Question

In: Physics

A communications satellite is in a synchronous orbit that is 3.47

A communications satellite is in a synchronous orbit that is 3.47

Solutions

Expert Solution

Note that the distance to be travelled is from A to B, then B to C.          
          
Cutting the figure into 2 right triangles, and using Pythagorean theorem,          
          
AB = AC =    1745009.943   m  
          

Thus, the total distance is the sum of AB and Ac,          
          
d =    3490019.885   m  
          

Thus, as light travels at c = 3E8 m/s, then t = d/c:          
          
t =    0.0116    s   [ANSWER]


Related Solutions

A communications satellite is in a synchronous orbit that is 3.09×107m directly above the equator. The...
A communications satellite is in a synchronous orbit that is 3.09×107m directly above the equator. The satellite is located midway between Quito, Equador, and Belem, Brazil, two cities almost on the equator that are separated by a distance of 3.48×106m. Calculate the time it takes for a telephone call to go by way of satellite between these cities. Ignore the curvature of the earth.
A communications satellite is in a synchronous orbit that is 3.61×107 m directly above the equator....
A communications satellite is in a synchronous orbit that is 3.61×107 m directly above the equator. The satellite is located midway between Quito, Equador, and Belem, Brazil, two cities almost on the equator that are separated by a distance of 3.87×106 m. Calculate the time it takes for a telephone call to go by way of satellite between these cities. Ignore the curvature of the earth.
Let’s say you wanted to have a communications satellite orbit the Moon so that it stayed...
Let’s say you wanted to have a communications satellite orbit the Moon so that it stayed exactly above one point of the Moon’s equator (similar to a geosynchronous here on Earth). What would the linear speed and lunar altitude of your communications satellite be?
A geosynchronous communications satellite in stationary orbit is used to relay radio messages from one point...
A geosynchronous communications satellite in stationary orbit is used to relay radio messages from one point to another around the earth’s curved surface. Apply Newton’s second law to the satellite to find its altitude in meters above the earth’s surface. RE = 6.38 × 106 m, mE = 5.97 × 1024 kg, and G = 6.674 × 10−11 Nm2/kg2.
Explain two reasons behind the commercial failure of Low Earth Orbit satellite mobile communications companies (such...
Explain two reasons behind the commercial failure of Low Earth Orbit satellite mobile communications companies (such as Iridium).
A satellite is put into an elliptical orbit around the Earth. When the satellite is at...
A satellite is put into an elliptical orbit around the Earth. When the satellite is at its perigee, its nearest point to the Earth, its height above the ground is hp=215.0 km,hp=215.0 km, and it is moving with a speed of vp=8.450 km/s.vp=8.450 km/s. The gravitational constant GG equals 6.67×10−11 m3⋅kg−1⋅s−26.67×10−11 m3·kg−1·s−2 and the mass of Earth equals 5.972×1024 kg.5.972×1024 kg. When the satellite reaches its apogee, at its farthest point from the Earth, what is its height haha above...
A satellite is put into an elliptical orbit around the Earth. When the satellite is at...
A satellite is put into an elliptical orbit around the Earth. When the satellite is at its perigee, its nearest point to the Earth, its height above the ground is ℎp=215.0 km, and it is moving with a speed of ?p=8.850 km/s. The gravitational constant ? equals 6.67×10−11 m3·kg−1·s−2 and the mass of Earth equals 5.972×1024 kg. When the satellite reaches its apogee, at its farthest point from the Earth, what is its height ℎa above the ground? For this...
A satellite is put into an elliptical orbit around the Earth. When the satellite is at...
A satellite is put into an elliptical orbit around the Earth. When the satellite is at its perigee, its nearest point to the Earth, its height above the ground is hp=215.0 km,hp=215.0 km, and it is moving with a speed of vp=8.450 km/s.vp=8.450 km/s. The gravitational constant GG equals 6.67×10−11 m3⋅kg−1⋅s−26.67×10−11 m3·kg−1·s−2 and the mass of Earth equals 5.972×1024 kg.5.972×1024 kg. When the satellite reaches its apogee, at its farthest point from the Earth, what is its height haha above...
A satellite is in an elliptical orbit around the earth. The distance from the satellite to...
A satellite is in an elliptical orbit around the earth. The distance from the satellite to the center of the earth ranges from 7.2 Mm at perigee (where the speed is 8.0 km/s) to 9.9 Mm at apogee. 1. Assume the initial conditions are x = 0, y = 7.2 × 106 m, vx = 8.0×103 m/s, and vy = 0. Use python program to print its speed, distance from the earth, kinetic energy, potential energy, and total mechanical energy...
The physics of satellite motion around Jupiter. A 155 kg satellite is placed in orbit 6.00...
The physics of satellite motion around Jupiter. A 155 kg satellite is placed in orbit 6.00 x 105 m above the surface of Jupiter. The tangential speed of the satellite is 3.99 x 104 m/s. Jupiter has a mass of 1.90 x 1027 kg and a radius of 7.14 x 107 m. Determine the radius of motion of the satellite. Calculate the centripetal force acting on the satellite to maintain its orbit. What is the cause of this centripetal force?...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT