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 = GMeR+h


By putting values

v = 6.6310¹¹Nm²kg²610²kg6.410m+910m

v = 39.7810¹³Nm²kg¹6.410m+0.9910m

v = 39.7810¹³Nm²kg¹7.310m

v = 54.410m²s²

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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