Question

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

Answer 1

Explanation:

Use the formula, \bf{T=2\pi\sqrt{\frac{R^3}{GM_E}}}T=2π

GM

E

R

3

Here, T is time period, R is orbit radius

G is gravitational constant and M_EM

E

is mass of earth.

Given,

M_EM

E

= 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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