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

Answer 1

The correct option is C

$40h$

Step 1: Given data

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

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

Let initially the separation between the earth and the satellite is $r$

New separation $4r$

Step 2: Formula used

From Kepler's third law, the squares of the orbital periods are directly proportional to the cubes of the semi-major axes of their orbits that is,

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

Where, $T=$Time period of satellite, $r=$semi-major axis

Step 3: Calculation of the new time period

Let $T\text{'}$ is the new time period

$T{\text{'}}^2={\left(4r\right)}^3\dots \dots \dots \dots \dots \left(2\right)$

On dividing equation (1) by(2)

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

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

Substituting the value 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 C is the correct answer.