Correct Answer : Fixed Wireless
Explanation : Fixed Wireless is a type of communication that occurs between two fixed devices or between two fixed locations. An example of fixed wireless is communication between a computer and a Wi-Fi router. All other options are well implemented through satellites.
Correct Answer : 5.85 km/s
Explaination : Given,Distance between Earth and satellite (r) = 5250 + 6378= 11,628 kmStandard gravitational parameter of Earth (μ) = 398,600 km3/s2Since, gravitational attraction between Earth and satellite = centrifugal force of the satellite, we can derive,Tangential velocity of satellite (V) = (μ/r)1/2= (398,600/11,628)1/2= 5.85 km/s
Correct Answer : 42,647 km3/s2
Explaination : Given,Force of attraction between Earth and Mars (F) = 6.39 x 1023 NDistance between Earth and Mars (r) = 225 x 106 kmStandard Gravitational Parameter of Earth (μE) = 398,600 km3/s2Universal gravitational constant (G) = 6.67408 × 10-11 m3/(kg.s2)From Newton’s law of gravitation,Mass of Mars (m) = (Fr2)/μE= (6.39*1023*(225*106)2)/(398,600*109)= 6.39 x 1023 kgStandard Gravitational Parameter of Earth (μM) = G.m= 6.67408*10-11*6.39*1023= 42,647 km3/s2
Correct Answer : 1.771 x 1026 N
Explaination : Given,Standard gravitational parameter of earth (μ) = 398,600 km3/s2Mass of moon (m) = 7.348 x 1022 kgDistance between earth and moon (r) = 384,400 kmForce of attraction (F) = (μ.m) / (r)2= (398,600 x 7.348 x 1022) / (384,400)2= 1.982 x 1017 N
Correct Answer : 9.78 m/s2
Explaination : Given,Acceleration due to gravity at MSL (g0) = 9.81 m/s2Radius of Earth (RE) = 6378 kmHeight above the surface (h) = 10 kmAcceleration due to gravity at 10 km (g) = g0 / (1 + h / RE)2= 9.81/(1+10/6378)2= 9.78 m/s2
Correct Answer : 5.85 m/s2
Explaination : Given,Thrust (T) = 31 x 106 NMass of rocket (m) = 1.98 x 106 kgResultant Force (F) = T-mg= 31*106-(1.98*106*9.81)= 11.57 x 106 NFrom Newton’s second law,Resultant instantaneous acceleration (a) = F/m=(11.57*106)/(1.98*106)= 5.85 m/s2
Correct Answer : N-body problem
Explanation : When we have o predict and analyze the motion of multiple celestial objects in the sky as a result of their gravitational forces acting on each other, we use N-body problem to solve the dynamics.
Correct Answer : False
Explanation : In N-body problems, where there’s mutual forces between several celestial bodies, the net external force and torque are both zero due to the Newton’s third law.
Correct Answer : 6N
Explanation : In case of N-body problem, there are 3N second order differential equations which results in 6N motion variables.
Correct Answer : 3N
Explanation : The N-body problem has 3N degrees of freedom which makes use of 3N second order differential equations for finding the position of the particle.
Correct Answer : Mass of one of the bodies is negligible
Explanation : In case of restricted three-body problem, it is a simplified version of three-body problem where it is assumed that one of the masses is almost negligible compared to the other two. This m1 and m2 move in Keplerian orbit which is unaffected by m3.
Correct Answer : 10-3 g’s
Explaination : There are certain variations introduced due to the earth’ oblateness effect. This force is of the order 10-3 g’s. Newton’s law is only applicable for spherical objects.
Correct Answer : One-body problem
Explanation : The simplest case of the N-body problem is a case which has only one body. In the entire space, there lies only one body with mass m. It experiences no force, acceleration and moves with a constant velocity.
Correct Answer : Closed form solution does not exist
Explanation : N-body problems are usually chaotic for initial conditions and require numerical method to compute the result. Analytical results cannot be obtained wince there no closed form solution existing. So far, two body problems and restricted 3-body problem can be solved.
