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Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationEnergy & PotentialEasy2 minPYQ_2020

PhysicsEasysingle choice

A particle is projected with velocity $k v_{e}$ in the vertically upward direction from the ground into space ( $v_{e}$ is escape velocity and $k < 1$ ). The maximum height from the centre of the earth to which it can reach is

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Answer:

Solution:

Total energy of surface = Total energy at $h$ height

$\Rightarrow \frac{1}{2} m {(k v_{e})}^{2} + \{\frac{- G M m}{R}\} = 0 + \{\frac{- G M m}{R + h}\}$

$\Rightarrow k^{2} \frac{1}{2} m {v_{e}}^{2} + (\frac{- G M m}{R}) = (\frac{- G M m}{R + h})$

$\Rightarrow k^{2} \frac{1}{2} m (\frac{2 G M}{R}) - (\frac{G M m}{R}) = - (\frac{G M m}{R + h})$

$\Rightarrow \frac{1}{R} - \frac{k^{2}}{R} = \frac{1}{R + h}$

$\Rightarrow \frac{1 - k^{2}}{R} = \frac{1}{R + h}$

$\Rightarrow R + h = \frac{R}{1 - k^{2}}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:GravitationSubtopic:Energy & Potential

⏱ 2mℹ️ Source: PYQ_2020

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