Question

A satellite is revolving in a circular equatorial orbit of radius \[R=2\times {{10}^{4}}km\] from east to west. Calculate the interval after which it will appear at the same equatorial town. Given that the radius of the earth \[=6400km\]and g(acceleration due to gravity)\[~=10m~{{s}^{-2}}\].

Answer 1

Hint:In order to calculate the interval after which it will appear at the same equatorial town, we need to use some formula from the binding energy and satellites.

Complete step by step answer:
Let be the actual angular velocity of the satellite from east to west and be the angular speed of the earth(west to east).
Then,
\[{\omega _{relative}} = \omega - ( - {\omega _e}) = \omega + {\omega _e}\]
\[ \Rightarrow \omega = {\omega _{rel}} - {\omega _e}\]
By the dynamics of circular motion,
\[\dfrac{GMm}{{{R}^{2}}}=m{{\omega }^{2}}R\]
\[~{{\omega }^{2}}=\dfrac{g{{R}^{2}}}{{{R}^{3}}}\] (since \[GM = gR_e^3\])
\[~\Rightarrow \omega =\sqrt{\dfrac{gR_{e}^{3}}{{{R}^{3}}}}\]
\[~\Rightarrow {{\omega }_{rel}}=\sqrt{\dfrac{gR_{e}^{3}}{{{R}^{3}}}}+{{\omega }_{e}}\]
\[~\Rightarrow {{\omega }_{rel}}=\sqrt{\dfrac{10\times {{6.4}^{2}}\times {{10}^{12}}}{{{2}^{3}}\times {{10}^{21}}}}+7.27\times {{10}^{-5}}\]
$\left( \text{since }~{{\omega }_{e}}=\dfrac{2\pi }{86400}=7.27\times {{10}^{-5}} \right)$
\[\Rightarrow {{\omega }_{rel}}=22.6\times {{10}^{-5}}+7.27\times {{10}^{-5}}=30\times {{10}^{-5}}rad{{s}^{-1}}\]
So,
Interval \[=\dfrac{2\pi }{{{\omega }_{rel}}}=\dfrac{2\pi }{30\times {{10}^{-5}}}=2.09\times {{10}^{4}}s=5h48min\]
Thus, the correct answer to this question is \[5h48min\].

Additional Information:
The binding energy of a satellite can be defined as the minimum amount of energy required to be supplied to it in order to free the satellite from the gravitational influence of the planet (i.e. in order to take the satellite from the orbit to a point at infinity). When the satellite is orbiting around the earth it possesses two types of mechanical energy. The kinetic energy due to its orbital motion and potential energy due to its position in the gravitational field of the earth.The total energy of a satellite is just the sum of its gravitational potential and kinetic energies. Assuming that mechanical energy is conserved (which it is for the most part in the vacuum of space), the sum of the kinetic and potential energies of the satellite would remain constant.

Note:While solving this question, we should be aware of the different types of formula used here. The formulae are basically from binding energy and satellites and must use the values provided in the question.