Question

Time period of a satellite revolving around a planet in an orbit of radius R is T. Periodic time of a satellite moving in an orbit of radius 9R will be-
A. 27T
B. 81T
C. 729T
D. 3T

Answer 1

Hint: Using the third law of Kepler’s, we will solve this answer further. The mathematical equation derived from the law is ${T^2} \propto {R^3}$. In 1687 Isaac Newton demonstrated that relationships such as Kepler's would extend to a good approximation in the Solar System, as a consequence of his own laws of motion and universal gravitation law. Refer to the solution below for further explanation.
Formula used: ${\left[ {\dfrac{{{T_2}}}{{{T_1}}}} \right]^2} = {\left[ {\dfrac{{{R_2}}}{{{R_1}}}} \right]^3}$

Complete Step-by-Step solution:
${T_1}$ = Time period of the satellite revolving around the planet is given as T.
${R_1}$ = The radius of the orbit is given as R.
${R_2}$ = The new radius of the orbit of which we have to find the time period is given in the question as 9R.
${T_2}$ = The new time period which we have to find.
As we know, according to the Kepler’s third law, ${T^2} \propto {R^3}$
$ \Rightarrow {\left[ {\dfrac{{{T_2}}}{{{T_1}}}} \right]^2} = {\left[ {\dfrac{{{R_2}}}{{{R_1}}}} \right]^3}$
Putting the values, we will get-
$
   \Rightarrow {\left[ {\dfrac{{{T_2}}}{{{T_1}}}} \right]^2} = {\left[ {\dfrac{{{R_2}}}{{{R_1}}}} \right]^3} \\
    \\
   \Rightarrow {\left[ {\dfrac{{{T_2}}}{T}} \right]^2} = {\left[ {\dfrac{{9R}}{R}} \right]^3} \\
    \\
   \Rightarrow {\left[ {{T_2}} \right]^2} = {\left[ {\dfrac{{9R}}{R}} \right]^3} \times {T^2} \\
    \\
   \Rightarrow {\left[ {{T_2}} \right]^2} = \dfrac{{729{R^3}}}{{{R^3}}} \times {T^2} \\
    \\
   \Rightarrow {T_2} = \sqrt {729 \times {T^2}} \\
    \\
    \\
   \Rightarrow {T_2} = 27T \\
$
Hence, option A is the correct option.

Note: Kepler's planetary motion laws are three mathematical laws that explain the movement of planets around the Sun, written between 1609 and 1619 by Johannes Kepler. Kepler’s third law states that- The squares ratio of any two planet's cycles is equal to the cube ratio of their average distances from the earth.