What is the true weight of an object in a geostationary satellite that weighed exactly \[10.0N\] at the north pole ?
Answer
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Hint: A geostationary satellite is an earth-orbiting satellite, placed at an altitude of approximately 35,800 kilometers (\[35800km\]) directly over the equator, that revolves in the same direction the earth rotates and at this altitude, one orbit takes a day equal to 24 hours, the same length of time as the earth requires to rotate once on its axis.
Complete answer:
The term geostationary comes from the fact that such a satellite appears nearly stationary in the sky as seen by a ground-based observer.
The new global mobile communications network uses geostationary satellites.
Let \[W = mg\] is the weight at the north pole ,
Also we know that the radius of the earth is assumed to be \[6400km\].
\[ \Rightarrow R = 6400km\].
And \[W' = Mg'\] is the true weight in a geostationary satellite.
\[ \Rightarrow m{g'} = \dfrac{{mg}}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}\] ………..eq (1)
\[ \Rightarrow \dfrac{h}{R} = \dfrac{{35800km}}{{6400km}}\]
\[ \Rightarrow \dfrac{h}{R} = 5.593\]
Putting the value in eq (1) we get,
\[ \Rightarrow m{g'} = \dfrac{{mg}}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}\]
\[ \Rightarrow m{g'} = \dfrac{{10}}{{{{(1 + 5.593)}^2}}}\]
Solving the equation we get
\[ \Rightarrow m{g'} = 0.225N\]
Hence the true weight of an object in a geostationary satellite that weighed exactly \[10.0N\] at the north pole is \[0.225N\].
Note: Geostationary satellites have two limitations:-
First, Because the orbital zone is an extremely narrow ring in the plane of the equator and the number of satellites that can be maintained in geostationary orbits without mutual conflict (or even collision) is limited.
Second, The distance that an electromagnetic signal ( EM Signal ) must travel to and from a geostationary satellite is a minimum of 71,600 kilometers or equal to 44,600 miles. Thus the latency of at least 240 milliseconds is introduced when an EM signal, traveling at 300,000 kilometers per second or 186,000 miles per seconds makes a round trip from the surface to the satellite and back.
Complete answer:
The term geostationary comes from the fact that such a satellite appears nearly stationary in the sky as seen by a ground-based observer.
The new global mobile communications network uses geostationary satellites.
Let \[W = mg\] is the weight at the north pole ,
Also we know that the radius of the earth is assumed to be \[6400km\].
\[ \Rightarrow R = 6400km\].
And \[W' = Mg'\] is the true weight in a geostationary satellite.
\[ \Rightarrow m{g'} = \dfrac{{mg}}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}\] ………..eq (1)
\[ \Rightarrow \dfrac{h}{R} = \dfrac{{35800km}}{{6400km}}\]
\[ \Rightarrow \dfrac{h}{R} = 5.593\]
Putting the value in eq (1) we get,
\[ \Rightarrow m{g'} = \dfrac{{mg}}{{{{\left( {1 + \dfrac{h}{R}} \right)}^2}}}\]
\[ \Rightarrow m{g'} = \dfrac{{10}}{{{{(1 + 5.593)}^2}}}\]
Solving the equation we get
\[ \Rightarrow m{g'} = 0.225N\]
Hence the true weight of an object in a geostationary satellite that weighed exactly \[10.0N\] at the north pole is \[0.225N\].
Note: Geostationary satellites have two limitations:-
First, Because the orbital zone is an extremely narrow ring in the plane of the equator and the number of satellites that can be maintained in geostationary orbits without mutual conflict (or even collision) is limited.
Second, The distance that an electromagnetic signal ( EM Signal ) must travel to and from a geostationary satellite is a minimum of 71,600 kilometers or equal to 44,600 miles. Thus the latency of at least 240 milliseconds is introduced when an EM signal, traveling at 300,000 kilometers per second or 186,000 miles per seconds makes a round trip from the surface to the satellite and back.
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