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If density of earth increased 4 times and its radius becomes half of then out weight will be
$ g \alpha \rho R $
A man can jump to a height of 1.5 m on a planet A what is the height ne may be able to jump on another planet whose density and radius are respectively one- quater and one- third that of planet A
$ H = { V^2 \over 2 g } \Rightarrow H \alpha { 1 \over g } \Rightarrow {H_B \over H_A } = { g_A \over g_B} $ Now $ g_B = { g_A \over 12} as g \alpha \rho R $ $ \therefore { H_B \over H_A } = { g_A \over g_B} =12 \Rightarrow H_B = 12 H_A = 12 \times 1.5 = 18 m $
If the value of ‘g’ acceleration due to gravity, at earth surface fis $10ms^{–2}$. its value in $ms^{–2}$ at the center of earth, which is assumed to be a sphere of Radius ‘R’ meter and uniform density is
At the center of the Earth, the value of gravitational acceleration (g) is zero. This is because the gravitational forces exerted by the mass of the Earth in all directions balance out, resulting in no net gravitational force. Therefore, the acceleration due to gravity at the center is zero.
Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If the ratio of densities of earth $ \rho_e $ and moon $ \rho_m $ is $ { \rho_e \over \rho_m } = 5/3 $then radius of moon Rm in terms of Re will be ...............
$ g = {4 \over 3 } \pi \rho GR \Rightarrow g \alpha \rho R \Rightarrow { g_e \over g_m } = { \rho_e \over \rho_m } . { R_e \over R_m } $ $ \Rightarrow { 6 \over 1 } = { 5\ over 3 } . { Re \over Rm } $ $ \Rightarrow Rm = { 5 \over 18 } Re $
The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from the surface of the earth is = ............... (g = acceleration due to gravity at the surface of earth)
$ g' = \left( { R \over R + 2R } \right) = g /9 $
A spherical planet has a mass Mp and diameter Dp A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity, equal to
Gravitational attraction fore on particle B $ F_g = { GMp m \over (Dp /2 )^2 } $ Acceleration of paritcle due to particle due to gravity $ a = {Fg \over m } = { 4GMp \over Dp^2 } $
In a gravitational field, at a point where the gravitational potential is zero
$ I = - { dv \over dx } $
The mass of the earth is $6.00 \times 10^{24} kg $ and that of the moon is $7.40 \times 10^{22} kg $. The constant of gravitation$ G = 6.67 \tmies 10^{-11} Nm^2 kg^{–2}$. The potential energy of the system is $ -7.79 \times 10^{28} Joules$. the mean distance between the earth and moon is = meter.
$ U = - { GMm \over r } $ $ 7.79 \times 10^{28} = { 6.67 \times 10^{11} \times 7 \times 10^{22} \times 6 \times 10^{24} \over r } $ $ \therefore r = 3.8 \times 10^8 m $
The masses and radii of earth and moon are $M_1$, $R_1$ and $M_2$, $ R_2$
respectively. Their centres are d distance of apart. The minimum velocity with which a particle of mass m should be projected from a point midway between their centres so that it esacapes to infinity is...............
$ V = - { GM_1 \over d/2 } + -{GM_2 \over d/2 } $ $ Now P.E = mv = -{2 GM \over d} (m_1 + m_2 ) (m --> mass of particle) $ K.E = P.E $ \Rightarrow { 1 \over 2 } mv^2 = { 2GM \over d } (m_1 + m_2 ) $ $ \Rightarrow v = 2 \sqrt { G (M_1 + M_2 ) \over d } $
A rocket is launched with velocity $10 kms^{-1}$. If radius of earth is R then maximum height attained by it will be = ..............
If the body is projected with velocity $ \upsilon ( \upsilon \lt Ve ) $ then height up to $ h = { R \over { Ve^2 \over V^2} -1 }= { R \over \left( { 11.2 \over 10} \right)^2} =4 R (approx ) $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.