The radius of the earth's orbit is 1.50 \times 10^{11} m and that of pluto is 5.93\times 10^{12}...


The radius of the earth's orbit is {eq}1.50 \times 10^{11} {/eq} m and that of pluto is {eq}5.93\times 10^{12} {/eq} m. The star that this planet orbit is identical to our sun. What is the orbital period of this planet?

Orbital period:

In the astronomical analysis, when the heavenly object revolves around the orbit of another heavenly body, then the time taken by the object in the duration of revolution is called the orbital period of the heavenly object.

Answer and Explanation:

Given data:

  • The radius of earth's orbit is, {eq}r = 1.50 \times {10^{11}}\;{\rm{m}} {/eq}.
  • The radius of orbit of Pluto is, {eq}R = 5.93 \times {10^{12}}\;{\rm{m}} {/eq}.

Since, the star at pluto is identical to that of sun.

Then expression for the orbital time period of pluto is given as,

{eq}T = \sqrt {\dfrac{{4{\pi ^2} \times {R^3}}}{{G{M_S}}}} {/eq}

Here, G is the gravitational constant, its value is {eq}6.67 \times {10^{ - 11}}\;{{\rm{m}}^{\rm{3}}}{\rm{/kg}} \cdot {{\rm{s}}^{\rm{2}}} {/eq} and {eq}{M_S} {/eq} is the mass of sun, its value is {eq}1.989 \times {10^{30}}\;{\rm{kg}} {/eq}.

Solve by substituting the value in the expression as,

{eq}\begin{align*} T &= \sqrt {\dfrac{{4{\pi ^2} \times {R^3}}}{{G{M_S}}}} \\ T &= \sqrt {\dfrac{{4 \times {\pi ^2} \times {{\left( {5.93 \times {{10}^{12}}} \right)}^3}}}{{6.67 \times {{10}^{ - 11}} \times 1.989 \times {{10}^{30}}}}} \\ T &= 7.87 \times {10^9}\;{\rm{seconds}} \end{align*} {/eq}

Therefore, the orbital period of the planet is {eq}7.87 \times {10^9}\;{\rm{seconds}} {/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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