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A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth is
- a)5
- b)10
- c)6√2
- d)6 / √2
Correct answer is option 'C'. Can you explain this answer?
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A geostationary satellite is orbiting the earth at a height of 5R abo...
According to Kepler’s third law T2 ∝ r3/2
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A geostationary satellite is orbiting the earth at a height of 5R abo...
The time period of a satellite is the time it takes for the satellite to complete one full orbit around the Earth. The time period can be calculated using Kepler's Third Law of Planetary Motion, which states that the square of the time period is directly proportional to the cube of the semi-major axis of the satellite's orbit.
In this question, we are given that a geostationary satellite is orbiting the Earth at a height of 5R above the surface of the Earth, where R is the radius of the Earth. To calculate the time period of another satellite orbiting at a height of 2R from the surface of the Earth, we can use the fact that the semi-major axis of the orbit is equal to the sum of the radius of the Earth and the height of the satellite above the surface.
Let's calculate the semi-major axis for the geostationary satellite first:
Semi-major axis = Radius of the Earth + Height of the satellite
= R + 5R
= 6R
Now, let's calculate the semi-major axis for the other satellite:
Semi-major axis = Radius of the Earth + Height of the satellite
= R + 2R
= 3R
According to Kepler's Third Law, we have:
(T1)^2/(T2)^2 = (a1)^3/(a2)^3
Substituting the values:
(T1)^2/(T2)^2 = (6R)^3/(3R)^3
(T1)^2/(T2)^2 = 6^3/3^3
(T1)^2/(T2)^2 = 216/27
(T1)^2/(T2)^2 = 8
Taking the square root of both sides:
T1/T2 = √8
T1/T2 = 2√2
Since we know that the time period of the geostationary satellite (T1) is 24 hours, we can calculate the time period of the other satellite (T2):
24/T2 = 2√2
Solving for T2:
T2 = 24/(2√2)
T2 = 12/√2
T2 = 6√2
Therefore, the time period of the satellite at a height of 2R from the surface of the Earth is 6√2 hours, which corresponds to option C.
In this question, we are given that a geostationary satellite is orbiting the Earth at a height of 5R above the surface of the Earth, where R is the radius of the Earth. To calculate the time period of another satellite orbiting at a height of 2R from the surface of the Earth, we can use the fact that the semi-major axis of the orbit is equal to the sum of the radius of the Earth and the height of the satellite above the surface.
Let's calculate the semi-major axis for the geostationary satellite first:
Semi-major axis = Radius of the Earth + Height of the satellite
= R + 5R
= 6R
Now, let's calculate the semi-major axis for the other satellite:
Semi-major axis = Radius of the Earth + Height of the satellite
= R + 2R
= 3R
According to Kepler's Third Law, we have:
(T1)^2/(T2)^2 = (a1)^3/(a2)^3
Substituting the values:
(T1)^2/(T2)^2 = (6R)^3/(3R)^3
(T1)^2/(T2)^2 = 6^3/3^3
(T1)^2/(T2)^2 = 216/27
(T1)^2/(T2)^2 = 8
Taking the square root of both sides:
T1/T2 = √8
T1/T2 = 2√2
Since we know that the time period of the geostationary satellite (T1) is 24 hours, we can calculate the time period of the other satellite (T2):
24/T2 = 2√2
Solving for T2:
T2 = 24/(2√2)
T2 = 12/√2
T2 = 6√2
Therefore, the time period of the satellite at a height of 2R from the surface of the Earth is 6√2 hours, which corresponds to option C.
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A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth. The time period of another satellite in hours at a height of 2R from the surface of the earth isa)5b)10c)6√2d)6 / √2Correct answer is option 'C'. Can you explain this answer?
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