The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.
Given: The escape speed of a projectile on the earth’s surface is 11 .2 kms –1 .
Total energy of the projectile at the Earth’s surface is given as,
E i = 1 2 m v p 2 − 1 2 m v esc 2
Where, v esc is the escape velocity of the projectile, v p is the velocity of the projection of the projectile and m is the mass of particle.
Total energy of the projectile far away from the Earth’s surface is given as,
E f = 1 2 m v f 2
Where, v f is the velocity of the projectile far away from the earth.
From the law of conservation of energy,
E i = E f 1 2 m v p 2 − 1 2 m v esc 2 = 1 2 m v f 2
Solving further for v f , we get
v f = ( v p 2 − v esc 2 ) = ( 3 v esc ) 2 − v esc 2 = 8 v esc 2 = 8 v esc
By substituting the given values in the above expression, we get
v f = 8 × 11.2 = 31.7 kms -1
Thus, speed of the body far away from the earth is 31 .7 kms -1 .
Given: The escape speed of a projectile on the earth’s surface is
Total energy of the projectile at the Earth’s surface is given as,
Where,
Total energy of the projectile far away from the Earth’s surface is given as,
Where,
From the law of conservation of energy,
Solving further for
By substituting the given values in the above expression, we get
Thus, speed of the body far away from the earth is
