Circular orbit — velocity, period, energy: Exercises

Figure — Circular orbit — velocity, period, energy

The picture above is the whole toolkit: every quantity ( $v, T, K, U, E$ ) is a function of the single number $r$ , the distance from the center. Read it as "pick $r$ , read off everything."

Level 1 — Recognition

Goal: pull the right formula off the shelf and plug in. No algebraic rearranging.

L1.1

A satellite orbits Earth at radius $r = 7.00 \times 1 0^{6} m$ . Find its orbital speed $v$ .

Recall Solution L1.1

WHAT: Use $v = GM / r$ directly — we know $r$ , we want $v$ . $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )}{7.00 \times 1 0 ^{6}} = 5.691 \times 1 0^{7}$ $v \approx 7.54 \times 1 0^{3} m/s = 7.54 km/s .$ Why no satellite mass? The mass $m$ cancels in $GM m / r^{2} = m v^{2} / r$ , so speed never depends on $m$ .

L1.2

The same satellite has mass $m = 1200 kg$ . Find its total mechanical energy $E$ .

Recall Solution L1.2

WHAT: Use $E = - \frac{GM m}{2 r}$ . Here $m$ does matter (energy scales with mass). $E = - \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1200 )}{2 ( 7.00 \times 1 0 ^{6} )}$ $E \approx - 3.42 \times 1 0^{10} J .$ Why negative? Negative total energy means bound — the satellite is trapped, you'd have to add energy to free it.

L1.3

Find the period $T$ of the satellite at $r = 7.00 \times 1 0^{6} m$ .

Recall Solution L1.3

WHAT: Use $T = 2 π r^{3} / GM$ — we want a time from a radius. $T = 2 π \frac{( 7.00 \times 1 0 ^{6} ) ^{3}}{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} )} = 2 π 8.610 \times 1 0^{6}$ $T \approx 2 π (2934) \approx 5.83 \times 1 0^{3} s \approx 97 min .$ Check: $T = 2 π r / v = 2 π (7.00 \times 1 0^{6}) /7540 \approx 5.83 \times 1 0^{3} s$ ✓ — the two routes agree.

Level 2 — Application

Goal: rearrange a formula, or chain two formulas together.

L2.1

A satellite orbits the Moon ( $M = 7.35 \times 1 0^{22} kg$ , radius $1.74 \times 1 0^{6} m$ ) at altitude $100 km$ . Find its speed.

Recall Solution L2.1

Step 1 — build the radius. $r = 1.74 \times 1 0^{6} + 1.00 \times 1 0^{5} = 1.84 \times 1 0^{6} m$ . Step 2 — plug into $v = GM / r$ with the Moon's $M$ : $v = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 7.35 \times 1 0 ^{22} )}{1.84 \times 1 0 ^{6}} = 2.666 \times 1 0^{6} \approx 1.63 \times 1 0^{3} m/s .$ About $1.63 km/s$ — much slower than Earth orbit because the Moon's $M$ is tiny.

L2.2

Invert Kepler: an Earth satellite has period $T = 2.00 hours = 7200 s$ . Find its orbital radius $r$ .

Recall Solution L2.2

WHAT: We know $T$ , want $r$ . Solve $T^{2} = \frac{4 π ^{2}}{GM} r^{3}$ for $r$ : $r = (\frac{GM T ^{2}}{4 π ^{2}})^{1/3} .$ Why cube-root? $T^{2} \propto r^{3}$ , so to undo the cube we take the $1/3$ power. $r = (\frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 7200 ) ^{2}}{4 π ^{2}})^{1/3} = (5.229 \times 1 0^{20})^{1/3}$ $r \approx 8.06 \times 1 0^{6} m .$

L2.3

Show that the kinetic energy of any circular orbit equals $- E$ , and verify numerically for the L1 satellite ( $m = 1200 kg$ , $r = 7.00 \times 1 0^{6}$ ).

Recall Solution L2.3

Algebra: $K = \frac{GM m}{2 r}$ and $E = - \frac{GM m}{2 r}$ , so $K = - E$ exactly (the Virial Theorem for $1/ r$ gravity). Numbers: $K = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 1200 )}{2 ( 7.00 \times 1 0 ^{6} )} \approx 3.42 \times 1 0^{10} J$ . And from L1.2, $E \approx - 3.42 \times 1 0^{10} J$ , so indeed $K = - E$ . ✓

Level 3 — Analysis

Goal: compare two orbits, reason about ratios and directions of change.

L3.1

Satellite A orbits at $r$ ; satellite B orbits at $4 r$ around the same planet. Find the ratios $v_{B} / v_{A}$ and $T_{B} / T_{A}$ .

Recall Solution L3.1

Speed ratio. $v \propto r^{- 1/2}$ , so $\frac{v _{B}}{v _{A}} = (\frac{4 r}{r})^{- 1/2} = 4^{- 1/2} = \frac{1}{2}$ . B moves half as fast. Period ratio. $T \propto r^{3/2}$ , so $\frac{T _{B}}{T _{A}} = 4^{3/2} = 8$ . B takes 8× longer per lap. Sanity: B is 4× farther but only half the speed, so time $\propto r / v = 4/ (1/2) = 8$ ✓.

L3.2

By what factor does the total energy $E$ change when you move a satellite from $r$ to $4 r$ ? Does it go up or down?

Recall Solution L3.2

$E \propto - 1/ r$ . So $\frac{E _{B}}{E _{A}} = \frac{- 1/ ( 4 r )}{- 1/ r} = \frac{1}{4}$ . $E_{B}$ is $\frac{1}{4}$ of $E_{A}$ in magnitude, but both are negative, so $E_{B}$ is less negative — energy has risen (moved toward zero). The higher orbit has more total energy even though it moves slower.

L3.3

A satellite is in a circular orbit. Its speed is measured to be exactly $2$ times the local circular speed at that radius. What is its total energy, and what happens to it?

Recall Solution L3.3

WHAT: At the same $r$ , circular speed gives $v_{c}^{2} = GM / r$ . Here $v^{2} = 2 v_{c}^{2} = 2 GM / r$ . Kinetic: $K = \frac{1}{2} m v^{2} = \frac{GM m}{r}$ . Potential (unchanged, depends only on $r$ ): $U = - \frac{GM m}{r}$ . Total: $E = K + U = \frac{GM m}{r} - \frac{GM m}{r} = 0$ . $E = 0$ is exactly the escape boundary — the body is no longer bound. This is why Escape Velocity satisfies $v_{esc} = 2 v_{c i r c}$ .

Level 4 — Synthesis

Goal: combine energy bookkeeping, Kepler, and definitions into a multi-step plan.

L4.1 — Energy to raise an orbit

Move a $2000 kg$ satellite from a circular orbit at $r_{1} = 7.00 \times 1 0^{6} m$ to a circular orbit at $r_{2} = 1.20 \times 1 0^{7} m$ around Earth. How much energy must be added?

Recall Solution L4.1

WHAT: Energy added $= E_{2} - E_{1}$ , the difference of total orbital energies. $Δ E = - \frac{GM m}{2 r _{2}} - (- \frac{GM m}{2 r _{1}}) = \frac{GM m}{2} (\frac{1}{r _{1}} - \frac{1}{r _{2}}) .$ Why this form? Higher orbit has less negative $E$ , so $Δ E > 0$ : raising an orbit costs energy. $Δ E = \frac{( 6.674 \times 1 0 ^{- 11} ) ( 5.97 \times 1 0 ^{24} ) ( 2000 )}{2} (\frac{1}{7.00 \times 1 0 ^{6}} - \frac{1}{1.20 \times 1 0 ^{7}}) .$ The prefactor $\frac{GM m}{2} = 3.984 \times 1 0^{17}$ . The bracket $= 1.4286 \times 1 0^{- 7} - 8.333 \times 1 0^{- 8} = 5.952 \times 1 0^{- 8}$ . $Δ E \approx (3.984 \times 1 0^{17}) (5.952 \times 1 0^{- 8}) \approx 2.37 \times 1 0^{10} J .$

L4.2 — From surface launch to orbit

Starting from rest on Earth's surface (radius $R_{\oplus}$ ), how much energy per kilogram is needed to reach a circular orbit at $r = 7.00 \times 1 0^{6} m$ ? (Ignore Earth's rotation and air.)

Recall Solution L4.2

WHAT: Compare total energy per unit mass in the two states. On the surface at rest: $E_{i} / m = - \frac{GM}{R _{\oplus}}$ (only potential, no kinetic). In orbit: $E_{f} / m = - \frac{GM}{2 r}$ . $\frac{Δ E}{m} = - \frac{GM}{2 r} + \frac{GM}{R _{\oplus}} = GM (\frac{1}{R _{\oplus}} - \frac{1}{2 r}) .$ $GM = (6.674 \times 1 0^{- 11}) (5.97 \times 1 0^{24}) = 3.984 \times 1 0^{14}$ . Bracket $= \frac{1}{6.37 \times 1 0 ^{6}} - \frac{1}{1.40 \times 1 0 ^{7}} = 1.5699 \times 1 0^{- 7} - 7.143 \times 1 0^{- 8} = 8.556 \times 1 0^{- 8}$ . $\frac{Δ E}{m} \approx (3.984 \times 1 0^{14}) (8.556 \times 1 0^{- 8}) \approx 3.41 \times 1 0^{7} J/kg .$ About $34 MJ$ per kilogram — the fundamental energy cost of low Earth orbit.

L4.3 — Density of a planet from a grazing orbit

A satellite skims the surface of a planet ( $r \approx R$ , the planet radius) with period $T$ . Show the period depends only on the planet's density $ρ$ , and compute $T$ for $ρ = 5510 kg/m^{3}$ (Earth's mean density).

Recall Solution L4.3

Step 1 — Kepler at $r = R$ : $T^{2} = \frac{4 π ^{2} R ^{3}}{GM}$ . Step 2 — replace $M$ by density. For a sphere $M = ρ \cdot \frac{4}{3} π R^{3}$ . Substitute: $T^{2} = \frac{4 π ^{2} R ^{3}}{G \cdot ρ \frac{4}{3} π R ^{3}} = \frac{4 π ^{2}}{\frac{4}{3} π Gρ} = \frac{3 π}{Gρ} .$ Why did $R$ vanish? Both the numerator $R^{3}$ and the mass's $R^{3}$ cancel — the grazing period knows nothing about size, only density. $T = \frac{3 π}{Gρ} = \frac{3 π}{( 6.674 \times 1 0 ^{- 11} ) ( 5510 )} = 2.562 \times 1 0^{7} \approx 5.06 \times 1 0^{3} s \approx 84 min .$ This is the famous "minimum orbital period" — every rocky planet of Earth-like density has a grazing period near 84 minutes.

Level 5 — Mastery

Goal: derive a new relation, or handle a limiting/degenerate case with care.

L5.1 — The virial theorem, verified

For a circular orbit, prove that the time-averaged kinetic energy satisfies $⟨ K ⟩ = - \frac{1}{2} ⟨ U ⟩$ , and that $E = - K$ . (For a circle every instant is the same, so time-averages are just the constant values.)

Recall Solution L5.1

Constants (circular ⇒ no averaging needed): $K = \frac{GM m}{2 r}, U = - \frac{GM m}{r} .$ Check virial: $- \frac{1}{2} U = - \frac{1}{2} (- \frac{GM m}{r}) = \frac{GM m}{2 r} = K$ ✓. Check $E = - K$ : $E = K + U = \frac{GM m}{2 r} - \frac{GM m}{r} = - \frac{GM m}{2 r} = - K$ ✓. These are the Virial Theorem relations $2 K + U = 0$ and $E = \frac{1}{2} U = - K$ , holding exactly for the inverse-square force.

L5.2 — Vis-viva sanity check

The general vis-viva equation gives speed at distance $r$ on an orbit with semi-major axis $a$ (half the long diameter of the ellipse): $v^{2} = GM (\frac{2}{r} - \frac{1}{a}) .$ Show that setting $a = r$ (a circle, where the semi-major axis equals the constant radius) recovers $v = GM / r$ .

Recall Solution L5.2

WHAT: For a circle every point is at the same distance, so $a = r$ . Substitute: $v^{2} = GM (\frac{2}{r} - \frac{1}{r}) = GM \cdot \frac{1}{r} = \frac{GM}{r} .$ $v = \frac{GM}{r} . ✓$ The circular formula is the special case of vis-viva with zero eccentricity. This is why the parent note's speed drops straight out of the general orbit law.

L5.3 — Degenerate limit: $r \to \infty$

Examine what happens to $v$ , $T$ , and $E$ as the orbit radius grows without bound. Interpret each limit physically.

Recall Solution L5.3

Speed: $v = GM / r \to 0$ . Infinitely large orbits are traversed infinitely slowly — gravity is too weak to demand any speed.
Period: $T = 2 π r^{3} / GM \to \infty$ . One lap takes forever.
Energy: $E = - GM m / (2 r) \to 0^{-}$ . The orbit approaches the escape boundary $E = 0$ from below: infinitely far, barely bound, at rest. Add any tiny bit more energy and the orbit becomes open (hyperbolic, $E > 0$ ) and the body leaves forever.

The other extreme, $r \to R_{\oplus}$ (grazing): $v$ is maximal, $T$ minimal (~84 min from L4.3), $E$ most negative — the most tightly bound circular orbit that still clears the surface.

Recall Master self-test (cover the answers)

Circular speed at $r = 7.00 \times 1 0^{6}$ m about Earth ::: $\approx 7.54 km/s$ Period at that radius ::: $\approx 5.83 \times 1 0^{3} s \approx 97 min$ Speed ratio for orbits at $r$ and $4 r$ ::: $v_{4 r} / v_{r} = \frac{1}{2}$ Period ratio for orbits at $r$ and $4 r$ ::: $T_{4 r} / T_{r} = 8$ Speed that makes $E = 0$ at radius $r$ ::: $2 v_{c i r c}$ (escape) Grazing-orbit period depends only on ::: planet density $ρ$ , via $T = 3 π / Gρ$ Vis-viva with $a = r$ gives ::: $v = GM / r$ , the circular case

Connections

3.2.13 Circular orbit — velocity, period, energy (Hinglish) — parent topic (Hinglish companion).
Newton's Law of Universal Gravitation — the force behind every formula here.
Centripetal Force and Uniform Circular Motion — the $m v^{2} / r$ requirement.
Kepler's Three Laws — L2.2, L4.3 invert $T^{2} \propto r^{3}$ .
Gravitational Potential Energy — the $U = - GM m / r$ used in every energy problem.
Escape Velocity — the $E = 0$ boundary met in L3.3 and L5.3.
Elliptical Orbits and the Vis-viva Equation — L5.2's general law.
Virial Theorem — L2.3 and L5.1's $E = - K$ , $U = 2 E$ .