3.10 An earth-orbiting satellite has a period of 14 h and a perigee radius of 10,000 km. At time t = 10 h after perigee passage, determine (a) The radial position. (b) The speed. (c) The radial component of the velocity. {Ans : (a) 42 356 km: (b) 2.303 km/s: (c) –1.271 km/s}

3.10 An earth-orbiting satellite has a period of 14 h and a perigee radius of 10,000 km. At time t = 10 h after perigee passage, determine (a) The radial position. (b) The speed. (c) The radial component of the velocity. {Ans : (a) 42 356 km: (b) 2.303 km/s: (c) –1.271 km/s}

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Question

3.10 An earth-orbiting satellite has a period of 14 h and a perigee radius of 10,000 km. At time
t = 10 h after perigee passage, determine
(a) The radial position.
(b) The speed.
(c) The radial component of the velocity.
{Ans.: (a) 42,356 km; (b) 2.303 km/s; (c) -1.271 km/s}

Expert Solution

Concept and Principle

The radial position of a satellite in an elliptical orbit is given by,

$r = \frac{a (1 - e^{2})}{1 + e c o s θ}$

Here a is the length of the semi-major axis, e is the eccentricity and θ path angle.

Speed of the satellite will be given by,

$\frac{v^{2}}{2} - \frac{μ}{r} = \frac{- μ}{2 a}$

Here r is the radial position, μ is the standard gravitational parameter and a is the length of the semi-major axis.

The radial component of the speed will be given by,

$v_{r} = \frac{μ}{h} e s i n θ$

Here h is the specific relative angular momentum of the orbit, μ is the standard gravitational parameter, e is the eccentricity, and θ path angle.

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