# 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}.

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