A communication satellite is launched at 42000 km above Earth. Find its orbital speed. Ans. (2876 m s⁻¹)


Given:  

Height of the Communication Satellite  = h =  42,000 km = = 42 ｘ 10³ｘ 10³ m = 42 ｘ 10⁶ m 

∴ Mass of the Earth = Mₑ = 6 ｘ 10²⁴ kg
∴ Radius of the Earth = Rₑ = 6.4 ｘ 10⁶ m
∴ Gravitational Constant = G = 6.67 ｘ 10⁻¹¹N m² kg⁻²

To Find:
Orbital speed of Satellite = Vₒ = ? meters

Solution:   

We have the formula for the orbital speed of the Satellite

Vₒ = `\sqrt \frac {G Mₑ}{Rₑ + h}`

by putting the values we have

Vₒ = `\sqrt \frac {6.67 ｘ 10⁻¹¹N m² kg⁻² ｘ 6 ｘ 10²⁴ kg}{6.4 ｘ 10⁶ m + 42 ｘ 10⁶ m }`

Vₒ = `\sqrt \frac {6.67 ｘ 6 ｘ 10⁻¹¹⁺²⁴ }{48.4 ｘ 10⁶}` m s⁻¹

Vₒ = `\sqrt \frac {6.67 ｘ 6 ｘ 10⁻¹¹⁺²⁴⁻⁶ }{48.4}`m s⁻¹

Vₒ = `\sqrt { 0.827 ｘ 10⁷ }`m s⁻¹

Vₒ = `\sqrt { 8.27 ｘ 10⁶ }`m s⁻¹

Vₒ = 2.8775ｘ 10³ m s⁻¹

or

Vₒ = 2877.5 m s⁻¹

or

Vₒ = 2,876 m s⁻¹

Thus the orbital speed Communications satellite will be 7430 m s⁻¹ when launched at 42,000 km above Earth. 

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