Answer:
The period of the satellite is 12.05 s.
Explanation:
The period of the satellite can be determined by;
T = [tex]\sqrt{\frac{4\pi ^{2} r^{2} }{GM} }[/tex]
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 [tex]10^{6}[/tex]) = 38.28 x [tex]10^{6}[/tex] m, M = 5.98 x [tex]10^{24}[/tex] kg, G = 6.673 x [tex]10^{-11}[/tex] N[tex]m^{2}[/tex]/ [tex]kg^{2}[/tex].
Therefore,
T = [tex]\sqrt{\frac{4*(\frac{22}{7} )^{2} (38.28*10^{6}) ^{2} }{6.673*10^{-11} *5.98*10^{24} } }[/tex]
= [tex]\sqrt{\frac{5.79*10^{16} }{3.99*10^{14} } }[/tex]
= [tex]\sqrt{145.113}[/tex]
= 12.0463
T = 12.05 s
The period of the satellite is 12.05 s.
Period of the satellite is 13,410 seconds.
Given that;
r = 6RM = 5.98 × 10²⁴ kg
G = 6.67 × 10⁻¹¹ m³/kg
R = 6.38 × 10⁶ m
Computation:
[tex]V = \sqrt{\frac{GM}{r + R} } \\\\V = \sqrt{\frac{6.67 \times 10^{-11} \times5.98 \times10^{24}}{7 \times 6.38 \times10^6} } \\\\V = 2,988 m/s \\\\time\ period = \frac{2\times \pi \times R }{V} \\\\Time\ period = \frac{2\times \3.14 \times 6.38 \times10^6 }{2988}\\ Time\ period = 13,410 \ Seconds[/tex]
A span of time during which something happens or is expected to happen. hours. a period of endless duration downtime. a period of time during which something isn't working.
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