Irodov 1.200: Circular Speed Alone Fixes the Period – Amit & Deepali Sinha Foundation

Problem Irodov 1.200

A planet travels in a circular orbit about the Sun at a constant speed $v = 34.9$ km/s, measured in the heliocentric (Sun-centred, non-rotating) reference frame. Find the orbital period $T$ .

A planet of mass m in a circular orbit of radius r about the Sun of mass M, moving with speed v tangent to the circle. — A planet of mass $m$ in a circular orbit of radius $r$ about the Sun of mass $M$ , moving with speed $\vect v$ tangent to the circle.

Given

$v = 34.9\text{ km/s}$

Find

T

Solution

Step 1Gravity supplies the centripetal force

For a circular orbit of radius $r$ , the Sun’s gravitational pull on the planet equals exactly the centripetal force needed to hold it on that circle. Writing Newton’s law of gravitation against Newton’s second law for uniform circular motion gives one equation linking $r$ , $v$ and the Sun’s mass $M$ .

\frac{\gamma M m}{r^2} = \frac{m v^2}{r} \;\Longrightarrow\; \gamma M = v^2 r

r = \dfrac{\gamma M}{v^2}

(1)

Step 2Period from the circumference and the speed

The period is the time to complete one circumference at the constant speed $v$ ; substituting the radius from (1) eliminates $r$ in favor of the given speed and the Sun’s mass.

T = \frac{2\pi r}{v} = \frac{2\pi}{v}\left(\frac{\gamma M}{v^2}\right)

The standard values of the gravitational constant and the Sun’s mass, together with the given speed, are

\gamma = 6.67\times10^{-11}\ \mathrm{N\,m^2/kg^2}, \qquad M = 1.97\times10^{30}\ \mathrm{kg}, \qquad v = 3.49\times10^{4}\ \mathrm{m/s},

and substituting them gives

T = \frac{2\pi(6.67\times10^{-11})(1.97\times10^{30})}{(3.49\times10^4)^3}\ \mathrm{s} = 1.94\times10^{7}\ \mathrm{s}.

Converting to days,

1.94\times10^{7}\ \mathrm{s}\times\frac{1\ \mathrm{d}}{8.64\times10^{4}\ \mathrm{s}} = 225\ \mathrm{d}.

T = \dfrac{2\pi\gamma M}{v^3} = 225\text{ days}

(2)

Check

Dimensionally $[\gamma][M]/[v]^3 = (\mathrm{m^3\,kg^{-1}\,s^{-2}})(\mathrm{kg})/(\mathrm{m^3\,s^{-3}}) = \mathrm{s}$ , a time, as required. Also, since $r=\gamma M/v^2$ shrinks as $v$ grows, a faster circular orbit is always a tighter one, consistent with the inner planets, which move faster, orbiting closer to the Sun than the outer ones.

Remark

The planet’s own mass canceling here is the same fact behind Galileo’s observation that all falling bodies accelerate alike: in circular orbital motion, so long as the orbiting body is light enough to leave the Sun fixed, the gravitational pull (proportional to the orbiting body’s mass) and the centripetal requirement (also proportional to that mass) cancel identically, so the dynamics depends only on the central mass. That is also why orbital mechanics almost always works with the single product $\gamma M$ , the Sun’s standard gravitational parameter, known to far higher precision than $\gamma$ and $M$ measured separately.