Kinetic energy

The kinetic energy needed to project a body of mass m from the Earth's surface out of the Earth's field is?

R R is the radius of the Earth. g g is acceleration due to gravity

1 2 \dfrac {1}{2} mgR 1 8 \dfrac {1}{8} mgR 1 4 \dfrac {1}{4} mgR mgR 1 16 \dfrac {1}{16} mgR

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1 solution

The escape velocity required is 2 g R \sqrt{2gR} .

Hence the required Kinetic Energy is:

K = 1 2 m v 2 = 1 2 m ( 2 g R ) = m g R K = \dfrac{1}{2}mv^2 = \dfrac{1}{2} m (2gR) = \boxed{mgR}

A tougher par would be by considering variable g

Aditya Kumar - 5 years, 10 months ago

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