Question

If the radius of moon’s orbit is 60 times the earth’s radius and period of revolution of moon is 27.3 days, then the radius of the orbit of moon is
(A) \[3.86 \times {10^8}\,m\]
(B) \[3.86 \times {10^6}\,m\]
(C) \[3.86 \times {10^4}\,m\]
(D) 3.86 km

Answer 1

Hint: Use the formula for period of revolution for any heavenly body. Express the radius of earth in terms of radius of moon’s orbit as given in the question. After substituting it in the formula, you’ll get the desired answer.

Formula used:
The period of revolution of any heavenly body around its host planet or star is given as,
\[T = 2\pi \sqrt {\dfrac{{{r^3}}}{{g{R^2}}}} \]
Here, T is the period in second, r is the radius of orbit of that body in meter, g is the acceleration due to gravity and R is the radius of earth in meter.

Complete step by step answer:
We know the period of revolution of moon around earth is given as,
\[T = 2\pi \sqrt {\dfrac{{{r^3}}}{{g{R^2}}}} \]…… (1)
Here, T is the period in second, r is the radius of orbit of the moon in meters, g is the acceleration due to gravity and R is the radius of earth in meters.
We have given that the radius of the moon's orbit is 60 times the radius of earth. Therefore,
\[r = 60R\]
\[ \Rightarrow R = \dfrac{r}{{60}}\]
We substitute \[R = \dfrac{r}{{60}}\] in equation (1).
\[T = 2\pi \sqrt {\dfrac{{{r^3}}}{{g{{\left( {\dfrac{r}{{60}}} \right)}^2}}}} \]
\[ \Rightarrow T = 2\pi \sqrt {\dfrac{{3600r}}{g}} \]
Squaring the above equation, we get,
\[{T^2} = 4{\pi ^2}\left( {\dfrac{{3600r}}{g}} \right)\]
\[ \Rightarrow r = \dfrac{{{T^2}g}}{{3600 \times 4{\pi ^2}}}\]
We substitute 27.3 days for T and \[9.8\,m/{s^2}\] for g in the above equation.
\[r = \dfrac{{{{\left( {27.3 \times 24 \times 60 \times 60} \right)}^2}\left( {9.8} \right)}}{{3600 \times 4{{\left( {3.14} \right)}^2}}}\]
\[ \therefore r = 3.86 \times {10^8}\,m\]

Therefore, the radius of the moon's orbit is \[3.86 \times {10^8}\,m\].So, the correct answer is option (A).

Note:To use the formula (1), the period should be in seconds and distance should be in meters. There are other formulae derived from this formula if the period is given in years. In formula (1), the radius of the orbit r is actually the semi-major axis. For nearly circular orbits, we generally take the average radius as the semi-major axis.