Question

A satellite circles the earth in an orbit whose radius is six times the earth's radius. The earth's mass is 5.98 1024 kg, and its radius is 6.38 106 m. What is the period of the satellite

Answer 1

Answer:

The period of the satellite is 12.05 s.

Explanation:

The period of the satellite can be determined by;

T = $\sqrt{\frac{4\pi ^{2} r^{2} }{GM} }$

where: T is the period, r is the radius of the orbit, G is the gravitation constant and M is the mass of the Earth.

Given that: r = 6 x (6.38 x $10^{6}$) = 38.28 x $10^{6}$ m, M = 5.98 x  $10^{24}$ kg, G = 6.673 x $10^{-11}$ N$m^{2}$/ $kg^{2}$.

Therefore,

T = $\sqrt{\frac{4*(\frac{22}{7} )^{2} (38.28*10^{6}) ^{2} }{6.673*10^{-11} *5.98*10^{24} } }$

  = $\sqrt{\frac{5.79*10^{16} }{3.99*10^{14} } }$

  = $\sqrt{145.113}$

  = 12.0463

T = 12.05 s

The period of the satellite is 12.05 s.