Question

How much time a satellite in an orbit at height 35780 km above the 's surface would take, if the mass of the would have been 4 times its original mass

Answer 1

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

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Answer 2

Use the formula, \bf{T=2\pi\sqrt{\frac{R^3}{GM_E}}}T=2πGME​R3​​ 
Here, T is time period, R is orbit radius 
G is gravitational constant and M_EME​ is mass of earth.

Given, 
M_EME​ = 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