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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

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Explanation

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.

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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 ...............

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Explanation

$ 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 $

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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)

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Explanation

$ g' = \left( { R \over R + 2R } \right) = g /9 $

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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

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Explanation

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 } $

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In a gravitational field, at a point where the gravitational potential is zero

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Explanation

$ I = - { dv \over dx } $

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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.

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Explanation

$ 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 $

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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...............

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Explanation

$ 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 } $

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A rocket is launched with velocity $10 kms^{-1}$. If radius of earth is R then maximum height attained by it will be = ..............

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Explanation

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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What is the intensity of gravitational field at the center of spherical shell

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Explanation

The intensity of the gravitational field inside a spherical shell is zero. This can be understood by applying the shell theorem, which states that a uniform spherical shell of mass exerts no net gravitational force on a particle located inside it. Hence, the gravitational field intensity at the center of the spherical shell is zero.

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A body of mass m kg starts falling from a point 2R above the earth’s surface. Its K.E. when it has fallen to a point ‘R’ above the Earth’s surface = ..................... [R - Radius of Earth, M-mass of Earth G-Gravitational constant]

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Explanation

$ P.E U = - { GMm \over r } = { GMm \over R+ h } $ $ U _ { initial } = - { GMm \over 3R } and U_{ final } = -{ GMm \over 2r } $ $ loss of P.E = gain in K.E = { GMm \over 2 R } - { GMm \over 3R } = { GMm \over 6R } $

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