Classical-Mechanics · Unit 11 · Video 3 · Interactive Practice

Gravity's Two Orbits: The Satellite That Never Moves and the Stars That Circle Nothing

IKey Formulas

Formula	Name	What it takes
$\dfrac{G m_e}{r_s^2} = \omega^2 r_s$	Gravity as the centripetal force	$m_s$ cancels
$r_s = \left(\dfrac{G m_e}{\omega^2}\right)^{1/3},\qquad \omega = \dfrac{2\pi}{T}$	Geostationary radius	Period and $m_e$
$\omega^2 = \dfrac{G(m_1 + m_2)}{s^3}$	Binary angular velocity	$s = r_1 + r_2$
$T = \sqrt{\dfrac{4\pi^2 s^3}{G(m_1 + m_2)}}$	Binary period	Both masses

Key Insight: A body's own mass always cancels out of its own radial equation — but its partner's mass never does. What survives in a two-body orbit is the total mass $m_1 + m_2$ .

IIFinding the Geostationary Radius

Exactly one orbit radius gives a period of one sidereal day, $T = 86164\ \text{s}$ .

Orbit radius r 3.00 Rₑ

IIITwo Stars, One Angular Velocity

Both stars sweep the same angle in the same time, so their radii split the separation in inverse proportion to their masses.

m₁ 1.0 m₂ 3.0 phase 0°

IVWhen the Single-Mass Formula Lies

How large must $m_2$ grow before the period built from $m_1$ alone stops being the truth?

Guess q = m₂/m₁ 0.250

💡 Jupiter is the Sun's heaviest companion at $q \approx 9.5 \times 10^{-4}$ , so the single-mass period is wrong by only about $0.05\%$ — the honest two-body law hides in plain sight.

VQuiz Questions

Problem 1 · Radius of a 12-Hour Orbit

Given: a satellite in uniform circular orbit about the Earth with period $T = 4.32 \times 10^{4}\ \text{s}$ , with $G = 6.67 \times 10^{-11}$ and $m_e = 5.98 \times 10^{24}\ \text{kg}$ — find the orbit radius $r$ .

✅ Correct! With

\omega = 2\pi/T = 1.454 \times 10^{-4}\ \text{s}^{-1}

, the cube root of

Gm_e/\omega^2

gives

r = 2.66 \times 10^{7}

❌ That is the geostationary radius. It belongs to

T = 86164

s, not to

T = 4.32 \times 10^{4}

s — rerun the cube root with the new

\omega

❌ Wrong power of the period.

r \propto T^{2/3}

, not

T^{1/3}

: halving

T

scales

r

2^{-2/3} = 0.631

, not by

2^{-1/3} = 0.794

❌ The period is not proportional to the radius. Cubing in

r_s^3 = Gm_e/\omega^2

means

r \propto T^{2/3}

, so halving

T

does not halve

r

❌ Not quite. Use

r = (Gm_e/\omega^2)^{1/3}

with

\omega = 2\pi/T

; the cube root of

1.885 \times 10^{22}\ \text{m}^3

is the answer.

Show solution

Gravity supplies the centripetal force, and the satellite's own mass cancels:

\frac{G m_s m_e}{r^2} = m_s \omega^2 r \quad\Longrightarrow\quad r^3 = \frac{G m_e}{\omega^2}

Step 1 — angular velocity:

\omega = \frac{2\pi}{T} = \frac{2\pi}{4.32 \times 10^{4}} = 1.454 \times 10^{-4}\ \text{s}^{-1}

Step 2 — cube the radius:

r^3 = \frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})}{(1.454 \times 10^{-4})^2} = \frac{3.99 \times 10^{14}}{2.115 \times 10^{-8}} = 1.885 \times 10^{22}\ \text{m}^3

Step 3 — take the cube root:

r = (1.885 \times 10^{22})^{1/3} = 2.66 \times 10^{7}\ \text{m}

As a check, $r \propto T^{2/3}$ : this period is half the sidereal day, so $r = (4.22 \times 10^{7})(2^{-2/3}) = (4.22 \times 10^{7})(0.631) = 2.66 \times 10^{7}$ m.

Problem 2 · Radius or Altitude?

Given: the geostationary orbit radius $r_s = 4.22 \times 10^{7}\ \text{m}$ and the Earth's mean radius $R_e = 6.37 \times 10^{6}\ \text{m}$ — find the satellite's altitude above the ground.

✅ Correct!

r_s

is measured from the Earth's center, so the altitude is

r_s - R_e = 3.58 \times 10^{7}

m, about

36{,}000

km.

❌ That is the orbit radius, not the altitude.

r_s

starts at the center of the Earth because that is where the gravitational force points — subtract

R_e

to reach the ground.

❌ Check the sign. The ground is inside the orbit, so the surface is at

R_e

and the altitude is

r_s - R_e

, not

r_s + R_e

❌ Check the powers of ten.

4.22 \times 10^{7} - 0.637 \times 10^{7} = 3.58 \times 10^{7}

m, ten times your value.

❌ Not quite. Altitude

= r_s - R_e

, with both distances measured from the Earth's center.

Show solution

The origin of the whole derivation sits at the Earth's center, so $r_s$ is a distance from the center, not from the surface. Line up the powers of ten before subtracting:

h = r_s - R_e = 4.22 \times 10^{7} - 0.637 \times 10^{7} = 3.58 \times 10^{7}\ \text{m}

In Earth radii the same statement reads $6.62\,R_e - 1.00\,R_e = 5.62\,R_e$ , and $3.58 \times 10^{7}\ \text{m} \approx 36{,}000$ km — the familiar figure quoted for the geostationary belt.

Problem 3 · A Double Star, End to End

Given: two stars in uniform circular motion about their common center with $m_1 = 2.0 \times 10^{30}\ \text{kg}$ , $m_2 = 6.0 \times 10^{30}\ \text{kg}$ and a fixed separation $s = 3.0 \times 10^{11}\ \text{m}$ ( $G = 6.67 \times 10^{-11}$ ) — find star 1's orbit radius $r_1$ and the period $T$ .

What is $r_1$ ?

What is $T$ ?

✅ Correct! The lighter star swings on the wider circle,

r_1 = 2.25 \times 10^{11}

m, and the period built from the total mass is

4.47 \times 10^{7}

❌ The ratio is inverted. Newton's Second Law on star 1 gives

r_1 = Gm_2/(\omega^2 s^2)

— star 1's radius is set by its partner's mass, so the lighter star travels farther.

❌ That is the symmetric answer.

r_1 = r_2 = s/2

only when

m_1 = m_2

; here the masses differ by a factor of 3.

❌ Not quite. Use

m_1 r_1 = m_2 r_2

together with

s = r_1 + r_2

, which gives

r_1 = s\,m_2/(m_1 + m_2)

❌ Wrong distance in the cube. The gravitational force depends on the full separation

s

, not on

r_1

— the period formula cubes

s

❌ You dropped a mass. That is

2\pi\sqrt{s^3/(Gm_2)}

; the honest two-body result uses the sum

m_1 + m_2 = 8.0 \times 10^{30}

kg.

❌ Almost — the $2\pi$ is missing.

\sqrt{s^3/(GM)} = 7.11 \times 10^{6}

s is

1/\omega

; the period is

T = 2\pi/\omega

❌ Not quite. Use

T = \sqrt{4\pi^2 s^3 /\big(G(m_1+m_2)\big)}

with

s^3 = 2.7 \times 10^{34}\ \text{m}^3

Show solution

Step 1 — the radii. Newton's Second Law along each star's radial direction gives $r_1 = Gm_2/(\omega^2 s^2)$ and $r_2 = Gm_1/(\omega^2 s^2)$ , so $m_1 r_1 = m_2 r_2$ . With $s = r_1 + r_2$ :

r_1 = s\,\frac{m_2}{m_1 + m_2} = (3.0 \times 10^{11})\frac{6.0 \times 10^{30}}{8.0 \times 10^{30}} = 2.25 \times 10^{11}\ \text{m}

(and $r_2 = 0.75 \times 10^{11}$ m, so $r_1 + r_2 = s\ \checkmark$ ).

Step 2 — the angular velocity. Adding the two radius equations gives

\omega^2 = \frac{G(m_1 + m_2)}{s^3} = \frac{(6.67 \times 10^{-11})(8.0 \times 10^{30})}{2.7 \times 10^{34}} = 1.977 \times 10^{-14}\ \text{s}^{-2}

Step 3 — the period.

T = \frac{2\pi}{\omega} = \sqrt{\frac{4\pi^2 s^3}{G(m_1+m_2)}} = 2\pi\sqrt{\frac{2.7 \times 10^{34}}{5.336 \times 10^{20}}} = 2\pi (7.11 \times 10^{6}) = 4.47 \times 10^{7}\ \text{s}

That is about $1.4$ years.

Problem 4 · Two Stars vs. One

Given: two identical stars of mass $m$ orbit their common center with separation $s$ . Compare their period with the period of a negligibly light planet circling a single star of mass $m$ at radius $s$ — which statement is true?

✅ Correct! Doubling the mass that steers the orbit from

m

2m

raises

\omega^2

2

, so

T

drops by

\sqrt{2}

❌ Wrong direction. More total mass means a stronger pull, a larger

\omega

, and therefore a shorter period.

❌ You applied the factor to the wrong quantity. The total mass doubles, so

\omega^2

doubles — but

T \propto 1/\sqrt{M}

, so the period falls by

\sqrt{2}

, not by

2

❌ Not quite. Compare

T = \sqrt{4\pi^2 s^3/(G\cdot 2m)}

with

T = \sqrt{4\pi^2 s^3/(G m)}

— the separation is the same, only the mass in the denominator differs.

Show solution

Binary. With $m_1 = m_2 = m$ the two-body result gives

T_{\text{bin}} = \sqrt{\frac{4\pi^2 s^3}{G(m + m)}} = \sqrt{\frac{4\pi^2 s^3}{2Gm}}

Planet around one star. A negligible mass at radius $s$ obeys the single-mass law:

T_{\text{one}} = \sqrt{\frac{4\pi^2 s^3}{G m}}

Ratio.

\frac{T_{\text{bin}}}{T_{\text{one}}} = \sqrt{\frac{Gm}{2Gm}} = \frac{1}{\sqrt{2}} \approx 0.707

The binary is faster because both masses pull. The single-mass law is the special case $m_2 \ll m_1$ , where the second mass is small enough to ignore and the common center falls inside the large body.

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