Question

Given that ${T^2} = k{R^3}$, express the constant k of the given relation in days kilometres. Given, $k = {10^{ - 13}}{\rm{ }}{{\rm{s}}^2}{{\rm{m}}^{ - 3}}$. The moon is at a distance of $3.84 \times {10^5}{\rm{ km}}$ from the earth. Obtain its time period of revolution in days.

Answer 1

The time period of revolution of moon is to be evaluated using the given expression ${T^2} = k{R^3}$. We will rewrite the given values of k and R in such a way that the final expression for revolution will come out in days.

Complete step by step answer:
It is given that the time period of revolution of moon at a distance of $3.84 \times {10^5}{\rm{ km}}$ from the earth is given by the relation ${T^2} = k{R^3}$, where R is the distance between moon and the earth.
We are required to calculate the time period of revolution in days so rewriting the value of k.
We know that one day has twenty four hours, one hour has sixty minutes and one minute has sixty seconds. Therefore, seconds in one day are given as:

1{\rm{ day }} = \left( {24 \times 60 \times 60} \right){\rm{ s}}\\\ \Rightarrow 1{\rm{ s }} = \dfrac{1}{{\left( {24 \times 60 \times 60} \right)}}{\rm{ day}} \end{array}$$ Substitute $$\dfrac{1}{{\left( {24 \times 60 \times 60} \right)}}{\rm{ day}}$$ for $$1{\rm{ s}}$$ in the given value of k. $$\begin{array}{c} k = {10^{ - 13}}{\left[ {\dfrac{{{\rm{1 day}}}}{{\left( {24 \times 60 \times 60} \right){\rm{ s}}}}{\rm{ }}} \right]^2}{\left[ {\dfrac{{1000{\rm{ m}}}}{{1{\rm{ km}}}}} \right]^3}{{\rm{s}}^2}{{\rm{m}}^{ - 3}}\\\ = 1.339 \times {10^{ - 14}}{\rm{ da}}{{\rm{y}}^2}{\rm{k}}{{\rm{m}}^{ - 3}} \end{array}$$ Time period of revolution of moon. $${T^2} = k{R^3}$$……(1) Substitute $$1.339 \times {10^{ - 14}}{\rm{ da}}{{\rm{y}}^2}{\rm{k}}{{\rm{m}}^{ - 3}}$$ for k in equation (1). $${T^2} = \left[ {1.339 \times {{10}^{ - 14}}{\rm{ da}}{{\rm{y}}^2}{\rm{k}}{{\rm{m}}^{ - 3}}} \right]{\left[ {3.84 \times {{10}^5}{\rm{ km}}} \right]^3}$$ Taking the square root of the above expression. $$\begin{array}{c} T = {\left[ {1.339 \times {{10}^{ - 14}}{\rm{ da}}{{\rm{y}}^2}{\rm{k}}{{\rm{m}}^{ - 3}}} \right]^{\dfrac{1}{2}}}{\left[ {3.84 \times {{10}^5}{\rm{ km}}} \right]^{\dfrac{3}{2}}}\\\ = 27.53{\rm{ days}} \end{array}$$ **Note:** Alternate method: We can also obtain the time period of revolution by substituting the values of k and R in the given expression of time period after necessary rearrangement. The rearrangement includes expressing the dimension of k in the form of day which is the unit of time and metre which is the unit of distance. And expressing the unit of distance of the moon from the earth in the form of metre so that the final value of the time period of revolution will be obtained in a number of days.