Question

The time period of a satellite of the earth is $5h$. If the separation between the earth and the satellite is increased to $4$ times the previous value, the new time period will become

• A. $40h$
• B. $20h$
• C. $10h$
• D. $80h$

Answer 1

The correct option is A

$40h$

Step 1: Given Data:

Time period$\left(T\right)=5h$

The distance between earth and satellite is increased by $4times$

Let the original separation distance be$R$

New separation distance $R\text{'}=4R$

Step 2: Formula Used:

From Kepler's third law, the square of the orbital periods of planets is directly proportional to the cube of the semi-major axis of their orbital that is

$T^2\propto R^3\dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

Where, $T=$Time period, $R=$a distance of semi- major axis

Step 3: Calculation of the new time period:

Let $T\text{'}$ is the new time period and $R\text{'}$ be the new separation distance then from equation (1),

$$\begin{gathered} T^{\text{'}2}\alpha {\left(R\text{'}\right)}^3 \\ {T}^{\text{'}2}\alpha {\left(4R\right)}^3\dots \dots \dots \dots \dots \dots \dots \left(2\right) \end{gathered}$$

On dividing equation (1) by equation (2)

So,

$\begin{array}{rcl}{\left(\frac{T}{T\text{'}}\right)}^2& =& {\left(\frac{R}{R\text{'}}\right)}^3\\ \left(\frac{T}{T\text{'}}\right)& =& {\left(\frac{R}{R\text{'}}\right)}^{\frac{3}{2}}\\ \left(\frac{T}{T\text{'}}\right)& =& {\left(\frac{1}{4}\right)}^{\frac{3}{2}}\\ \left(\frac{T\text{'}}{T}\right)& =& {\left(4\right)}^{\frac{3}{2}}\\ T\text{'}& =& 8\times T\\ T\text{'}& =& 8T\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(3\right)\end{array}$

The time period of a satellite of the earth is $T=5h$

Substituting the value of $T=5h$ in equation (3)

$\begin{array}{rcl}T\text{'}& =& 8T\\ & =& 8\times 5\\ & =& 40h\end{array}$

Thus, the new time period will be $40h$.

Hence option A is the correct answer..