how much time a satellite in an Orbit at HYD 35780 km above earth surface would take if the mass of the earth would have been 4 times its original mass
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Use the formula,
Here, T is time period, R is orbit radius
G is gravitational constant and is mass of earth.
Given,
= 4 × original mass of earth,
= 4 × 5.92 × 10²⁴ kg
Orbit radius , R = Earth's radius + height of satellite from Earth's surface
= 6400 km + 35780 km
= 42180 km = 4.2180 × 10⁷ m
G = 6.67 × 10⁻¹¹ Nm²/Kg²
Now, T = 2π√{(4.2180 × 10⁷)³/(6.67 × 10⁻¹¹) × (4 × 5.92 × 10²⁴)}
= 43309.8 sec
So, T = 43309.8/3600 ≈ 12 hrs
Hence, time to complete revolution by satellite is 12 hrs
hope it will help you
Here, T is time period, R is orbit radius
G is gravitational constant and is mass of earth.
Given,
= 4 × original mass of earth,
= 4 × 5.92 × 10²⁴ kg
Orbit radius , R = Earth's radius + height of satellite from Earth's surface
= 6400 km + 35780 km
= 42180 km = 4.2180 × 10⁷ m
G = 6.67 × 10⁻¹¹ Nm²/Kg²
Now, T = 2π√{(4.2180 × 10⁷)³/(6.67 × 10⁻¹¹) × (4 × 5.92 × 10²⁴)}
= 43309.8 sec
So, T = 43309.8/3600 ≈ 12 hrs
Hence, time to complete revolution by satellite is 12 hrs
hope it will help you
ankushpal93:
thanks bro
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