Respuesta :
From Kepler's 3rd law, time period of satellite orbiting around planet in a circular orbit is given by,
T² = 4π²r³/GM
M= mass of the Earth = 6×10²⁴ Kg.
∴ T² = 4×3.14²×(2.56×10⁷)³/6.67×10⁻¹¹×6×10²⁴
T² = 1653339719
∴ T = 40661.28 seconds.
T² = 4π²r³/GM
M= mass of the Earth = 6×10²⁴ Kg.
∴ T² = 4×3.14²×(2.56×10⁷)³/6.67×10⁻¹¹×6×10²⁴
T² = 1653339719
∴ T = 40661.28 seconds.
Its period will be T = 40661.28 sec
What is orbital velocity ?
The orbital velocity of a satellite is defined as the minimum velocity required to place or maintain a satellite in a given orbit.
since , v0 : v (orbital velocity) = [tex]\sqrt{(GM)/R}[/tex]
T = 2πR / v0
T = 2πR / [tex]\sqrt{(GM)/R}[/tex]
[tex]T^{2}[/tex] = 4[tex]\pi ^{2}[/tex][tex]R^{3}[/tex] / GM
M= mass of the Earth = 6×10²⁴ Kg.
T² = 4×3.14²×(2.56×10⁷)³/6.67×10⁻¹¹×6×10²⁴
T = 40661.28 seconds.
Its period will be T = 40661.28 sec
learn more about orbital velocity
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