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Physics - Gravitation Question with Solution | TestHub

PhysicsGravitationEnergy & PotentialEasy2 minPYQ_2020

PhysicsEasysingle choice

A body is projected up with a velocity equal to ${\frac{3}{4}}^{t h}$ of the escape velocity from the surface of the earth. The height it reaches from the center of the earth is

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Answer:

Solution:

$v = \frac{3}{4} v_{e}$

$K . E . = \frac{1}{2} m v^{2} = \frac{1}{2} m {(\frac{3}{4} v_{e})}^{2} = \frac{9}{32} m v_{e}^{2} = \frac{9}{32} m (\frac{2 G M}{R})$

$\frac{9}{16} \frac{G M m}{R} - \frac{G M m}{R} = - \frac{G M m}{r}$

$r = \frac{16 R}{7}$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:GravitationSubtopic:Energy & Potential

⏱ 2mℹ️ Source: PYQ_2020

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