A particle of mass ‘m’ is projected with a velocity v=kVe(k < 1) from the surface of the earth. (Ve=escape velocity) The maximum height above the surface reached by the particle is
\(\frac{Rk^2}{1-k^2}\)
\(R(\frac{k}{1-k})^2\)
\(R(\frac{k}{1+k})^2\)
\(\frac{R^2k}{1+k}\)
The Correct Option is A
Solution and Explanation
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Concepts Used:
Escape Speed
Escape speed is the minimum speed, which is required by the object to escape from the gravitational influence of a plannet. Escape speed for Earth’s surface is 11,186 m/sec.
The formula for escape speed is given below:
ve = (2GM / r)1/2
where ,
ve = Escape Velocity
G = Universal Gravitational Constant
M = Mass of the body to be escaped from
r = Distance from the centre of the mass