Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationEnergy & PotentialEasy2 minPYQ_2020

PhysicsEasysingle choice

The escape velocity for a planet from the surface is $v_{e}$ . A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be

Options:

Answer:

Solution:

Taking the potential at a large distance from the planet as zero, the potential at the centre of the planet $= - \frac{3 GM}{2 R}$ .

$\frac{1}{2} {mv}^{2} = m [0 - (- \frac{3 GM}{2 R})]$

or $v^{2} = \frac{3 GM}{2 R} = 3 Rg$

$\Rightarrow v^{2} = \frac{3}{2} (2 RG) = \frac{3}{2} v_{e}^{2} (∵ v_{e} = \sqrt{2 R G})$

$\Rightarrow v = \sqrt{1.5} v_{e}$

Subject:PhysicsTopic:GravitationSubtopic:Energy & Potential

⏱ 2mℹ️ Source: PYQ_2020

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