<p>The period of oscillation of a near-Earth satellite (neglecting atmospheric e

Question

<p>The period of oscillation of a near-Earth satellite (neglecting atmospheric effects) is 84.3 min. What is the period of a near-Moon satellite? (<em>R_E</em>&#160;= 6.37 x 10⁶ m; <em>R_M&#160;</em>= 1.74 x 106 m; <em>M_E&#160;= </em>5.98 x 10²⁴ kg; <em>M_M</em>&#160;= 7.36 x 10²²&#160;kg.)</p>
<p>a. 6.03 x 10^-3&#160;min.</p>
<p>b. 0.713 min.</p>
<p>c. 84.3 min.</p>
<p><strong>d. 108 min.</strong></p>
<p>e. 140 min.</p>
<p><strong>(d)</strong>&#160;is the answer. Please show and explain why and how it is the answer.</p>

Explanation / Answer

The period of the near-Earth satellite is TE = 84.3 min
The radius of the Earth is RE = 6.37*106 m The radius of the moon is RM = 1.74*106 m The mass of the Earth is ME = 5.98*1024 kg The mass of the moon is MM = 7.36*1022 kg Let TM be the period of the near-Moon satellite The period of the satellite in the circular orbit is given by T2 = 42R3/GM G= 6.67*10-11 Nm2/kg2 is the universal gravitational constant T2M/R3 = 42/G T2M/R3 = constant T2M/R3 = 42/G T2M/R3 = constant Therefore TE2ME/RE3 = TM2MM/RM3 (84.3 min)2(5.98*1024 kg)/(6.37*106 m)3 = TM2(7.36*1022 kg)/(1.74*106 m)3 TM = 108 min TM = 108 min

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