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The orbital velocity of a body at height H above the surface of the Earth is 36% of that near to the surface of the earth of radius R if the escape velocity at the surface of the Earth is 11.2 km per second then its value at the height H will be?
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The orbital velocity of a body at height H above the surface of the Ea...
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The orbital velocity of a body at height H above the surface of the Ea...
Understanding Orbital Velocity
To determine the escape velocity at height H above the Earth's surface, we first need to establish some key concepts about orbital and escape velocities.
Orbital Velocity Near the Surface
- The escape velocity at the surface of the Earth is given as 11.2 km/s.
- The orbital velocity (v) near the Earth's surface can be expressed as:
v = (1/√2) * escape velocity
v ≈ (1/√2) * 11.2 km/s ≈ 7.9 km/s.
Orbital Velocity at Height H
- According to the problem, the orbital velocity at height H is 36% of that near the surface:
v_H = 0.36 * v ≈ 0.36 * 7.9 km/s ≈ 2.84 km/s.
Escape Velocity at Height H
- The escape velocity (V_escape) at any height can be derived from the escape velocity at the surface using the formula:
V_escape(H) = V_escape(surface) * √(R / (R + H)),
where R is the radius of the Earth.
- Given that the escape velocity at the surface is 11.2 km/s, we can express the escape velocity at height H in terms of this value.
Calculating Escape Velocity at Height H
- Since we have already established that orbital velocity at height H is 2.84 km/s, we can apply the relationship:
V_escape(H) = 2 * orbital velocity at height H.
- Therefore,
V_escape(H) = 2 * 2.84 km/s ≈ 5.68 km/s.
Conclusion
The escape velocity at height H above the Earth's surface is approximately 5.68 km/s. This value reflects the reduced gravitational influence at that height compared to the surface.
To determine the escape velocity at height H above the Earth's surface, we first need to establish some key concepts about orbital and escape velocities.
Orbital Velocity Near the Surface
- The escape velocity at the surface of the Earth is given as 11.2 km/s.
- The orbital velocity (v) near the Earth's surface can be expressed as:
v = (1/√2) * escape velocity
v ≈ (1/√2) * 11.2 km/s ≈ 7.9 km/s.
Orbital Velocity at Height H
- According to the problem, the orbital velocity at height H is 36% of that near the surface:
v_H = 0.36 * v ≈ 0.36 * 7.9 km/s ≈ 2.84 km/s.
Escape Velocity at Height H
- The escape velocity (V_escape) at any height can be derived from the escape velocity at the surface using the formula:
V_escape(H) = V_escape(surface) * √(R / (R + H)),
where R is the radius of the Earth.
- Given that the escape velocity at the surface is 11.2 km/s, we can express the escape velocity at height H in terms of this value.
Calculating Escape Velocity at Height H
- Since we have already established that orbital velocity at height H is 2.84 km/s, we can apply the relationship:
V_escape(H) = 2 * orbital velocity at height H.
- Therefore,
V_escape(H) = 2 * 2.84 km/s ≈ 5.68 km/s.
Conclusion
The escape velocity at height H above the Earth's surface is approximately 5.68 km/s. This value reflects the reduced gravitational influence at that height compared to the surface.
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The orbital velocity of a body at height H above the surface of the Earth is 36% of that near to the surface of the earth of radius R if the escape velocity at the surface of the Earth is 11.2 km per second then its value at the height H will be?
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The orbital velocity of a body at height H above the surface of the Earth is 36% of that near to the surface of the earth of radius R if the escape velocity at the surface of the Earth is 11.2 km per second then its value at the height H will be? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The orbital velocity of a body at height H above the surface of the Earth is 36% of that near to the surface of the earth of radius R if the escape velocity at the surface of the Earth is 11.2 km per second then its value at the height H will be? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The orbital velocity of a body at height H above the surface of the Earth is 36% of that near to the surface of the earth of radius R if the escape velocity at the surface of the Earth is 11.2 km per second then its value at the height H will be?.
The orbital velocity of a body at height H above the surface of the Earth is 36% of that near to the surface of the earth of radius R if the escape velocity at the surface of the Earth is 11.2 km per second then its value at the height H will be? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about The orbital velocity of a body at height H above the surface of the Earth is 36% of that near to the surface of the earth of radius R if the escape velocity at the surface of the Earth is 11.2 km per second then its value at the height H will be? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The orbital velocity of a body at height H above the surface of the Earth is 36% of that near to the surface of the earth of radius R if the escape velocity at the surface of the Earth is 11.2 km per second then its value at the height H will be?.
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