A polar satellite is launched at 850 km above Earth. Find its orbital speed. Ans. (7431 ms⁻¹)


Given:  

Height of the Satellite  = h =  850 Km = = 0.85 ｘ 10³ｘ 10³ m = 0.85 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 10⁻¹¹ N 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 10⁻¹¹ N m² kg⁻² ｘ 6 ｘ 10²⁴ kg}{6.4 ｘ 10⁶ m + 0.85 ｘ 10⁶ m }`

Vₒ = `\sqrt \frac {6.67 ｘ 6 ｘ 10⁻¹¹⁺²⁴ }{7.25 ｘ 10⁶}` m s⁻¹

Vₒ = `\sqrt \frac {6.67 ｘ 6 ｘ 10⁻¹¹⁺²⁴⁻⁶ }{7.25}`m s⁻¹

Vₒ = `\sqrt { 5.52 ｘ 10⁷ }`m s⁻¹

Vₒ = `\sqrt { 552 ｘ 10⁶ }`m s⁻¹

Vₒ = 0.742967ｘ 10³ m s⁻¹

or

Vₒ = 7429.67 m s⁻¹

or

Vₒ = 7430 m s⁻¹

Thus the orbital speed polar satellite will be 7430 ms⁻¹ when launched at 850 km above Earth. 

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