
A satellite having time period same as that of the earth’s rotation about its own
Axis is orbiting the earth at a height 8R above the surface of earth, where, R is
Radius of earth. What will the time period of another satellite at a height 3.5R
From the surface of earth?
Answer
583.8k+ views
Hint: The GSO is populated at present by under 170 dynamic satellites, by more than 110 non-practical articles and by an obscure number of flotsam and jetsam. The quantity of radio communication Space Stations informed to the IFRB, in the coordinating process and with advance distribution is 457.
Complete answer:
So for finding the time period for the next satellite from the surface will be:
\[\dfrac{T2}{T1}={{\left( \dfrac{R2}{R1} \right)}^{\dfrac{3}{2}}}\]
\[T2=24{{\left( \dfrac{2.5R}{7R} \right)}^{\dfrac{3}{2}}}\]
\[T2=6\sqrt{2}h\]
The formula for time is: \[T\text{ }(period)\text{ }=\dfrac{1}{f\text{ }\left( frequency \right)}\]
\[\lambda \text{ }=\text{ }\dfrac{c}{f}\] = wave speed c (m/s) / frequency f (Hz).
The unit hertz (Hz) was once called cps = cycles per second.
The streamlined depiction of the GSO with a round Earth as the main drawing in the body is lacking and particularly deluding. Annoyances by the fascination of the Moon and the Sun and by the other minor characteristic powers must be considered to show up at a reasonable depiction of the particular circle.
Since the ordinary powers don't concede to a geostationary satellite, but human-made station-keeping is essential to keep up the satellite at or close to its ostensible orbital position. The nearness of the non-practical items and flotsam and jetsam in the GSO is a wellspring of the conceivable threat to dynamic satellites. The likelihood of an impact is little by and large yet might be significantly higher at packed longitudes. The likelihood of crash will increment later on and will be significant for massive space structures.
Note:
Logical grounds underlying sway claims over fragments of the GSO are talked about. Peaceful accords on confining the measure of non-useful articles and trash and on systematic evaluation of dormant satellites into removal circles are crucial for a proceeding with utilization of the advantages of the GSO.
Complete answer:
So for finding the time period for the next satellite from the surface will be:
\[\dfrac{T2}{T1}={{\left( \dfrac{R2}{R1} \right)}^{\dfrac{3}{2}}}\]
\[T2=24{{\left( \dfrac{2.5R}{7R} \right)}^{\dfrac{3}{2}}}\]
\[T2=6\sqrt{2}h\]
The formula for time is: \[T\text{ }(period)\text{ }=\dfrac{1}{f\text{ }\left( frequency \right)}\]
\[\lambda \text{ }=\text{ }\dfrac{c}{f}\] = wave speed c (m/s) / frequency f (Hz).
The unit hertz (Hz) was once called cps = cycles per second.
The streamlined depiction of the GSO with a round Earth as the main drawing in the body is lacking and particularly deluding. Annoyances by the fascination of the Moon and the Sun and by the other minor characteristic powers must be considered to show up at a reasonable depiction of the particular circle.
Since the ordinary powers don't concede to a geostationary satellite, but human-made station-keeping is essential to keep up the satellite at or close to its ostensible orbital position. The nearness of the non-practical items and flotsam and jetsam in the GSO is a wellspring of the conceivable threat to dynamic satellites. The likelihood of an impact is little by and large yet might be significantly higher at packed longitudes. The likelihood of crash will increment later on and will be significant for massive space structures.
Note:
Logical grounds underlying sway claims over fragments of the GSO are talked about. Peaceful accords on confining the measure of non-useful articles and trash and on systematic evaluation of dormant satellites into removal circles are crucial for a proceeding with utilization of the advantages of the GSO.
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