Irodov 1.201: Kepler's Third Law Places Jupiter – Amit & Deepali Sinha Foundation

Problem Irodov 1.201

Jupiter takes 12 times as long as the Earth to complete one orbit around the Sun. Treating both orbits as circles centered on the Sun, find:

the ratio of Jupiter’s orbital radius to the Earth’s;
Jupiter’s orbital speed and acceleration, both in the heliocentric (Sun-centred, non-rotating) reference frame.

The Earth E and Jupiter J on their circular orbits about the Sun M, each with its orbital velocity v_E, v_J along its circle. The two radii are not drawn to scale. — The Earth $E$ and Jupiter $J$ on their circular orbits about the Sun $M$ , each with its orbital velocity $\vect v_E$ , $\vect v_J$ along its circle. The two radii are not drawn to scale.

Given

$T_J = 12\,T_E$

Find

(a)

r_J/r_E

, (b)

v_J

a_J

Solution

Step 1Kepler’s third law from the circular-orbit balance

For any planet of mass $m$ in a circular orbit of radius $r$ about the Sun, gravity supplies the centripetal force; writing the speed as the circumference over the period, $v = 2\pi r/T$ , and eliminating $v$ gives a relation between $T$ and $r$ that holds, with the same constant, for every planet orbiting the same Sun.

\frac{\gamma M m}{r^2} = \frac{m}{r}\left(\frac{2\pi r}{T}\right)^2 \;\Longrightarrow\; \gamma M = \frac{4\pi^2 r^3}{T^2}

T^2 = \dfrac{4\pi^2}{\gamma M}r^3

(1)

Step 2Distance ratio from the period ratio

Since $\gamma M$ is the same Sun for both planets, applying (1) to Earth and to Jupiter and dividing one by the other cancels that constant, leaving the radius ratio in terms of the given period ratio alone.

\left(\frac{T_J}{T_E}\right)^2 = \left(\frac{r_J}{r_E}\right)^3

\frac{r_J}{r_E} = \left(\frac{T_J}{T_E}\right)^{2/3} = 12^{2/3}

\dfrac{r_J}{r_E} = 12^{2/3} = 5.2

(2)

Step 3Jupiter’s orbital speed

The orbital speed is the circumference covered in one period. Writing Jupiter’s radius and period through the Earth’s own orbital data turns this into a number.

v_J = \frac{2\pi r_J}{T_J}

The Earth’s orbital radius and period are $r_E = 1.50\times10^{11}\ \mathrm{m}$ and $T_E = 3.156\times10^{7}\ \mathrm{s}$ , so with $r_J = 12^{2/3}r_E$ and $T_J = 12\,T_E$ , the radius evaluates first,

r_J = 12^{2/3}\times1.50\times10^{11}\ \mathrm m = 7.86\times10^{11}\ \mathrm m, \tag{3}

and then the speed,

v_J = \frac{2\pi\left(7.86\times10^{11}\ \mathrm m\right)}{12\times3.156\times10^{7}\ \mathrm s} = 1.30\times10^{4}\ \mathrm{m/s}.

v_J = \dfrac{2\pi r_E}{12^{1/3}T_E} = 13\text{ km/s}

(4)

Step 4Jupiter’s centripetal acceleration

In uniform circular motion the centripetal acceleration is $v^2/r$ ; the speed (4) and the radius (3) give the number directly.

a_J = \frac{v_J^2}{r_J}

a_J = \frac{(1.30\times10^{4}\ \mathrm{m/s})^2}{7.86\times10^{11}\ \mathrm m} = 2.2\times10^{-4}\ \mathrm{m/s^2}.

a_J = \dfrac{v_J^2}{r_J} = 2.2\times10^{-4}\text{ m/s}^2

(5)

Check

The same acceleration follows from the inverse square law directly, without ever going through $v_J$ : $\gamma M/r_J^2 = (6.67\times10^{-11})(1.97\times10^{30})/(7.86\times10^{11})^2 = 2.1\times10^{-4}\ \mathrm{m/s^2}$ , agreeing with the $2.2\times10^{-4}\ \mathrm{m/s^2}$ of (5) to about two percent. That gap is not rounding: it is the mild inconsistency between the tabulated $\gamma M_\odot$ and the Earth’s own tabulated $r_E$ , $T_E$ (the two imply values of $\gamma M_\odot$ that differ by about 1.8 percent), and both routes still express the same centripetal requirement.

Remark

Steps 1 and 2 never needed a value for $\gamma$ or the Sun’s mass, only the given period ratio: any two planets sharing the same Sun can be compared this way without knowing $\gamma$ or the Sun’s mass individually, because both cancel from the ratio. Only when an absolute distance or speed is wanted, as in Steps 3 and 4, does the Earth’s own orbit have to supply the missing scale, which is exactly how the Earth’s orbit serves as the meter stick for the rest of the solar system.