# A particle is projected upward from the surface of earth(radius R) with a K.E. equal to half the minimum value needed for it to escape from earth's gravity. To what height does it rises above the surface of earth?

1
by manojSen878

2014-12-22T02:15:55+05:30

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Total energy of the particle on surface of Earth
= KE + PE = KE - G Me m / Re

KE needed for escaping from the gravity of Earth = G Me m /Re.  Then the total energy of the particle will be zero. Hence, it will escape from the gravitational field.

Energy given to the particle = 1/2 * G Me m / Re

Total energy of the particle on surface of Earth = KE + PE = -1/2 G Me m / Re

Total energy of a particle in an orbit (radius R) around Earth has a total energy =
- G Me m / R + 1/2 G Me m /R  = - 1/2 * G Me m / R

Comparing the two above equations and as per the principle of conservation of energy,    The radius of the orbit = R = Re.

hence, the particle stays on the surface of Earth.