^{24}=3.982\times 10^{14}$; times $4.2\times 10^5=1.673\times 10^{20}$. Divide by $6.78\times 10^6$: $2.467\times 10^{13}$. $\boxed{U\approx -2.47\times 10^{13}\text{J}}$ Negative, as every bound object must be.

WHY energy conservation: gravity is conservative, so $K_i+U_i=K_f+U_f$.

• Start: at rest ($K_i=0$) at $r\to \infty$ ($U_i=0$). So total energy $E=0$.
• End: at $r=R_{\oplus }$, $U_f=-GMm/R$, $K_f=\frac{1}{2}m{v}^2$.

Setting $E=0$: $\frac{1}{2}m{v}^2-\frac{GMm}{R}=0\Rightarrow v=\sqrt{\frac{2GM}{R}}.$ This is exactly the escape speed — falling in from infinity is the time-reverse of just barely escaping. For Earth: $v=\sqrt{\frac{2(6.67\times 10^{-11})(5.97\times 10^{24})}{6.37\times 10^6}}\approx 1.12\times 10^4\text{m/s}=\boxed{11.2\text{km/s}}.$

Exact: energy conservation between surface ($r=R$) and top ($r=R+h$, at rest): $\frac{1}{2}m{v}^2-\frac{GMm}{R}=0-\frac{GMm}{R+h}\Rightarrow v=\sqrt{2GM\left(\frac{1}{R}-\frac{1}{R+h}\right)}.$ With $R=6.37\times 10^6$, $R+h=6.67\times 10^6$: $\frac{1}{R}-\frac{1}{R+h}=1.5699\times 10^{-7}-1.4993\times 10^{-7}=7.06\times 10^{-9}.$ $v=\sqrt{2(3.982\times 10^{14})(7.06\times 10^{-9})}=\sqrt{5.62\times 10^6}\approx 2.37\times 10^3\text{m/s}.$ Approximate: $v=\sqrt{2gh}=\sqrt{2(9.8)(3\times 10^5)}=\sqrt{5.88\times 10^6}\approx 2.42\times 10^3$ m/s. $\boxed{v_{\text{exact}}\approx 2.37\text{km/s},v_{mgh}\approx 2.42\text{km/s}}$ The $mgh$ answer is ~$2\%$ too high because it pretends $g$ stays $9.8$ all the way up, but $g$ actually weakens with height, so less speed is really needed.

WHY $E=-GMm/2r$: for a circular orbit, gravity supplies the centripetal force, forcing $K=\frac{1}{2}|U|$, so total energy is $E=K+U=-GMm/2r$ (parent note & Kepler's Laws & Orbital Energy). Energy needed $=E_2-E_1$: $\Delta E=-\frac{GMm}{2r_2}-\left(-\frac{GMm}{2r_1}\right)=\frac{GMm}{2}\left(\frac{1}{r_1}-\frac{1}{r_2}\right).$ $GMm=(6.67\times 10^{-11})(5.97\times 10^{24})(1200)=4.778\times 10^{17}$. $\frac{1}{r_1}-\frac{1}{r_2}=1.4286\times 10^{-7}-1.0\times 10^{-7}=4.286\times 10^{-8}.$ $\Delta E=\frac{1}{2}(4.778\times 10^{17})(4.286\times 10^{-8})\approx 1.02\times 10^{10}\text{J}.$ $\boxed{\Delta E\approx 1.02\times 10^{10}\text{J}>0}$ Positive: you must add energy to climb to a higher orbit — matches the parent's Forecast-then-Verify. ✔

WHAT: radial fall, energy conserved between $r_1=R+500$ km and $r_2=R+100$ km, starting at rest. $0-\frac{GMm}{r_1}=\frac{1}{2}m{v}^2-\frac{GMm}{r_2}\Rightarrow v=\sqrt{2GM\left(\frac{1}{r_2}-\frac{1}{r_1}\right)}.$ $r_1=6.87\times 10^6$, $r_2=6.47\times 10^6$. $\frac{1}{r_2}-\frac{1}{r_1}=1.5456\times 10^{-7}-1.4556\times 10^{-7}=9.00\times 10^{-9}.$ $v=\sqrt{2(3.982\times 10^{14})(9.00\times 10^{-9})}=\sqrt{7.168\times 10^6}\approx 2.68\times 10^3\text{m/s}.$ $\boxed{v\approx 2.68\text{km/s}}$ (A quick $mgh$ estimate with $g\approx 9.8$ over $400$ km gives $\sqrt{2\cdot 9.8\cdot 4\times 10^5}\approx 2.80$ km/s — a few percent high because $g$ is smaller up there.)

Step 1 — factor out the exact form: $\Delta U=\frac{GMm}{R^2}\cdot \frac{h}{1+h/R}=mgh\cdot \frac{1}{1+h/R},\text{since }g=\frac{GM}{R^2}.$ WHY this shape: it isolates the pure $mgh$ and a correction factor $\frac{1}{1+h/R}$. Step 2 — take the limit: as $h/R\to 0$, the factor $\to 1$, giving $\Delta U\to mgh$. That is the schoolbook formula, recovered. Step 3 — error size: the true value is smaller than $mgh$ by the factor $\frac{1}{1+h/R}$. Fractional error $\approx \frac{h}{R}$ (from $\frac{1}{1+x}\approx 1-x$). At $h=100$ km: $\frac{h}{R}=\frac{1.0\times 10^5}{6.37\times 10^6}\approx 0.0157=\boxed{1.6\%}.$ So $mgh$ overestimates by about $1.6\%$ at 100 km — consistent with the L3 comparisons. See Conservative Forces and Potential Energy and Work-Energy Theorem.

Step 1 — force balance. Gravity is the centripetal force: $\frac{GMm}{r^2}=\frac{m{v}^2}{r}\Rightarrow m{v}^2=\frac{GMm}{r}.$ Step 2 — kinetic energy. $K=\frac{1}{2}m{v}^2=\frac{GMm}{2r}$. Step 3 — compare to $U$. Since $U=-\frac{GMm}{r}$, we get $K=-\frac{1}{2}U$. ✔ Step 4 — total energy. $E=K+U=\frac{GMm}{2r}-\frac{GMm}{r}=-\frac{GMm}{2r}=\frac{1}{2}U=-K.$ Interpretation: the kinetic energy is exactly half the depth of the well. This is the virial theorem for an inverse-square force. Physically: a bound orbit sits halfway "up" from the bottom in energy terms, which is why $E<0$ (bound) yet the object still moves.

Sign of $U$ at larger $r$? ::: Increases toward $0$ (less negative). $r$ in $U=-GMm/r$ is measured from…? ::: The planet's centre, so $r=R+h$. Impact speed falling from infinity equals…? ::: The escape speed, $\sqrt{2GM/R}$. Total energy of a circular orbit? ::: $E=-GMm/2r=\frac{1}{2}U=-K$. Fractional error of $mgh$ at height $h$? ::: About $h/R$ (it over-estimates).