A satellite orbits the earth with perigee radius of 7,500 km and apogee radius of 77,000 km, what is its eccentricity?

0.251

0.412

0.822

0.99

Correct Answer : 0.822

Explanation : Apogee radius (r_a) = 77,000 km
Perigee radius (r_p) = 7,500 km
Eccentricity (e) = (r_a – r_p) / (r_a + r_p)
= (77,000 – 7,500) / (77,000 + 7,500)
= 0.822

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