UNIVERSE
SATELLITESICY BODIES
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Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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If V1 is the escape velocity and V0 is the orbital velocity of a satellite close to the surface of the Earth, then the relationship between V1 and V0 is
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Explanation:
Detailed explanation-1: -If Ve is escape velocity and V0 is orbital velocity of a satellite for orbit close to the earth’s surface, then these are related by: v0 = √2 v.
Detailed explanation-2: -The escape velocity is a multiplicative factor greater than the orbital velocity: v = sqrt(2GM/r) (escape velocity = sqrt(2) * orbital velocity) If an object achieves escape velocity, it leaves its orbit and is no longer gravitationally bound to the object it was orbiting.
Detailed explanation-3: -Relationship Between Escape And Orbital Velocity It shows that escape velocity is √2 times greater than orbital velocity.
Detailed explanation-4: -The orbital speed for an Earth satellite near the surface of the Earth is 7 km/sec.
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