Cal (captain of the Harbinger)
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The formula for escape velocity comprises of a constant, G, which we refer to as the universal gravitational constant. The value of it is = 6.673 × 10-11 N . m2 / kg2. The unit for escape velocity is meters per second (m/s). Escape velocity = \(\sqrt{\frac{2 (gravitational constant) (mass of the planet of moon) }{radius of the planet or moon}}\)
as m 1. Ans: C. Escape velocity, v e = √2gR. It is independent of the mass of the particle. Thus, it will depend on m 0. Q2. The earth retains its atmosphere. This is due to: the special shape of the earth. the escape velocity which is greater than the mean speed of the atmospheric molecules.
The escape velocity from the earth is about 11 k m s − 1. The escape velocity from a planet having twice the radius and the same mean density as the earth, is The escape velocity from a planet having twice the radius and the same mean density as the earth, is
Assertion :The value of escape velocity from the surface of earth at 30 o and 60 o is v 1 = 2 v e, v 2 = 2 / 3 v e. Reason: The value of escape velocity is independent of angle of projection. Reason: The value of escape velocity is independent of angle of projection.
The escape velocity for a body projected vertically upwards from the surface of the earth is 11.2 k m s − 1. It the body is projected in a direction making an angle of 45 ∘ with vertical, the escape velocity will be
Escape velocity of a particle depends on its mass m as. A particle of mass M at rest decays into two particles of masses m1 and m2 having non-zero velocities. The ratio of the De-Broglie wavelengths of the particles λ1: λ2. Two trucks of mass M each are moving in opposite direction on adjacent parallel tracks with same velocity u.
Escape velocity is defined as the lowest velocity a body must have in order to escape the gravitational attraction of a particular planet or other object. Let say, a body is revolving in an orbit of radius r around a planet of mass M .
The escape velocity ' v ' of a body depends upon (1) the acceleration due to gravity (g) of the planet and (2) the radius (R) of the planet. Use the method of dimensions to obtain a relation between v , g , and R .
50 %. Solution. Verified by Toppr. Suppose the distance between moon and earth is r. Its current orbital velocity, therefore is V o = √GM e r; and the velocity it requires to escape from this distance is V e = √2GM e r, This implies V e = √2V o. therefore to escape the percentage increase in velocity will be: V e −V o V o ×100= √2− ...
Escape velocity from a planet is ve. If its mass is increased to 8 times and its radius is increased to 2 times, then the new escape velocity would be (1) ve (2) Vēve (3) 2V6 (4) 2V2v 100. One-fourth length of a spring of force constant 'k'is
