Correct Answer : 7.08 km/s
Explaination : Polar coordinates are represented as (r, θ)Therefore, r = 1200 kmθ(t) = 0.0059t radvT = r(dθ(t)/dt)= 1200*0.0059= 7.08 km/s
Correct Answer : 3.47 km/s
Explaination : Standard gravitational parameter (μ) = 398,600 km3/s2Specific angular momentum (h) = 57,112 km2/sEccentricity for a parabolic orbit (e) = 1True anomaly (θ) = 0.52 radRadial velocity can then be given by:vr = (μ/h)(esinθ)= (398,600/57,112)(1*sin(0.52 rad))= 3.47 km/s
Correct Answer : 6.98 km/s
Explaination : Given,Standard gravitational parameter (μ) = 398,600 km3/s2Specific angular momentum (h) = 57,112 km2/sEccentricity for a circular orbit (e) = 0Tangential velocity can then be given by:vT = (μ/h)(1 + ecosθ)= (398,600/57,112)(1 + 0)= 6.98 km/s
Correct Answer : vis-viva
Explanation : Vis-viva equation is the other name for energy equation for orbits. It is Latin for “living force”. The equation describes the conservation of energy of each orbit. Kinetic energy and Gravitational energy compensate for each other to conserve the overall energy.
Correct Answer : -30.3 km3/s2
Explaination : Given,Standard gravitational parameter (μ) = 398,600 km3/s2Radius of earth (RE) = 6378 kmSemi-major axis (a) = (525 + 6250 + 6378)/2= 6576.5 kmSpecific energy (ε) = -μ/(2a)= -398,600/(2*6576.5)= -30.3 km3/s2
Correct Answer : True
Explanation : True. Kinematic properties of the moving body relative to the stationary body can be calculated by assuming the bodies as point masses concentrated at their respective center of masses. Despite the bodies being rigid bodies, knowledge of point mass dynamics is all that is required. Rigid body dynamics is applied to calculate more complex data which is beyond the purpose of a two-body problem.
Correct Answer : 22.62 x 106 km2
Explaination : Semi-major axis (a) = 9,000 kmSemi-minor axis (b) = 4,000 kmArea swept (at t = T/5) (A)=?Due to similarity,A/(T/3) = πab/T || Area of ellipse = πabA = πab/5= π*9,000*4,000/5= 22.62 x 106 km2
Correct Answer : 1.633 km/s
Explaination : Standard gravitational parameter (μ) = 4903 km3/s2Radius of moon (RM) = 1738 kmAltitude from surface of moon (z) = 100 kmVelocity of satellite (v) = (μ/(RM + z))1/2= (4903/(1738+100))1/2= 1.633 km/s
Correct Answer : 2931.8 s
Explaination : Perigee distance (rp) = 6378 + 400 = 6778 kmApogee distance (ra) = 6378 + 900 = 7278 kmSemi-major axis (a) = (rp + ra)/2= (6778 + 7278)/2= 7028 kmTime period of the orbit (T) = 2πa3/2/μ1/2= 2π*70283/2/398,6001/2= 5863.53 sTime taken to travel from perigee to apogee = T/2 = 5863.53/2 = 2931.8 s
Correct Answer : 72,876.32 km2/s
Explaination : Velocity of satellite (v) = 10.7 km/sFlight path angle (γ) =15°Satellite distance (r) = 7048 kmTangential velocity (vt) = vcos(γ) = 10.7*cos(15°) = 10.34 km/sSpecific angular momentum (h) = rvt = 7048*10.34 = 72,876.32 km2/s
Correct Answer : h and e
Explaination : h and e (specific angular momentum and eccentricity) are the orbital constants among all the letters in the orbit equation. μ (standard gravitational parameter) is a constant that is independent of the orbit and depends on the mass of the larger body. Whereas, true anomaly (θ) is not constant and is dependent on the spacecraft/satellite position relative to perigee line.
Correct Answer : Curvilinear
Explanation : Curvilinear motion because all orbits/trajectories have a radius of curvature. Or in other words, all orbits are conic section and therefore curved. Rectilinear motion could be useful if a satellite moved in a straight line. Rotary motion also rotates, but it rotates along its own axis. Though most celestial bodies have rotary motion, it also does not apply to point masses (which is how bodies are assumed in two body problems). Brownian motion is a random motion of particles and has application in quantum world.
Correct Answer : Tends to zero
Explanation : The satellite while orbiting an attracting body has negative potential energy and its maximum potential energy tends to zero when the radius tends to infinity.
Correct Answer : Eccentricity
Explanation : In a system with two point masses m1, m2 having mutual gravitational force acting on each other, the center of mass lies somewhere along the line joining the two masses. The center of mass of this system is non accelerating having a constant velocity.
Correct Answer : 0
Explanation : When the equation of motion is derived for a to-body problem, the bodies experience gravitational force only. All the other forces and torques are assumed to be zero.
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.
Correct Answer : Vernal equinox
Explanation : Vernal equinox lies on the celestial equator which is the outward projection of Earth’s equator on the celestial sphere. This is the origin for measuring the longitude which is done is degrees from east to west.
Correct Answer : Nadir
Explanation : Nadir is the point on the imaginary celestial sphere which is directly below the observer. It is directly opposite to the Zenith. Usually when the astronauts carry out spacewalks, they refer to nadir which is the downward view of the satellite in the orbit.
Correct Answer : Zenith
Explanation : Zenith is the point on the imaginary celestial sphere which is directly overhead the observer. It is the highest point on the sphere and this is the point when the shadows appear to be at its smallest when Sun is at the zenith.
Correct Answer : Declination
Explanation : On the celestial sphere, latitude are nothing but the declination since the positive latitudes run from equator to the north pole, and the negative values run from equator to the south pole. These are measure in degrees.
Correct Answer : Ephemeris
Explanation : The coordinates of celestial body which includes stars, comets, asteroid, planets etc. are known as ephemeris when its coordinates are expressed as a function of time. The ephemeris depends on epoch or vernal equinox at that time.
Correct Answer : East to West
Explanation : Due to the Earth’s rotation which is in the direction of East to West, the celestial sphere also rotates East to West as it is just a fictious sphere with Earth’s centre as its own.
Correct Answer : Retrograde
Explanation : Inclination is the measure of angle between the orbital plane of the satellite and the equatorial plane of the orbit it revolves around. If the inclination angle is 0 degrees, it’s called a prograde orbit. When the inclination angle is 90 degrees, it is a polar orbit and when angle is 180 degrees it is known as retrograde orbit. In this particular orbit, the satellite moves in the opposite direction to the rotation of the planet around which it is orbiting.
Correct Answer : 6
Explanation : In total, there are six orbital elements defined. Out of these 2 parameters: eccentricity and angular momentum are used to define the orbit. True anomaly is used to find the location of the object in the orbit. Three additional Euler angles are used to describe the orientation of the orbit. These are- inclination, argument of perigee and right ascension of the ascending node.
Correct Answer : Circular
Explanation : For the circular orbits, the eccentricity is zero and there is no defined perigee. Thus, true anomaly and argument of perigee are not defined for the circular orbits.
Correct Answer : Satellite’s location in the orbit
Explanation : True anomaly is the angle measured in the direction of satellite motion, from perigee to the satellite’s location. The value varies from 0° to 360° and it is undetermined when the eccentricity is zero (i.e. for circular orbits).
Correct Answer : Orbit’s orientation in the orbital plane
Explanation : Argument of perigee is the angle measured in the direction of satellite motion, from the ascending node to the perigee. The value varies from 0° to 360° and it is undetermined when i = 0° or 180° or when the eccentricity is zero (i.e. for circular orbits).
Correct Answer : 52°
Explanation : The orbital inclination of ISS is 52° and it orbits around the Earth from west to east. 98° is the orbital inclination of Mapping and 0° is for Geostationary satellites.
Correct Answer : 1 ≤ e
Explaination : Eccentricity is the ratio of half the foci separation to the semi-major axis. For closed orbits, the eccentricity varies from 0 ≤ e ≤ 1 and foe the open orbits 1 ≤ e. Parabolic and Hyperbolic orbits are considered as open orbits as the body flies off to infinity eventually and does not return to the same angular position.
Correct Answer : Undefined
Explanation : Argument of perigee is the angle measured in the direction of the satellite’s motion from the ascending node to the perigee. When inclination is zero, there are no nodes thus its value is undefined.
Correct Answer : Time used as reference for mentioning orbital elements
Explanation : Epoch is used to refer to a particular time at the moment when the astronomical quantities such as the orbital elements are defined.
Correct Answer : Semi-major axis
Explanation : Out of the six Keplerian orbital elements, semi-major axis (a) helps in determining the size of the conic orbit. It is defined as the half of the long axis of the ellipse. The orbital period and energy depend on the orbit size.
Explanation : Eccentricity helps in determining the shape of the conic orbit. It is defined as the ratio of half the foci separation to the semi-major axis. For closed orbit, eccentricity ranges from 0 to 1 and for open orbits it is greater than 1.
Correct Answer : u0 = ω + v0
Explaination : Argument of periapsis ω is the angle formed between the ascending node and periapsis point measured in the direction of the object’s motion. Whereas, argument of latitude u0 at epoch is the angle between the ascending node and radius vector to the satellite at some time t and true anomaly at epoch v0 is the angle between periapsis and the position of satellite at a particular time t called epoch.They are related by:u0 = ω + v0
Correct Answer : False
Explanation : Keplerian orbits have a fundamental assumption that the only influenece on the orbit is gravitational force of the attracting body and it ha s aspherical potential field. Due to this the orbital elements do not change as a function of time.Although in real life, potentional field is not precisely sphrecial and there may be few changes in the orbital elements with time which is often neglected.
Correct Answer : 4
Explanation : The angle between the earth’s equitorial plane and orbital plane is known as inclination. Based on the different inclination angles, there are four types of orbits present.* Equitorial orbit : Angle of inclination is 0 or 180 deg.* Polar orbit : Angle of inclination is 90 deg.* Prograde orbit : Angle of inclincation is between 0 and 90 deg.* Retrograde orbit : Angle of inclination is between 90 and 180 deg.
Correct Answer : 0.44
Explanation : Semi-major axis (a) = 6580 kmSemi-minor axis (b) = 5910 kmEccentricity (e) = (1 – (b/a)2)1/2= (1-(5910/6580)2)1/2= 0.44
Correct Answer : 0.691
Explaination : Specific energy (ε) = -10.29 km2/s2Specific angular momentum (h) = 10,112 km2/sStandard gravitational parameter (μ) = 398,600 km3/s2Using the e-h-ε relatione = (1+(2εh)/μ)1/2= (1+(2*-10.29*10,112)/398,600)1/2= 0.691
Correct Answer : Parabolic
Explanation : Parabolic orbits are formed at the exact escape velocity of a planet. Elliptical and circular orbits are formed when the satellite’s velocity is less than the escape velocity. Whereas, at velocities higher than the escape velocity, hyperbolic orbits are formed.
Correct Answer : Elliptical
Explanation : Elliptical orbits have an eccentricity greater than zero and less than one. Most planets and orbits orbiting around the sun are elliptical in shape.
Correct Answer : 0.2098
Explanation : The eccentricity of an orbit can be determined as follows:e = (Apoapsis distance – Periapsis distance) / (Apoapsis distance + Periapsis distance)e = (10,378 – 6778) / (10,378 + 6778)e = 0.2098
Correct Answer : 5250 km
Explanation : Semi-major axis of an ellipse is half of the major axis. Major axis can be represented as a sum of perigee and apogee distances. Therefore,Semi-major axis = (2500 + 8000) / 2 = 5,250 km
Correct Answer : Heliocentric-Ecliptic System
Explanation : Heliocentric-Ecliptic system is a system which has sun as its center or origin. Fundamental plane of operation of this coordinate system is ecliptic plane of our solar system. Most of the planets along with the sun lie on this plane.
Correct Answer : Argument of periapsis
Explanation : Argument of periapsis is defined as the angle between ascending node and periapsis. The angle between vernal equinox and ascending node is longitude of the ascending node. The angle between earth’s equatorial plane unit vector and satellite’s angular momentum vector is inclination. True anomaly is the angle between perigee and spacecraft position relative to center of the earth.
Correct Answer : 0.822
Explanation : Apogee radius (ra) = 77,000 kmPerigee radius (rp) = 7,500 kmEccentricity (e) = (ra – rp) / (ra + rp)= (77,000 – 7,500) / (77,000 + 7,500)= 0.822
Correct Answer : True-anomaly-averaged radius
Explanation : True-anomaly-averaged radius is the average distance between satellite and earth for one complete orbit. To find the average distance between the earth and a satellite, the total true anomaly angle of 2π is divided into n equal segments with Δθ, and a sum of satellite radiuses is calculated from every nth interval. The sum is then divided by n. To make the solution easier, the sum is written as an integral and solved.
Correct Answer : 64.16°
Correct Answer : 52,910 km
Explanation : Apogee radius (ra) = 100,000 kmEccentricity (e) = 0.89Semi-major axis (a) = ra / (1 + e)= 100,000 / (1 + 0.89)= 52,910 km
Correct Answer : 0.648 km/s
Explaination : Gravitational Parameter (μ) = 398,600 km3/s2Semi-major axis (a) = 50,000 kmEccentricity (e) = 0.9From orbit equation for ellipse,Specific angular momentum (h) = (μa(1-e2))1/2= (398,600*50,000*(1-0.92))1/2= 61,536 km2/sApogee radius (ra) = a(1 + e)= 50,000 (1 + 0.9)= 95,000 kmApogee velocity (va) = h/ra= 61,536/95,000= 0.6477 km/s
Correct Answer : 40,450 km
Explanation : Apogee radius (ra) = 73,000 kmPerigee radius (rp) = 7,900 kmSemi-major axis (a) = (ra + rp) / 2= (73,000 + 7,900) / 2= 40,450 km
Correct Answer : 20.518 hours
Explaination : Apogee radius (ra) = 69,000 kmPerigee radius (rp) = 7,100 kmGravitational Parameter (μ) = 398,600 k3/s2Semi-major axis (a) = (ra + rp)/2= (69,000 + 7,100)/2= 38,050 kmTime period (T) = (2π/μ1/2)a3/2= (2π/398,6001/2)*38,0503/2= 73,865.77 s= 20.518 hours
Correct Answer : -4.927 km2/s2
Explaination : Apogee radius (ra) = 73,000 kmPerigee radius (rp) = 7,900 kmGravitational Parameter (μ) = 398,600 km3/s2Semi-major axis (a) = (ra + rp) / 2= (73,000 + 7,900) / 2= 40,450 kmSpecific Energy (ε) = -μ/(2a) = -398,600*(2*40,450)= -4.927 km2/s2
Correct Answer : 12.307 km/s
Explanation : Gravitational Parameter (μ) = 398,600 km3/s2Semi-major axis (a) = 50,000 kmEccentricity (e) = 0.9From orbit equation for ellipse,Specific angular momentum (h) = (μa(1 – e2))1/2= (398,600*50,000*(1-0.92))1/2= 61,536 km2/sPerigee radius (rp) = a(1-e)= 50,000 (1-0.9)= 5,000 kmPerigee velocity (vp) = h/rp= 61,536/5,000= 12.307 km/s
Correct Answer : 0.89
Explanation : Perigee radius (rp) = 6,000 kmSemi-major axis (a) = 52,910 kmEccentricity (e) = 1 – rp / a= 1 – 6,000 / 52,910= 0.89
Correct Answer : equal to
Explanation : Tangential velocity is exactly equal to the spacecraft’s velocity at apogee. True anomaly at the apogee is 180° and as the radial velocity is dependent on the true anomaly as a multiple of sin θ, it should be zero. Therefore, at this instant the tangential velocity and spacecraft’s velocity are the same.
Correct Answer : 0°
Explanation : Flight path angle (γ) = atan (esinθ/(1 + ecosθ))At perigee, true anomaly (θ) = 0°, this implies sinθ = 0γ = atan (0) = 0°, therefore 0° is the correct answer.
Explanation : Tangential velocity is equal to the satellite velocity in a circular orbit since both the velocities travel along the same direction throughout the orbit. Therefore, there is no radial component. There is acceleration along the radial direction though which keeps the satellite in orbit without falling back to earth.
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.
Explanation : Since, the tangential velocity equals to the satellite velocity at any point across a circular orbit, the flight path angle is always zero. Regardless, of what the true anomaly or height of the satellite, in a circular orbit the flight path angle is zero.
Correct Answer : 24.41 km/s
Explaination : Initial velocity (vi) = 27 km/sGravitational Parameter (μ) = 42,828 km3/s2Radius of Satellite (r) = 3,389.5 + 3,000= 6,389.5 kmOrbit velocity (v) = (μ/r)1/2= (42,828/6,389.5)1/2= 2.589 km/sDelta-v = vi – v= 27 – 2.589= 24.41 km/s
Correct Answer : 9.599 km/s
Explaination : Specific angular momentum (h) = 47,862.73 km2/sGravitational parameter (μ) = 324,859 km3/s2Escape velocity (vesc) = 21/2(μ/h)= 21/2*(324,859/47,862.73)= 9.599 km/s
Correct Answer : inversely proportional to square root of
Explanation : Only for circular orbits is the velocity of a satellite inversely proportional to square root of the radius of orbit. In all the other cases, the relationship is not as straightforward due to the tangential and radial velocity components. It cannot be inversely proportional either, though v = h/r. Since h, is dependent on r. Only equation that best describes the relationship is v = (μ/r)1/2 and μ is a constant.
Correct Answer : 10.38 km/s
Explaination : Specific angular momentum (h) = 87,899 km2/sEccentricity (e) = 2.5Gravitational Parameter (μ) = 398,600 km3/s2Hyperbolic excess velocity (v∞) = (μ/h)(e2 – 1)1/2= (398,600/87,989)(2.52 – 1)1/2= 10.38 km/s
Correct Answer : 10,984 km
Explaination : Radius of satellite (r) = 36,613 kmStandard gravitational parameter (μ) = 398,600 km3/s2Eccentricity (e) = 1.2We know for hyperbolic trajectories,(Satellite speed)2 = (Hyperbolic Excess Speed)2 + (Escape Velocity)2v2 = v∞2 + vesc24v∞2 = v∞2 + 2μ/r3v∞2 = 2μ/rr = (2μ)/(3v∞2)We know,v∞2 = (μ/h)2(e2 – 1)Substituting back into equation for r, we get,r = (2/3)(h2/μ)(1/e2 – 1)Substituting orbit equation for perigee radius, we get,r = (2/3)(rp/(e – 1))Therefore, rearranging the equation to get rprp = 3r(e-1)/2= 3*36,613*(1.2-1)/2= 10,984 km
Correct Answer : 9.74 km/s
Explaination : Perigee speed (vp) = 14.3 km/sSatellite radius (r) = 6378 + 900= 7278 kmEscape velocity (vesc) = (2μ/r)1/2= (2*398,600/7278)1/2= 10.466 km/sHyperbolic excess speed (v∞) = (vp2 – vesc2)1/2= (14.32 – 10.4662)1/2= 9.744 km/s
Correct Answer : 145,560 km2/s
Explaination : Specific angular momentum = Perigee radius x Perigee velocity= 7278 x 20= 145,560 km2/s
Correct Answer : 9.562 km/s
Explanation : Perigee radius (rp) = 6778 kmEccentricity (e) = 2.3True anomaly (θ) = 100°From the orbit equation,Specific angular momentum (h) = [μrp(1 + e)]1/2= [398,600*6778*(1 + 2.3)]1/2= 94,422.69 km2/sRadial velocity (vr) = (μ/h)esin(θ)= (398,600/94,422.69)*2.3*sin(100°)= 9.562 km/s
Correct Answer : Characteristic
Explanation : Characteristic energy or specific excess energy of a hyperbolic orbit is the energy required to escape. Specific energy on its own is a measure of a spacecraft’s total energy divided by its mass. Characteristic energy times mass is excess energy of a spacecraft, but the question asks for energy divided by mass. While, dynamic energy is not a term used in orbital mechanics.
Correct Answer : 56.44°
Explanation : Eccentricity (e) = 1.2Turn angle = 2*asin(1/e)= 2*asin(1/1.2)= 56.44°
Correct Answer : 20,000 km
Explanation : Perigee radius (rp) = 9,800 kmPerigee radius (ra) = -49,800 kmWe know, for hyperbolas,rp = a(e – 1)ra = -a(e + 1)where, a and e are semi-major axis and eccentricity, respectivelyDividing equations mentioned above and substituting the given values, we get,9800/49,800 = (e – 1)/(e + 1)0.1968(e + 1) = e – 10.8032e = 1.1968e = 1.49Substituting back into perigee radius equation, we get,Semi-major axis (a) = rp/(e – 1)= 9800/(1.49 – 1)= 20,000 km
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.
Correct Answer : 830.66 km
Explaination : Eccentricity (e) = 1.3Specific Energy (ε) = 199.3 km2/s2Gravitational parameter (μ) = 398,600 km3/s2We know,Semi-major axis (a) = μ/(2ε)= 398,600/(2*199.3)= 1000 kmAiming radius (Δ) = a*(e2 – 1)1/2= 1,000*(1.32 – 1)1/2= 830.66 km
