Question

Two satellites A and B of equal mass move in the equatorial plane of the earth, close to the earth's surface. Satellite A moves in the same direction as that of the rotation of the earth while Satellite B moves in opposite direction.Calculate the ratio of kinetic energy of B to that of A in the reference frame fixed to the earth. ( Take g = 9.8 m s⁻² and radius of earth = 6.37 * 10^6 km).

Answer 1

$\Large{\text{\bf{Hey there!}}$

$\textsf{Let \omega_A and \omega_B be the absolute angular speeds of A and B.}\\\textsf{Since, they are in same orbit, their time period must be same.}\\i.e,\boxed{\omega_A=\omega_B}$

$\textsf{Considering,the dynamics of circular motion in cases,}\\\\m\omega_A^2 R =\dfrac{GMm}{R^2}\\\\\boxed{\omega_A = \sqrt{\dfrac{g}{R}}}~~~~~~~~~(\because GM = gR^2)\\\\\textsf {Similarly},\\\\\boxed{\omega_B = \sqrt{\dfrac{g}{R}}}$

$\sf {So,}\\\omega_A = \omega_B = \sqrt{\dfrac{9.8}{6.37 \times 10^6}} =124 \times 10^{-5} \ rad \ s^{-1}$

$\sf Now,\\\omega_{AE} = \omega_A - \omega_E = 124 \times 10^{-5} - 7.3 \times 10^{-5}\ rad \ s^{-1}\\\boxed{\bold{\omega_{AE}=116.7 \times 10^{-5} \ rad \ s^{-1}}}\\\\And,\\\omega_{BE} (\textsf{velocity of B relative to E}) \\=\omega_B - (-\omega_E)\\=\omega_B + \omega_E\\\boxed{\bold{\omega_{BE}= 131.1 \times 10^{-5} \ rad \ s^{-1}}}$

$\sf So,\\\\\dfrac{K_B}{K_A} = \dfrac{\dfrac{1}{2} m\omega ^2_{BE}r^2}{\dfrac{1}{2}m\omega^2_{AE}r^2} = \dfrac{131.3^2}{116.7^2} = 1.27\\\\Hence,\\\\\boxed{\boxed{\bold{\dfrac{K_B}{K_A}= 1.27}}}$

$\Large{\textsf{Hope It Helps You!}}$

Answer 2

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