Which of these is not an assumption considered while deriving the equation of motion?

The point masses never touch

The bodies ae considered to be point masses

The center of mass of two bodies does not lie in the center

No external force acts on the bodies apart from gravitational force

Correct Answer : The center of mass of two bodies does not lie in the center

Explanation : While deriving the equation of motion, there are three main assumptions made as follows :

• Bodies of mass m1 and m2 are spherical point masses.
• The point masses never touch.
• No external force other than gravitational force acts on the two bodies.

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