Question

The time period of a geo-stationary satellite is 24 h, at a height 6R_E (R_E is the radius of earth) from surface of earth. The time period of another satellite whose height is 2.5 R_Efrom surface will be [NEET (Odisha) 2019]

Moderate

Solution

From Kepler's third law, the time period of revolution of satellite around earth is
$T^{2} \propto r^{3} or T \propto r^{3 / 2}$ …(i)
where, r is the radius of satellite's orbit.
Here, $r_{1} = 6 R_{E} + R_{E}, T_{1} = 24 h$
$r_{2} = 2.5 R_{E} + R_{E}, T_{2} = ?$
So, from Eq. (i), we get
$\frac{T_{1}}{T_{2}} = {(\frac{r_{1}}{r_{2}})}^{3 / 2}$
$\frac{24}{T_{2}} = {(\frac{6 R_{E} + R_{E}}{2.5 R_{E} + R_{E}})}^{3 / 2} = {(\frac{7}{3.5})}^{3 / 2}$
$\Rightarrow T_{2} = \frac{24}{(2)^{3 / 2}} = \frac{24}{2 \sqrt{2}} = \frac{12}{\sqrt{2}} = 6 \sqrt{2} h$

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