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The escape velocity of an object from the earth depends upon the mass of earth (M), its mean density $(\rho)$, its radius (R) and gravitational constant (G), thus the formula for escape veloctiy is

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Explanation

The escape velocity ( u) can be derived from the relationship between gravitational force and the required kinetic energy to escape. Given the mass of the Earth (M), its mean density ( ho), its radius (R), and the gravitational constant (G), the formula is $ v = R \\sqrt{\\frac{8 \\pi}{3} G \\rho} $. This formula is derived from the standard escape velocity formula and the relation between density, mass, and volume.

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Two small and heavy sphere, each of mass M, are placed distance r apart on a horizontal surface the gravitational potential at a mid point on the line joining the center of spheres is

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Explanation

Gravitational potential of A at 0 = $ - { GM \over r/2 } = - { 2GM \over r } $

of B at 0 = $ - { GM \over r/2 } = - { 2GM \over r } $ Total potential at 0 = $- { 4GM \over r } $

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The escape velocity of a body from earth’s surface is Ve. The escape velocity of the same body from a height equal to 7 R from earth’s surface will be

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Explanation

$ Ve \alpha { 1\over \sqrt r } $ $ { V_1 \over V_2 } = \sqrt{ r_2 \over r_1 } = \sqrt { R+7R \over R } = 2 \sqrt 2 \Rightarrow V_2 = { V_1 \over 2 \sqrt 2 } $

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An artificial satellite is revolving round the earth in a circular orbit. its velocity is half the escape velocity. Its height from the earth surface is = km

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Explanation

$ v = \sqrt { GM \over R+ h } = {1 \over 2 } \sqrt { 2GM \over R } $ $ \therefore 4R = 2 (R+h )$ $ \therefore h = R = 6400 km$

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The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is..........

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Explanation

$ { V_p \over V_e } = \sqrt {Mp \over Me} \times { Re \over Rp } = \sqrt { 6 \times {1 \over 2 } } = \sqrt 3 $ $ \therefore Ve = \sqrt 3 Ve $

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There are two planets, the ratio of radius of two planets is k but the acceleration due to gravity of both planets are g what will be the ratio of their escape velocity.

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Explanation

$ \upsilon = \sqrt{2gR} \therefore {V1 \over V2 } = \sqrt {g1 \over g2 } .{R1\over R2} = \sqrt {gk} = (kg)^ {1/2} $

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The escape velocity of a body on the surface of the earth is 11.2 km/sec. If the mass of the earth is increases to twice its present value and the radius of the earth becomes half, the escape velocity becomes = _________$kms^{–1}$

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Explanation

$ \therefore Ve = \sqrt {2GM \over R } \therefore Ve \alpha \sqrt { M \over R } $ If M becomes double and R becomes half then escape velocity; becomes two times

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3 particle each of mass m are kept at vertices of an equilateral triangle of side L. The gravitational field at center due to these particies is ................

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Explanation

Due to three particles net intensity at the center $ \vec I = \vec I_A + \vec I_B + \vec I_C = 0 $ because out of these three intensities ARE equal in magnitude and between each other is 120’

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Escape velocity on the surface of earth is $11.2 kms^{–1}$ Escape velocity from a planet whose masse is the same as that of earth and radius 1/4 that of earth is =……….$ kms^{–1}$

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Explanation

$ Ve \alpha {1 \over \sqrt R } $ if R becomes 1/4 then Ve will be 2 times

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The velocity with which a projectile must be fired so that it escapes earth’s gravitational does not depend on ..................

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Explanation

Escape velocity does not depends upon the angle of projection

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