derive an expression for orbital velocity and relate it with the escape velocity
Answers
Explanation:
Gravitational force of attraction provides the centripetal force for the earth to revolve around sun.
Hence we have begin mathsize 12px style G fraction numerator M cross times m over denominator R squared end fraction space equals space 1 half space fraction numerator m cross times v squared over denominator R end fraction space............ left parenthesis 1 right parenthesis end style
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 begin mathsize 12px style v space equals space square root of fraction numerator G cross times M over denominator R end fraction end root end style
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
Explanation:
for a satellite revolving around the earth.... centripetal force must act on it. this centripetal force is provided by earth
therefore,
mv*2/r=GMeMs/Re+h
v*2=GMe/Re+h
v=square root of GMe/Re+h
where Me-mass of the earth
Re-radius of the earth