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 = √6.63x10⁻¹¹Nm²kg⁻²x6x10²⁴kg6.4x10⁶m+9x10⁵m
v = √39.78x10¹³Nm²kg⁻¹6.4x10⁶m+0.99x10⁶m
v = √39.78x10¹³Nm²kg⁻¹7.3x10⁶m
v = √54.4x10⁶m²s⁻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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