The $K . E$ . of a satellite in an orbit close to the surface of the earth is $E$ . Its minimum $K . E$ . so as to escape from the gravitational field of the earth is.

2 E

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4 E

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2 \sqrt{2} E

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\sqrt{2} E

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Solution

The correct option is A $2 E$
$E = m v_{o}^{2} / 2 = G M m / 2 R$
Minimum $K . E = - P . E = G M m / R$
Therefore, Minimum $K . E = 2 E$

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The K.E. of a satellite in an orbit close to the surface of the earth is E. Its minimum K.E. so as to escape from the gravitational field of the earth is.

The correct option is A 2EE=mv2o/2=GMm/2RMinimum K.E=−P.E=GMm/RTherefore, Minimum K.E=2E

The $K . E$ . of a satellite in an orbit close to the surface of the earth is $E$ . Its minimum $K . E$ . so as to escape from the gravitational field of the earth is.

The correct option is A $2 E$
$E = m v_{o}^{2} / 2 = G M m / 2 R$
Minimum $K . E = - P . E = G M m / R$
Therefore, Minimum $K . E = 2 E$