Question

What is the period of revolution of earth satellites? Ignore the height of the satellite above the surface of earth.
Given: $(1)$The value of gravitational acceleration$g=10m{{s}^{-2}}$.
$(2)$ Radius of the earth${{R}_{E}}=6400km$. Take$\pi =3.14$
$\begin{aligned} & A.85\operatorname{minutes} \\\ & B.156\operatorname{minutes} \\\ & C.83.73minutes \\\ & D.90minutes \\\ \end{aligned}$

Answer 1

Hint : We can just solve this question from the basic velocity equation. We need to find total distance travelled by the satellite which will be the circumference of the orbit. Time taken will be the time period that we need to find and velocity could be found from the root of the product of acceleration due to gravity and radius of earth.
Formula used: $T=2\pi \sqrt{\dfrac{{{R}_{E}}}{g}}$

Complete solution: Firstly let us look how the expression for time period $T=2\pi \sqrt{\dfrac{{{R}_{E}}}{g}}$ is obtained.
We are starting from the basic velocity equation which is$v=\dfrac{s}{t}$which in this case will be changed as$v=\dfrac{2\pi {{R}_{E}}}{T}$where, $2\pi {{R}_{E}}$ is the circumference of the orbit of the satellite and$T$is the time period.
Now, let us rearrange the equation as, $T=\dfrac{2\pi {{R}_{E}}}{v}$.
Now, the velocity$v$is given as $v=\sqrt{\dfrac{GM}{{{R}_{E}}}}$in which$G$ is specific gravity of earth and$M$is mass of earth. This equation could be modified using the expression given for acceleration due to gravity, which is $g=\dfrac{GM}{{{R}_{E}}^{2}}$. From this equation we can get $g{{R}_{E}}=\dfrac{GM}{{{R}_{E}}}$ which we can substitute in the equation for velocity of the satellite and obtain the equation as,
$v=\sqrt{g{{R}_{E}}}$
Now, we will substitute this equation in the equation for time period which will lead us to the equation,
$T=\dfrac{2\pi {{R}_{E}}}{\sqrt{g{{R}_{E}}}}$
By cancelling out the common terms we will arrive at $T=2\pi \sqrt{\dfrac{{{R}_{E}}}{g}}$.
Now, let us find the time period of satellite by using the parameters given which are,
${{R}_{E}}=6400km$ and $g=10m{{s}^{-2}}$.
$\Rightarrow T=2\pi \sqrt{\dfrac{{{R}_{E}}}{g}}=2\times 3.14\times \sqrt{\dfrac{6400}{10}}$
$\begin{aligned} & T=2\times 3.14\times 0.8\times {{10}^{3}} \\\ & T=5.024\times {{10}^{3}}=5024s \\\ & T=\dfrac{5024}{60}=83.73\min \\\ \end{aligned}$
So, the time period is found to be $T=83.73\min$. That means option c is correct.

Note: We can also solve this equation by directly using the formula given to find time period of a satellite at a height of$h$from earth surface which is given as $T=2\pi \sqrt{\dfrac{{{\left( {{R}_{E}}+h \right)}^{3}}}{g{{R}_{E}}^{2}}}$. But while using this we must be aware to remember the condition $h\ll {{R}_{E}}$ which will change the equation as we used to solve. i.e., $T=2\pi \sqrt{\dfrac{{{R}_{E}}^{3}}{g{{R}_{E}}^{2}}}=2\pi \sqrt{\dfrac{{{R}_{E}}}{g}}$.