GK
PHYSICS
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Question
[CLICK ON ANY CHOICE TO KNOW THE RIGHT ANSWER]
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174.86 m/s
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1.50 m/s
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5.53 x 10^-6 m/s
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8.29 x 10^-3 m/s
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Detailed explanation-1: -The escape velocity vesc is expressed as vesc = Square root of√2GMr, where G is the gravitational constant, M is the mass of the attracting mass, and r is the distance from the centre of that mass.
Detailed explanation-2: -Question– Suppose the radius of Earth is 6.38 × 106 m and the mass of the planet earth is 5.98 × 1024 kg. Find out the escape velocity from planet earth. Therefore, the escape velocity from Earth is 11 184 m/s, or approximately 11.2 km/s.
Detailed explanation-3: -Escape velocity is the minimum velocity required by a body to be projected to overcome the gravitational pull of the earth. It is the minimum velocity required by an object to escape the gravitational field that is, escape the land without ever falling back.
Detailed explanation-4: -Applying the law of conservation of energy., VE is escape velocity, M is the mass of the earth and R is the radius of earth. So, we have the formula for escape velocity as, VE=√2GMR.