Calculate the altitude at which an artificial satelliteorbiting in the equatorial plane can be synchronous satellite (i.e.,so it can remain above the same spot on the surface of the earth).(hint: consider gravitational force is balanced with the centrifugal force, so zero gravitystate).
Calculate the altitude at which an artificial satelliteorbiting in the equatorial plane can be synchronous satellite...
A satellite of mass 42.5 kg in geosynchronous orbit at an altitude of 3.58 ✕ 104 km above the Earth's surface remains above the same spot on the Earth. Assume its orbit is circular. Find the magnitude of the gravitational force exerted by the Earth on the satellite. Hint: The answer is not 417 N
Previous Question: A satellite is said to be in a “geosynchronous” orbit if it appears always to remain over the exact same spot on the rotating Earth. What is the period of this orbit? [enter your answer in hours.] (a) At what distance (in km) from the center of the Earth must such a satellite be placed into orbit? [Enter your answer in regular decimal notation instead of scientific notation, i.e. not 3.97 x 10^5, but 397000 .] (b) Why...
Telecommunication satellites travel in geosynchronous orbits. these satellites move in sync with the Earth surface and always stay above the same point on the surface of the earth. all satellites in geosynchronous orbit must be located in the equatorial plane. Please, explain why it has to stay in the equatorial plane. Please EXPLAIN why it has to stay in the equatorial plane. DRAW a simple diagram showing forces acting on such a satellite. Hint: For any stable orbit the centripetal...
A synchronous satellite, which always remains above the same point on a planet's equator, is put in orbit around Jupiter to study that planet's famous red spot. Jupiter rotates once every 9.84 h. Use the following data to find the altitude of the satellite above the surface of the planet. Jupiter has a mass of 1.90 1027 kg, and a mean radius of 6.99 107 m. Answer must be in km
An artificial satellite circling the Earth completes each orbit in 135 minutes. (a) Find the altitude of the satellite. m (b) What is the value of g at the location of this satellite? m/s2 (a) Find the magnitude of the gravitational force (in N) between a planet with mass 8.25 X 1024 kg and its moon, with mass 2.20 x 1022 kg, if the average distance between their centers is 2.20 x 108, m. N (b) What is the moon's...
46. W A synchronous satellite, which always remains above the same point on a planet's equator, is put in circular orbit around Jupiter to study that planet's famous red spot. Jupiter rotates once every 9.84 h. Use the data of Table 7.3 to find the altitude of the satellite.
2. Consider a system consisting of the Sun, Earth, and a satellite in a circular orbit about the Earth. (a) Plot the gravitational acceleration of the satellite due to Earth's gravity as a function of the altitude of the satellite as measured from the surface of the Earth. Scale your plot so that the altitude goes from 0 km to 104 km. (b) Assume that the satellite instantaneously lies on the line between the Earth and the Sun. Make two...
Problem 1 a. The apogee and perigee altitude (distance from the surface of Earth) of the Chandra X-rays Observatory are 139,200 km and 9,620 km (final orbit after 4th burn, Aug 7, 1999). Estimate its orbital period (in hours). Hint: the mass of Chandra is much smaller than the mass of Earth. b. Communications and weather satellites are often placed in geosynchronous “parking” orbits above Earth. These are the orbits where satellites can remain fixed above a specific point on...
Centripetal force (university physics 1): At what altitude should a geosynchronous satellite be placed? A geosynchronous orbit means the satellite stays above the same point on earth...so what is its orbital period
A “synchronous” satellite, which always re- mains above the same point on a planet’s equator, is put in orbit about a planet similar to Jupiter. This planet rotates once every 9 . 6 h, has a mass of 1 . 9 × 10 27 kg and a radius of 6 . 99 × 10 7 m. Given that G = 6 . 67 × 10 − 11 N m 2 / kg 2 , calculate how far above Jupiter’s SURFACE...