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Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationEnergy & PotentialEasy2 minPYQ_2020

PhysicsEasysingle choice

The escape velocity of a body on the surface of the earth is $11.2 km s^{- 1}$ . If the earth's mass increases to twice its present value and radius of the earth becomes half, the escape velocity becomes

Options:

Answer:

Solution:

$V_{e} = \sqrt{\frac{2 G M_{e}}{R_{e}}} = 11.2 km s^{- 1}$

Now $M^{'} = 2 M_{e}$ & $R^{'} = \frac{R_{e}}{2}$

So $V_{e}^{'} = \sqrt{\frac{2 G (2 M_{e})}{(\frac{R_{e}}{2})}} = 2 \sqrt{\frac{2 G M_{e}}{R_{e}}} = 2 \times 11.2$

$= 22.4 km s^{- 1}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:GravitationSubtopic:Energy & Potential

⏱ 2mℹ️ Source: PYQ_2020

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