What should be the orbiting speed to launch a satellite in a circular orbit 900 km above the surface of the Earth (Me = 6 ｘ 10²⁴ kg) and its radius 6400 km. (Ans: 7.4 km s⁻¹)



Data Given:


Height of the circular orbit = h = 900 km = 9 ｘ10⁵m

Mass of the earth = Me = 6 ｘ 10²⁴ kg

Radius of earth = Re = 6400 km = 6.4 ｘ10⁶m

Gravitational Constant = G = 6.63 ｘ 10⁻¹¹ N m² kg⁻² 


To Find:


Orbiting speed of satellite  = v = ?


Solution:


By using the formula for finding the speed for launching a satellite is 

v = `\sqrt frac {GM_e}{R + h}`


By putting values

v = `\sqrt frac {6.63 ｘ 10⁻¹¹ N m² kg⁻² ｘ 6 ｘ 10²⁴ kg}{6.4 ｘ10⁶ m + 9 ｘ10⁵m}`

v = `\sqrt frac {39.78 ｘ 10¹³ N m² kg⁻¹}{6.4 ｘ10⁶ m + 0.99 ｘ10⁶m}`

v = `\sqrt frac {39.78 ｘ 10¹³ N m² kg⁻¹}{7.3 ｘ10⁶ m }`

v = `\sqrt {54.4 ｘ 10⁶ m² s⁻²}`

or

v = 7.4 ｘ 10³ m s⁻¹ -----------Ans.

The orbital seed to launch a satellite at height 900 km above the earth's surface will be 7.4 ｘ 10³ m s⁻¹




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