A geosynchcronous satellite is one that stays above the same
point on the Earth, that is it rotates with the same angular
velocity than Earth. This is possible only if the satellite is
above a point on the equator. Determine:
a) The height above the Earth's surface such satellite must
orbit?
b) What is the satellite speed?
c) If the satellite is orbiting 200 km above Earth, what is the satellite speed?
( Remember G=6.67x10-11 N m2/kg2, MEarth = 5.98x1024 Kg, REarth = 6380 km)
A geosynchcronous satellite is one that stays above the same point on the Earth, that is...
Find the speed of a satellite in a circular orbit around the Earth with a radius 2.77 times the mean radius of the Earth. (Radius of Earth -6.37x103 km, mass of Earth 5.98x1024 kg, G - 6.67x10 11 Nm2/kg2.)
a satellite in a geosynchronous orbit remains above the same
point on Earth
2. A satellite in a geosynchronous orbit remains above the same point on Earth provided it orbits in the equatorial plane in the same direction as Earth's rotation. (a) Calculate the speed of a satellite in such an orbit. (b) Calculate the total energy of a satellite in such an orbit.
At what height above the earth is the free-fall acceleration 10 % of its value at the surface? Assume Rearth = 6.37 × 106 m. What is the speed of a satellite orbiting at that height? Assume Mearth = 5.98 × 1024 kg.
A satellite moves in orbit about the Earth at a height of 9500 km above the Earth's surface. Find: a) the velocity of the satellite; b) its period of rotation; c) the magnitude of the Earth's gravitational force on the satellite if it has a mass of 2500 kg; d) if the satellite's velocity is increased by 20%, at what height must it now orbit above the Earth? (Note: Changing the satellite's velocity also changes the satellite's period of rotation).
A geosynchronous satellite is placed above the equator and orbiting around the earth to facilitate communication around the world. (You may consider that the mass of the satellite is m, mass of the earth is M(6x1024 kg), center-to-center distance between the satellite and the earth is r, radius of the earth is R(6.4×106 m), and the universal gravitational constant G = 6.67x10-11 N.m2/kg2.) Satellite Earth Applying newton's 2nd law, write an equation describing the circular motion of the satellite. (4...
A 270 kg satellite is orbiting on a circular orbit 6180 km above the Earth's surface. Determine the speed of the satellite. (The mass of the Earth is 5.97×1024 kg, and the radius of the Earth is 6370 km.)
1) An astronaut is orbiting the Earth preparing to repair a satellite. The satellite is in a circular orbit 600 km above the Earth's surface, where the free-fall acceleration is 8.20 m/s2. Take the radius of the Earth as 6400 km. a. Determine the speed of the satellite. Express your answer in meters per second. b. Determine the time interval required to complete one orbit around the Earth. Express your answers in minutes 2) The pilot of an airplane notes that the compass indicates...
A geostationary satellite is a satellite located in an orbit such that it remains above the same point on the Earth’s surface. [Assume it takes 23 hours 56 minutes 4.09 seconds for the Earth to spin around once.] a) What is the angular velocity of such a satellite? b) What is the altitude of such a satellite? c) Calculate the period of a satellite orbiting 200km above the Earth.
A 180 kg satellite is orbiting on a circular orbit 6355 km above the Earth's surface. Determine the speed of the satellite. (The mass of the Earth is 5.97x1024 kg, and the radius of the Earth is 6370 km.) Submit Answer Tries 0/12
Find the height H of a geosynchronous satellite above the surface of the earth. You may well want to find the radius of the orbit R first. You may use the following constants: The universal gravitational constant G is 6.67×10−11Nm2/kg2. The mass of the earth is 5.98×1024kg. The mass of the satellite is 2.10×102kg. The radius of the earth is 6.38×106m. Give the height of the orbit above the surface in km to three significant digits.