Exercises — Orbital velocity for circular orbit — derivation

$v_o=\sqrt{\frac{GM}{r}}$

• $v_o$ — the constant tangential speed for a circular orbit.
• $G$ — the universal gravitational constant (same everywhere in the cosmos).
• $M$ — mass of the central body being orbited (e.g. Earth).
• $r$ — distance from the centre of $M$ to the satellite, $r=R+h$. Notice: the satellite's own mass $m$ is not in the formula — it cancelled during the derivation.

No. The satellite mass $m$ cancels in $\frac{GMm}{r^2}=\frac{m{v}_o^2}{r}$, so $v_o$ depends only on $M$ and $r$. A feather and a truck at the same radius orbit at identical speed.

Near the surface $r\approx R$, so use the shortcut $v_o=\sqrt{gR}$: $v_o=\sqrt{9.8\times 6.4\times 10^6}=\sqrt{6.272\times 10^7}\approx 7.9\times 10^3\text{m/s}=7.9\text{km/s}.$

$r=R+h=6.4\times 10^6+3.6\times 10^6=1.0\times 10^7\text{m}.$ $v_o=\sqrt{\frac{g{R}^2}{r}}=\sqrt{\frac{9.8\times (6.4\times 10^6{)}^2}{1.0\times 10^7}}=\sqrt{\frac{9.8\times 4.096\times 10^{13}}{1.0\times 10^7}}.$ $=\sqrt{4.014\times 10^7}\approx 6.34\times 10^3\text{m/s}\approx 6.3\text{km/s}.$

$v_o=\sqrt{\frac{GM}{r}}=\sqrt{\frac{6.67\times 10^{-11}\times 6.0\times 10^{24}}{1.0\times 10^7}}=\sqrt{\frac{4.002\times 10^{14}}{1.0\times 10^7}}.$ $=\sqrt{4.002\times 10^7}\approx 6.33\times 10^3\text{m/s}\approx 6.3\text{km/s}.✓$ Both routes agree because $GM=g{R}^2$ — same physics, two data sets.

$v_o\propto 1/\sqrt{r}$, so multiplying $r$ by $9$ multiplies $v_o$ by $1/\sqrt{9}=1/3$. $v_o(9r)=\sqrt{\frac{GM}{9r}}=\frac{1}{3}\sqrt{\frac{GM}{r}}=\frac{1}{3}{v}_o(r).$ The speed drops to one-third.

Since $v_o\propto 1/\sqrt{r}$: $\frac{v_A}{v_B}=\sqrt{\frac{r_B}{r_A}}=\sqrt{\frac{4.2\times 10^7}{7\times 10^6}}=\sqrt{6}\approx \mathrm{2.449.}$ A is about 2.45 times faster than B. (See figure — inner orbit races, outer ambles.)

Escape velocity is $v_{esc}=\sqrt{2GM/r}$ (see Escape Velocity). $\frac{v_{esc}}{v_o}=\frac{\sqrt{2GM/r}}{\sqrt{GM/r}}=\sqrt{2}\approx \mathrm{1.414.}$ Independent of $G,M,r$ — always exactly $\sqrt{2}$. To leave you need $41.4\%$ more speed than to circle.

Set the two expressions for $v_o$ equal: $\frac{2\pi r}{T}=\sqrt{\frac{GM}{r}}.$ Square both sides: $\frac{4{\pi }^2{r}^2}{T^2}=\frac{GM}{r}\Rightarrow T^2=\frac{4{\pi }^2}{GM}{r}^3.$ So $T^2\propto r^3$ — exactly Kepler's Third Law, with proportionality constant $4{\pi }^2/GM$.

Solve for $r$: $r={\left(\frac{GM{T}^2}{4{\pi }^2}\right)}^{1/3}.$ Numerator: $GM{T}^2=4.002\times 10^{14}\times (86400{)}^2=4.002\times 10^{14}\times 7.465\times 10^9=2.988\times 10^{24}$. Divide by $4{\pi }^2=39.48$: $7.568\times 10^{22}$. Cube root: $r=(7.568\times 10^{22}{)}^{1/3}\approx 4.23\times 10^7\text{m}$ ($\approx 42,300\text{km}$ from centre). Then: $v_o=\sqrt{\frac{GM}{r}}=\sqrt{\frac{4.002\times 10^{14}}{4.23\times 10^7}}=\sqrt{9.46\times 10^6}\approx 3.08\times 10^3\text{m/s}\approx 3.1\text{km/s}.$ This is the Geostationary Orbit radius — check: only about $3.1$ km/s, far slower than LEO's $7.9$ km/s, as $1/\sqrt{r}$ predicts.

$v_{esc}=\sqrt{2}{v}_o=1.414\times 7.9=11.17\text{km/s}$. $\Delta v=v_{esc}-v_o=11.17-7.9=3.27\text{km/s}.$ Fraction: $\Delta v/v_{esc}=3.27/11.17\approx 0.293$, about 29% more.

$v_o=\sqrt{\frac{4.002\times 10^{14}}{3.84\times 10^8}}=\sqrt{1.042\times 10^6}\approx 1.021\times 10^3\text{m/s}\approx 1.02\text{km/s}.$ Excellent match with the real value — a treated circular orbit captures the Moon's motion to within a percent.

$v_o=\sqrt{g_p{R}_p}=\sqrt{3.7\times 3.4\times 10^6}=\sqrt{1.258\times 10^7}\approx 3.55\times 10^3\text{m/s}\approx 3.5\text{km/s}.$ Lower gravity and smaller radius both pull the surface-orbit speed down versus Earth's 7.9 km/s.

(a) From Kepler: $r={\left(\frac{GM{T}^2}{4{\pi }^2}\right)}^{1/3}$. $GM{T}^2=4.002\times 10^{14}\times (5.4\times 10^3{)}^2=4.002\times 10^{14}\times 2.916\times 10^7=1.167\times 10^{22}$. Divide by $39.48$: $2.956\times 10^{20}$. Cube root: $r=(2.956\times 10^{20}{)}^{1/3}\approx 6.66\times 10^6\text{m}$. (b) $v_o=\frac{2\pi r}{T}=\frac{2\pi \times 6.66\times 10^6}{5.4\times 10^3}=\frac{4.184\times 10^7}{5.4\times 10^3}\approx 7.75\times 10^3\text{m/s}\approx 7.7\text{km/s}.$ (c) $h=r-R=6.66\times 10^6-6.4\times 10^6=2.6\times 10^5\text{m}\approx 260\text{km}$ — a realistic LEO.

• $v_o$ formula? ⇒ $v_o=\sqrt{GM/r}$.
• $v_{esc}/v_o$ at same radius? ⇒ $\sqrt{2}$.
• Kepler constant linking $T^2$ and $r^3$? ⇒ $4{\pi }^2/GM$.
• Geostationary radius (from centre)? ⇒ $\approx 4.23\times 10^7$ m.
• Moon's orbital speed? ⇒ $\approx 1.02$ km/s.