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        A particle is thrown with escape velocity v from the surface of Earth calculate its velocity at height 3R
one year ago

Arun
18691 Points
							Dear Raja V = sqrt (GM/2R) V = escape velocity/2V = 11.2/2V = 5.6 km/sec  RegardsArun (askIITians forum expert)

one year ago
Sunder Muthukumaran
28 Points
							At the instant , where the particle is at height 3R , following things are taken into concern
The total mechanical energy is conserved , that is

The KE required by the particle to reach height 3R , is equal to the GPE ( Gravitational potential energy) at the point 3R from the surface of earth

Therefore ,
The total height = height from surface + radius of earth
= 3R + R = 4R
Then , we have ;
KE = GPE
½mVA2  =  GMm/4R  ( VA = Velocity at 3R from surface )
Solving for VA ,
We get :  VA  = $\sqrt{2GM/4R}$
= $\sqrt{GM/2R}$ $(\sqrt{2} / \sqrt{2})$ ( Multiply and divide $\sqrt{2}$ )
=$\sqrt{\frac{2GM}{R}} ( \frac{1}{2})$

=$\frac{_{Ve}}{2}$ ( Ve  = Escape velocity of earth

=  $\frac{11.2 Km/s}{2}$

= 5.6 Km/s

one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions