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2014-11-21T04:21:27+05:30

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The orbital velocity of a satellite body and its escape velocity to escape out of Earth's gravitational field are equal.

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Let v
 be the orbital (linear) velocity of a body revolving around the Earth, in a circular orbit of radius R.

     Kinetic energy = 1/2 m v
₀²,
     Potential energy in Earth's gravitational field at distance R = - G Me m / R
     where,  m = mass of the body (or satellite),
               Me = mass of Earth,      G = Universal Gravitational Constant

The centripetal force for the body in the orbit is supplied by the gravitational force.  Hence,
 
                  m v
₀² / R = G Me m / R²          =>   v₀² = G Me / R  ---- equation 1

       =>  K.E. = 1/2 m v
₀² = G Me m / 2 R   =   - P.E./ 2 

      Total energy at radius R = KE+PE = - G Me M / 2 R     --- equation 2

Suppose now, this body is given an additional velocity v (perpendicular to the orbit and along the radius) such that it goes to a distance d from center of Earth.   Since the total mechanical energy is conserved by the gravitational force,  the energy at a distance d from the center of Earth is given by:

E=\frac{1}{2}mv_d^2-\frac{GM_em}{d}=\frac{1}{2}mv^2-\frac{GM_em}{2R}\\\\For\ d=>\infty,\ and\ v_{\infty}=nearly\ 0,\ minimum\ velocity\ v_e\ needed:\\\\ \frac{1}{2}mv_e^2=\frac{GM_em}{2R}\\\\v_e=\sqrt{\frac{GM_e}{R}}=v_0\\

The escape velocity of a satellite is the velocity (along radius) required to send it away into the space, just manages to travel to infinite distance.  It is equal to the orbital velocity.


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