What is the eccentricity of a GEO?

1.0

1.01

0.0

0.87

Correct Answer : 0.0

Explanation : A geostationary orbit (GEO) is an orbit over a planet in which a satellite has a fixed position over the planet. This makes the orbital shape of GEO exactly as the shape of the planet. Since, all planets are spherical in shape their cross-sections are circular. Hence, the GEO is a circular orbit with eccentricity of 0. It might be slightly elliptical but for the sake of simplicity it can be assumed to circular.

Recently Updated in Spaceflight Mechanics Questions

What is the semi-minor axis of a hyperbola with specific energy of 199.3 km²/s² and eccentricity of 1.3? Gravitational parameter is 398,600 km³/s².

704.31 km

830.66 km

1000 km

1200 km

Correct Answer : Option (B) : 830.66 km

Explaination : Eccentricity (e) = 1.3
Specific Energy (ε) = 199.3 km²/s²
Gravitational parameter (μ) = 398,600 km³/s²
We know,
Semi-major axis (a) = μ/(2ε)
= 398,600/(2*199.3)
= 1000 km
Aiming radius (Δ) = a*(e² – 1)^1/2
= 1,000*(1.3² – 1)^1/2
= 830.66 km

Hyperbolic excess velocity is reached when the distance tends to infinity.

True

False

Can Not Say

None of the above

Correct Answer : Option (A) : True

Explanation : True. Hyperbolic excess velocity is the relative velocity between helio-centered ellipse and earth’s orbit. It is achieved when a spacecraft accelerates to a speed more than escape velocity. Maintaining a proper hyperbolic excess is important to perform inter-planetary transfers.

For a hyperbolic trajectory, what is its semi-major axis if its perigee radius is 9,800 km and apogee radius is -49,800 km?

18,000 km

19,000 km

20,000 km

21,000 km

Correct Answer : Option (C) : 20,000 km

Explanation : Perigee radius (r_p) = 9,800 km
Perigee radius (r_a) = -49,800 km
We know, for hyperbolas,
r_p = a(e – 1)
r_a = -a(e + 1)
where, a and e are semi-major axis and eccentricity, respectively
Dividing equations mentioned above and substituting the given values, we get,
9800/49,800 = (e – 1)/(e + 1)
0.1968(e + 1) = e – 1
0.8032e = 1.1968
e = 1.49
Substituting back into perigee radius equation, we get,
Semi-major axis (a) = r_p/(e – 1)
= 9800/(1.49 – 1)
= 20,000 km