NEET Practice Questions (MCQs) with Answers & Solutions

$ \triangle U = U _2 - U_1 = { mgh \over 1+h/Re } = { mgRe \over 1 + { Re \over Re} } = { mgRe \over 2 } $ $ \therefore U_2 - ( - mgRe ) = { mgRe \over 2 } $ $ \therefore U_2 = - { 1 \over 2 } mgRe $

If body is projected with velocity Ï… (Ï… Ve) then height up to which it will rise but $ u = { ve \over 2 } $ $ \therefore h = { R \over \left( { Ve \over Ve/2 } \right)^2 -1 } $ $ { R \over 4-1} = R/3 $

change in potential energy in displacing a body from $r_1$ and $r_2$ is given by $ \triangle U = GMm \left[ { 1 \over r_1 } - { 1 \over r_2 } \right] = GMm \left[ { 1 \over 2R } - { 1 \over 3R } \right] = { GMm \over 6R} $

$ K.E = { GMm \over 2R } \therefore K.E \alpha { 1 \over R} $

this should be equal to escape velocity i.e = $ \sqrt { 2gR } $

Escape velocity does not depend on the mass of the projectile

$ { V_p \over V_e } = \sqrt { {g_p \over g_e } . { R_p \over R_e} } = \sqrt { 2 \times 2 } = 2 $ $ \Rightarrow V_e = 2 V_e = 2 \times 11.2 = 22.4 km^{-1} $

$Ve = \sqrt { 2GM \over R} = R \sqrt { 8/3 \pi Gp } $ $ \therefore Ve \alpha R if \rho constant $ since the planet is having double radius incomparision to earth therefore the escape velocity becomes twice i.e. $ 22 kms^{-1}$

The escape velocity ( u) from the Earth's gravitational field is given by the formula $v = \\sqrt{2gr}$. Here, g is the acceleration due to gravity and r is the radius of the Earth. This formula is derived from equating the kinetic energy required to escape Earth's gravity to the gravitational potential energy.

Because it does not depend on the mass of projectile