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Question : 18 of 25
Marks: +1, -0
The escape speed of a projectile on the earth’s surface is 11.2 km s−1\text{km s}^{-1}. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.
Solution:  
Using law of conservation of energy
12mv02=12mv2−12mvE2\frac{1}{2} m v_0^2 = \frac{1}{2} m v^2 - \frac{1}{2} m v_E^2
⇒v0=v2−vE2\Rightarrow v_0 = \sqrt{v^2 - v_E^2}
=(3vE)2−vE2=8vE= \sqrt{(3 v_E)^2 - v_E^2} = \sqrt{8} v_E [Given v=3vE][ \text{Given } v = 3 v_E ]
=8Ă—11.2=31.68 km s−1= \sqrt{8} \times 11.2 = 31.68 \text{ km s}^{-1}
Here, v0=v_0 = speed of projectile when far away from the earth
v=v = velocity of projection of the body, vE=v_E = escape velocity
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