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A remote sensing satellite of earth revolve in a circular orbit at a height of 0.25×10^6. M above the, surface of earth if earth's radius is 6.38×10^6. M and g =9.8ms then, orbital speed of the satellite?
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A remote sensing satellite of earth revolve in a circular orbit at a h...
Orbital Speed of a Satellite in Circular Orbit
Formula: V = √(GM/R)
where,
V = orbital speed of the satellite
G = universal gravitational constant (6.67 x 10^-11 Nm^2/kg^2)
M = mass of the Earth (5.97 x 10^24 kg)
R = radius of the satellite's orbit (6.38 x 10^6 m + 0.25 x 10^6 m = 6.63 x 10^6 m)
Calculations:
- G, M, and R are known, so we can plug them into the formula and solve for V:
V = √[(6.67 x 10^-11 Nm^2/kg^2) x (5.97 x 10^24 kg) / (6.63 x 10^6 m)]
- Simplifying the expression inside the square root:
V = √(3.986 x 10^14 m^3/s^2 / 6.63 x 10^6 m)
V = √(6.02 x 10^7 m^2/s^2)
- Taking the square root:
V = 7.76 x 10^3 m/s
Therefore, the orbital speed of the satellite is 7.76 x 10^3 m/s.
Explanation:
- The speed of a satellite in a circular orbit is determined by the balance between the gravitational force of the Earth and the centrifugal force caused by the satellite's motion.
- The formula V = √(GM/R) combines these two forces into a single equation.
- The mass of the Earth and the radius of the satellite's orbit are known values, while the universal gravitational constant is a constant of nature.
- Plugging these values into the formula allows us to calculate the orbital speed of the satellite.
- In this case, the satellite is in a low Earth orbit, which means it is relatively close to the Earth's surface and therefore has a high orbital speed.
- The value we calculated (7.76 x 10^3 m/s) is typical for a satellite in this type of orbit.
Formula: V = √(GM/R)
where,
V = orbital speed of the satellite
G = universal gravitational constant (6.67 x 10^-11 Nm^2/kg^2)
M = mass of the Earth (5.97 x 10^24 kg)
R = radius of the satellite's orbit (6.38 x 10^6 m + 0.25 x 10^6 m = 6.63 x 10^6 m)
Calculations:
- G, M, and R are known, so we can plug them into the formula and solve for V:
V = √[(6.67 x 10^-11 Nm^2/kg^2) x (5.97 x 10^24 kg) / (6.63 x 10^6 m)]
- Simplifying the expression inside the square root:
V = √(3.986 x 10^14 m^3/s^2 / 6.63 x 10^6 m)
V = √(6.02 x 10^7 m^2/s^2)
- Taking the square root:
V = 7.76 x 10^3 m/s
Therefore, the orbital speed of the satellite is 7.76 x 10^3 m/s.
Explanation:
- The speed of a satellite in a circular orbit is determined by the balance between the gravitational force of the Earth and the centrifugal force caused by the satellite's motion.
- The formula V = √(GM/R) combines these two forces into a single equation.
- The mass of the Earth and the radius of the satellite's orbit are known values, while the universal gravitational constant is a constant of nature.
- Plugging these values into the formula allows us to calculate the orbital speed of the satellite.
- In this case, the satellite is in a low Earth orbit, which means it is relatively close to the Earth's surface and therefore has a high orbital speed.
- The value we calculated (7.76 x 10^3 m/s) is typical for a satellite in this type of orbit.
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A remote sensing satellite of earth revolve in a circular orbit at a height of 0.25×10^6. M above the, surface of earth if earth's radius is 6.38×10^6. M and g =9.8ms then, orbital speed of the satellite?
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A remote sensing satellite of earth revolve in a circular orbit at a height of 0.25×10^6. M above the, surface of earth if earth's radius is 6.38×10^6. M and g =9.8ms then, orbital speed of the satellite? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A remote sensing satellite of earth revolve in a circular orbit at a height of 0.25×10^6. M above the, surface of earth if earth's radius is 6.38×10^6. M and g =9.8ms then, orbital speed of the satellite? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A remote sensing satellite of earth revolve in a circular orbit at a height of 0.25×10^6. M above the, surface of earth if earth's radius is 6.38×10^6. M and g =9.8ms then, orbital speed of the satellite?.
A remote sensing satellite of earth revolve in a circular orbit at a height of 0.25×10^6. M above the, surface of earth if earth's radius is 6.38×10^6. M and g =9.8ms then, orbital speed of the satellite? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A remote sensing satellite of earth revolve in a circular orbit at a height of 0.25×10^6. M above the, surface of earth if earth's radius is 6.38×10^6. M and g =9.8ms then, orbital speed of the satellite? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A remote sensing satellite of earth revolve in a circular orbit at a height of 0.25×10^6. M above the, surface of earth if earth's radius is 6.38×10^6. M and g =9.8ms then, orbital speed of the satellite?.
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