Question

A particle of mass $m$ is projected from the surface of earth with a speed $V_0$ ($V_0$ < escape velocity). The speed of the particle at a height h$=$R (radius of the earth) is

• A. $\sqrt{gR}$
• B. $\sqrt{V_0^2-2gR}$
• C. $\sqrt{V_0^2-gR}$
• D. none of these

Answer 1

The correct option is C $\sqrt{V_0^2-gR}$

According to the law of conservation of energy

$(U+K.E{)}_{surface}=(U+K.E{)}_{height}$

$=>\frac{-GMm}{R}+\frac{1}{2}{m}_1{V}_0^2=\frac{-GMm}{R+h}+\frac{1}{2}m{V}^2$

$=>-mgR+\frac{1}{2}{m}_1{V}_0^2=\frac{-mgR}{2}+\frac{1}{2}m{V}^2$ $(\cdot .\cdot h=R)$

$=>\frac{1}{2}m{V}^2=\frac{1}{2}m{V}_0^2-\frac{1}{2}mgR$

$=V^2=V_0^2-gR=V=\sqrt{V_0^2-gR}$