# Astronautics Questions and Answers – Types of Orbits

This set of Astronautics Multiple Choice Questions & Answers (MCQs) focuses on “Types of Orbits”.

1. The semi-major axis of an elliptical orbit may be considered as _______________________
a) the sum of periapsis and apoapsis
b) the difference between apoapsis and periapsis
c) the average of periapsis and apoapsis
d) the product of periapsis and apoapsis

Explanation: The semi-major axis ‘a’ is related to the periapsis ‘P’ and apoapsis ‘A’ through the following relationship:
$$a = \frac{P + A}{2}$$
This formula is identical to the equation for the average of two quantities.

2. Given a pre-defined non-zero inclination, which of the following quantities defines the orientation of an orbit around Earth?
a) Longitude of ascending node
b) Eccentricity
c) True anomaly
d) Semi-major axis

Explanation: The longitude of ascending node (or the ‘right ascension of ascending node’) is the angle between the prime meridian (also called the ‘First Point of Aries’ in the context of satellite navigation) and the ascending node (i.e., the point at which the orbit intersects the equatorial plane on the side of the orbit where the satellite moves North) measured anti-clockwise (when looking down on Earth from above the North Pole). For a given inclination, an orbit can have a variety of orientations depending on its right ascension.

3. The argument of perigee is measured in the plane of ____________
a) earth’s equator
b) the orbit
c) the ecliptic
d) earth’s rotational axis

Explanation: The argument of perigee is the angle between the ascending node and the perigee point measured in the plane of the orbit.

4. For an inclined circular orbit, which of the following orbital elements becomes invalid?
a) Argument of periapsis
b) Longitude of ascending node
c) True anomaly
d) Inclination

Explanation: For a circular orbit, an object is always a fixed distance away from the primary body, meaning that a periapsis and apoapsis do not exist in such cases.

5. An equatorial orbit cannot have _______________
a) an argument of periapsis
b) a periapsis
c) an apoapsis
d) a node

Explanation: A node is defined as a point at which an orbit crosses the equatorial plane of the primary body (Earth, say). For any orbit with a non-zero inclination, there exist two nodes – the ascending node and the descending node. If the orbit lies in the plane of the equator, however, the concept of a ‘node’ does not make sense due to the fact that the orbit coincides with the equatorial plane.

6. By fixing a non-zero inclination and defining the value of the longitude of ascending node for an orbit, the orientation of the orbit can still vary.
a) True
b) False

Explanation: Keeping the inclination and longitude of ascending node fixed at constant finite values, the orientation of the orbit in space can still be varied by changing the argument of periapsis, which is done by rotating the orbit in its own plane.

7. What is the approximate height of a circular orbit around Earth of time period 24 hours? (Assume the mass of Earth as 5.972 × 1024 kg and its radius as 6378 km).
a) 42,240 kilometers
b) 35,900 kilometers
c) 10,000 kilometers
d) 25,050 kilometers

Explanation: Given,
T = 24 hours = 24 x 60 x 60 seconds = 86400 seconds
M = 5.972 × 1024 kg
Substitute the values in the following equation:
r3 = $$\frac{GM}{4π^2}T^2$$
Finally, we get a value of r = 42,240 kilometers. Remember that this is the orbital radius, which is always measured from the center of Earth. To obtain the height of the orbit (measured from the surface of our planet), we need to subtract the radius of Earth from the calculated orbital radius. So, the height of the orbit above the ground, or the orbital height, is equal to (42,240 kilometers – 6378 kilometers), which is approximately 35,900 kilometers.

8. The difference between the perigee and apogee of an Earth orbit of semi-major axis 7378 kilometers and eccentricity 0.02 is __________________
a) 500 km
b) 900.88 km
c) 50.67 km
d) 295.12 km

Explanation: Given, a = 7378 km; e = 0.02. We have perigee radius, Rp=a(1-e), and apogee radius, Ra=a(1+e). Using these formulae, we get Rp as 7230.44 km and Ra as 7525.56 km. The difference between the two is Ra-RP, i.e., 295.12 km.

9. The lunar reconnaissance orbiter is a space probe presently in orbit around the Moon which was launched back in 2009. The orbiter is just 20 kilometers above the Moon’s surface when at its closest and 165 kilometers away when farthest. The semi-major axis and eccentricity of the orbit of the probe are __________________
a) a = 2729.5 km and e = 0.396
b) a = 2829.5 km and e = 0.0896
c) a = 1829.5 km and e = 0.0396
d) a = 1029.5 km and e = 0.496

Explanation: Given, RP = 1737 + 20 km = 1757 km
Ra = 1737 + 165 km = 1902 km
Semi-major axis is given by
a=(Ra+RP)/2

Substituting the given values,we get a = 1829.5 km

Eccentricity can be written as
e=(Ra-RP)/(Ra+RP)=(Ra-RP)/2a
We get e = 0.0396.

10. The true anomaly is the angle between the position of an object in an elliptical orbit and the _______________________
a) ascending node
b) periapsis
c) descending node
d) apoapsis

Explanation: Firstly, the location of an object in orbit is identified by the line connecting the center of the primary body to the object. Its position is then specified by the ‘true anomaly’, which is the angle between this line and the line of periapsis. For a circular orbit, the position of the object is given in terms of the ‘argument of latitude’, which is the angle between the ascending node and the object.

11. The primary direction, or axis, defined as the zero-point for longitude (0˚ longitude) with reference to which all longitudinal measurements are taken (like right ascension of ascending node, longitude of periapsis, etc.) points in the direction of ______________
a) the Moon
b) the Vernal Equinox
c) the Sun
d) Aries

Explanation: Our planet’s equatorial plane is offset by an angle of 23.5 degrees with respect to the plane in which it revolves around the sun, the so-called ‘ecliptic’. The ecliptic intersects the plane of Earth’s equator at two points on the surface. In the course of a year, the sun appears to pass through these points twice. To understand how this happens, it is important to know about the ’sub-solar point’ – the position on Earth’s surface at which the sun is directly overhead. At times when the Earth is tilted towards the sun (northern summer, southern winter), this sub-solar point lies in the Northern Hemisphere, and when tilted away from the sun (northern winter, southern summer), this point moves to the Southern Hemisphere. During its transition between the two hemispheres, the sub-solar point must cross the equator (which may be thought of as a divider that separates both hemispheres). This happens two times in a year: March and September. When it occurs, the phenomena is called an ‘equinox’. The September equinox is also called the ‘vernal equinox’, and indicates the onset of winter in the northern hemisphere. The direction of the Sun against the background of stars during the vernal equinox is taken as the reference for all longitudinal measurements.

12. The reference frame used as a basis for tracking and positioning space-based objects in Earth orbit is the __________________ frame
a) ECR
b) ECEF
c) ECI
d) Ecliptic

Explanation: As with any real object, the position and velocity of a spacecraft must be determined with respect to a pre-established coordinate system, or reference frame. Unlike the Earth Centered Earth Fixed (ECEF) or Earth Centered Rotational (ECR) frame that rotates with our planet, the Earth Centered Inertial (ECI) frame is not ‘attached’ to Earth and hence does not rotate with it. The ECEF (or ECR) frame is convenient for pinpointing terrestrial (ground-based) objects, while the ECI frame makes it easier to track orbiting bodies in near-Earth space.
NOTE: ECR is just another title for ECEF, and both fundamentally mean the exact same frame.

Sanfoundry Global Education & Learning Series – Astronautics.

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