In recent years, a number of nearby stars have been found to possess planets. Suppose, the...

Question:

In recent years, a number of nearby stars have been found to possess planets. Suppose, the orbital radius of such a planet is found to be 3.3*10{eq}^{11} {/eq} m, with a period of 1080 days. Find the mass of the star

Time Period

In the field of rotational physics, the time period is specified as the full time utilized for one complete cycle of rotation or revolution about a center of reference point. It is displayed in the conventional units of a second.

Answer and Explanation:

Given data

  • The orbital radius of the planet is: {eq}R = 3.3 \times {10^{11}}\;{\rm{m}} {/eq}.
  • The time period for the orbit is: {eq}T = 1080\;{\rm{days}} {/eq}.


The expression for the time period of the orbit according to the third law provided by kepler is,

{eq}{T^2} = \dfrac{{4{\pi ^2}{R^3}}}{{Gm}} {/eq}

Here, the gravitational constant is {eq}G {/eq} and the mass of the star is {eq}m {/eq}.

The value of the gravitational constant is {eq}6.67 \times {10^{ - 11}}\;{\rm{N}}{\rm{.}}{{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}} {/eq}.


Substitute the values.

{eq}\begin{align*} {\left( {1080\;{\rm{days}}\left( {\dfrac{{24\;{\rm{h}}}}{{1\;{\rm{day}}}}} \right)\left( {\dfrac{{3600\;{\rm{s}}}}{{1\;{\rm{h}}}}} \right)} \right)^2} &= \dfrac{{4{\pi ^2}{{\left( {3.3 \times {{10}^{11}}\;{\rm{m}}} \right)}^3}}}{{\left( {6.67 \times {{10}^{ - 11}}\;{\rm{N}}{\rm{.}}{{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^2}} \right)m}}\\ {\left( {1080\;{\rm{days}}\left( {\dfrac{{24\;{\rm{h}}}}{{1\;{\rm{day}}}}} \right)\left( {\dfrac{{3600\;{\rm{s}}}}{{1\;{\rm{h}}}}} \right)} \right)^2} &= \dfrac{{4{\pi ^2}{{\left( {3.3 \times {{10}^{11}}\;{\rm{m}}} \right)}^3}}}{{\left( {6.67 \times {{10}^{ - 11}}\;{\rm{N}}{\rm{.}}{{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^2}\left( {\dfrac{{1\;{\rm{kg}}{\rm{.m/}}{{\rm{s}}^{\rm{2}}}}}{{1\;{\rm{N}}}}} \right)} \right)m}}\\ m &= 2.44 \times {10^{30}}\;{\rm{kg}} \end{align*} {/eq}


Thus, the mass of the star is {eq}2.44 \times {10^{30}}\;{\rm{kg}} {/eq}.


Learn more about this topic:

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Kepler's Three Laws of Planetary Motion

from Basics of Astronomy

Chapter 22 / Lesson 12
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