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प्रश्न
Answer the following question in detail.
Define escape speed.
उत्तर
The minimum velocity with which a body should be thrown vertically upwards from the surface of the Earth so that it escapes the Earth’s gravitational field is called the escape velocity (ve) of the body.
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संबंधित प्रश्न
Answer the following question.
On which factors does the escape speed of a body from the surface of Earth depend?
Answer the following question in detail.
Derive an expression for the escape speed of an object from the surface of each.
Answer the following question in detail.
If the Earth were made of wood, the mass of wooden Earth would have been 10% as much as it is now (without change in its diameter). Calculate escape speed from the surface of this Earth.
A body is thrown from the surface of the earth velocity v `"m"/"s"`. The maximum height above the earth's surface upto which it will reach is ____________.
(R = radius of earth, g = acceleration due to gravity)
Earth has mass M1 and radius R1. Moon has mass M2 and radius R2. Distance between their centre is r. A body of mass M is placed on the line joining them at a distance `"r"/3` from centre of the earth. To project the mass M to escape to infinity, the minimum speed required is ______.
A body is projected vertically from the surface of the earth of radius 'R' with velocity equal to half of the escape velocity. The maximum height reached by the body is ______.
The escape velocity from earth is 11.2 km. Another planet is having mass 1000 times and radius 10 times that of earth, then escape velocity at that planet will be ____________.
A hole is drilled half way to the centre of the earth. A body weighs 300 N on the surface of the earth. How much will, it weigh at the bottom of the hole?
The magnitudes of the gravitational potentials at distances r1 and r2 from the centre of a uniform sphere of radius R and mass M are V1 and V2 respectively. Then ______.
The gravitational potential energy of a rocket of mass 200 kg at a distance 106 m from the centre of the earth is 3 × 108 J. The weight of the rocket at a distance 108 m from the centre of the earth is ______.
What should be the velocity of earth due to rotation about its own axis so that the weight at equator becomes `(3/5)^"th"` of initial value?
(Radius of earth on equator = 6400 km, `g=10m/s^2,cos0^circ=1`)
A body is projected vertically upwards from earth's surface. If velocity of projection is `(1/3)^"rd"` of escape velocity, then the height upto which a body rises is ______.
(R =radius of earth)
A planet has radius `1/4`th of the radius of earth and acceleration due to gravity double than that of the earth. Then the ratio of escape velocity on the surface of planet to that on the earth's surface will be ______.
Two particles of masses m and 9m are separated by a distance r. At a point on the line joining them the gravitational field is zero. The gravitational potential at that point is ______.
( G = universal constant of gravitation)
The escape velocity on the surface of the earth is 11.2 kms-1. If mass and radius of a planet is 4 and 2 times of the earth respectively, then what is the escape velocity from the planet?
The ratio of the radii of the planets P1 and P2 is k. The ratio of acceleration due to gravity is r. The ratio of the escape velocities from them will be ______.
