CBSE Class 11-science Answered
derive an expression for escape velocity and orbital velocity
Asked by vedavidyasvy | 30 Sep, 2018, 01:56: PM
Expert Answer
Gravitational force of attraction provides the centripetal force for the earth to revolve around sun.
Hence we have
where G is gravitational constant, M is mass of Sun, m is mass of earth, R is orbital radius of earth and v is orbital velocity.
from eqn.(1), we get
Escape velocity :-
Suppose the vertically projected object reach infinity. Let its speed at infinity is vf .
The energy of an object is sum of potential and kinetic energy.
Let W∞ denotes the gravitational potential energy of the object at infinity. The total energy of the projectile at infinity is
E(∞) = W∞ + (1/2)m×vf2 ......................(1)
If the object is thrown initially with a speed vi from earth surface, its initial total energy is
E(R) = (1/2)m×vi2 - [ G×M×m / R ] + W∞ ............... (2)
By energy conversion, (1) and (2) are same.
(1/2)m×vi2 - [ G×M×m / R ] = (1/2)m×vf2 ....................(3)
eqns.(1) and (2) are equated to get eqn.(3) by cancelling out W∞ on both sides
RHS of eqn.(3) is always greater than or equal to zero to make vi as escape velocity
hence we get
hence we get escape velocity as
Answered by Thiyagarajan K | 04 Oct, 2018, 01:02: AM
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