If a hypothetical planet in our Solar System had a sidereal period of 11.1 Earth years and a...


If a hypothetical planet in our Solar System had a sidereal period of {eq}11.1 {/eq} Earth years and a circular orbit, how far from the Sun in Astronomical Units would it be? (Show your work.)

Planetary Orbits

Planets follow elliptical orbits with the Sun in one of the ellipse's focus. A simple approximation to the planetary motion involves considering circular orbits with the Sun in the circumference's center. Under this approximation, the gravitational force of the Sun is directed to the center of the orbit, perpendicular to the trajectory. The only acceleration of the planet is centripetal rotating with a constant velocity whose direction changes continuously. The squared orbital period is proportional to the cube of the orbital radius.

Answer and Explanation:

The equation of motion for the planet in its circular orbit about the Sun is,

{eq}F_{G}=m_pa_c {/eq},


{eq}F_G=G\dfrac{m_pM_S}{d^2} {/eq},

is the gravitational force. The involved terms are the universal gravitational constant {eq}G=6.67 \times 10^11 \;\rm N\cdot m^2/kg^2 {/eq}, {eq}m_p {/eq} is the mass of the planet, {eq}M_S=1.99 \times 10^{30} \;\rm kg {/eq} the Sun's mass, and {eq}d {/eq} the radius of the circular orbit. In the right-hand side of equation (1) appear the product of the planet's mass and the centripetal acceleration,

{eq}a_c=\dfrac{v^2}{d} {/eq},

which is the quotient between the orbital speed and the radius.

Substituting these results,

{eq}G\dfrac{m_pM_S}{d^2}=m\dfrac{v^2}{d} {/eq}.

Now we use the expression for the sidereal period in the circular orbit,

{eq}T=\dfrac{2 \pi d}{v} \implies v=\dfrac{2 \pi d}{T} {/eq},

as the ratio between the length of the orbit and the velocity.


{eq}G\dfrac{m_pM_S}{d^2}=m_p\dfrac{4 \pi^2 d}{T^2} {/eq}.

Solving for the orbital radius,

{eq}d=\left(G \dfrac{M_S T^2}{4 \pi ^2} \right)^{1/3} {/eq}.

Introducing the numeric values,

{eq}d=\left(6.67 \times 10^{-11} \;\rm N\cdot m^2/kg^2 \cdot \dfrac{1.99 \times 10^{30} \;\rm kg \cdot (11.1 \;\rm years \cdot 365.25 \;\rm days/year \cdot 24 \;\rm h/day \cdot 3600 \;\rm s/h)^2}{4 \pi ^2} \right)^{1/3}=\boxed{7.44\ times 10^11 \;\rm m} {/eq}.

To convert this result to astronomical units we divide the radius by the distance from the Earth to the Sun,

{eq}d_{AU}=\dfrac{d}{r_{ES}}=\dfrac{7.44 \times 10^11 \;\rm m}{1.50 \times 10^{11} \;\rm m}=\boxed{4.96 \;\rm AU} {/eq}.

The distance from the planet to the Sun is {eq}4.96 \;\rm AU {/eq}.

Learn more about this topic:

Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12

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