Question

The time period of a satellite in a circular orbit of radius $R$ is $T$. The period of another satellite in a circular orbit of radius $9R$ is

• A. $3T$
• B. $9T$
• C. $27T$
• D. $12T$

Answer 1

The correct option is C

$27T$

Step 1: Given Data:

The circular orbital radius of a planet$=R$

The time period of a satellite$=T$

The circular orbital radius of another planet$=9R$

Let the time period of another satellite $=T\text{'}$

Step 2: Formula Used:

The velocity of a satellite orbiting in a circular path,

$v=\frac{2\pi R}{T}\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

Where $T=$Time period, $R=$Radius of planet

The velocity can also be given as

$v=\sqrt{\frac{GM}{R}}\dots \dots \dots \dots \dots \dots \dots \dots \left(2\right)$

Where $G=$gravitational constant, $M=$the mass of the planet

Equating equation (1) and equation (2),

$$\begin{gathered} \sqrt{\frac{GM}{R}}=\frac{2\pi R}{T} \\ T=\frac{2\pi R}{\sqrt{\frac{GM}{R}}} \end{gathered}$$

On squaring,

$T^2=4{\pi }^2\frac{R^3}{GM}\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(3\right)$

Step 3: Calculation of time period:

The time period for a satellite having a circular orbital radius $R$ from the equation (3)

$T^2\alpha R^3\dots \dots \dots \dots \dots \dots \left(4\right)$

The time period for another satellite having a circular orbital radius $9R$ from the equation (3)

$T{\text{'}}^2\propto {\left(9R\right)}^3\dots \dots \dots \dots \dots \left(5\right)$

Dividing equation (4) by equation (5)

$\begin{array}{rcl}{\left(\frac{T}{T\text{'}}\right)}^2& =& {\left(\frac{R}{R\text{'}}\right)}^3\\ \left(\frac{T}{T\text{'}}\right)& =& {\left(\frac{R}{9R}\right)}^{\frac{3}{2}}\\ \left(\frac{T}{T\text{'}}\right)& =& {\left(\frac{1}{9}\right)}^{\frac{3}{2}}\\ \left(\frac{T\text{'}}{T}\right)& =& {\left(9\right)}^{\frac{3}{2}}\\ T\text{'}& =& 27\times T\\ T\text{'}& =& 27T\end{array}$

Thus, the period of another satellite in a circular orbit of radius $9R$ is $27T$.

Hence option C is the correct answer.