PHy 214 CH 13 Satelite (Figure 1) Shows A A Geos.. A Satellite T Example 13.5 A
ID: 2031095 • Letter: P
Question
PHy 214 CH 13 Satelite (Figure 1) Shows A A Geos.. A Satellite T Example 13.5 A Geosynchronous Satellite Consider a satellite of mass m moving in a circular orbit around the Earth at a constant speed v and at an altitude h above the Earth's surface as illustrated in the figure (A) Determine the speed of the satellite in tems of G, h, Re (the radius of the Earth), and Mg (the mass of the Earth) (B) If the satellite is to be geosynchronous (that is appearing to remain over a fixed position on the Earth), how fast is it moving through space? SOLVE IT (A) Determine the speed of the satellite in terms of G, h, RE (the radius of the Earth), and Mg (the mass of the Earth). A satelite of mass m moving around the Earth in a diraular orbit of radius r with constant speed v. The only force acting on the satelite is the grevitationel force (Not drewn to scale.) Conceptualize Imagine the satellite moving around the Earth in a circular orbit under the influence of the gravitational force Categorize The satellite moves in a circular orbit at a constant speed. Therefore, we categorize the satellite as a particle in uniform circular motion as well as a particle under a net force Analyze The only external force acting on the satellite is the gravitational force from the Earth, which acts toward the center of the Earth and keeps the satellite in its circular orbit. Mgm Apply the particle under a net force and particle in uniform circular motion models to the satellite: ma-cr m Solve for v the center of the Earth to the satellite isr , noting that the distance r fromExplanation / Answer
According o the concept of the gravitation
here the satelisite moves in a circular orbit
so the centripetal force is devolped
centripetal force =gravitation attraction force
mv^2/r=GMem/r^2
mv^2/(R+h)=GMem/(R+h)^2
v^2=GMe/(R+h)
V={GMe/(R+h)}^1/2
mass of the earth Me=5.972*10^27 kg
height above the earth surface h=35.800 km
radius of the earth Re=6400 km
the distance b/w the satelitte and earth =>R+h=42200 km
now we find the orbital speed
the orbital speed V={6.67*10^-11*5.972*10^27/4.2200*10^7 }^1/2
=3.1*10^3 m/s
=3.1 km/s
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