UPSEE Solved Model Paper 6

Here, we have to find out the velocity of a principle on reaching the earth if it falls from infinity We know that Kinetic energy on the surface KE = $\frac{1}{2} m v^2$ Potential energy on the surface PE = $\frac{-GM}{R}$ Where: m - mass of the particle M - mass of the earth v - velocity G - universal gravitational constant R - radius of earth since the total enegy at infinity is zero KE + PE = 0 ∴ $\frac{1}{2} m v^2 + \frac{-GMm}{R}$ = 0 Or, $\frac{1}{2} m v^2$ = $\frac{GMm}{R}$ $v^2$ = $\frac{2GM}{R}$ ∴ Velocity , v = $\sqrt{\frac{2GM}{R}}$ And also we know that g is acceleration at the surface of earth And its equation on gravitation g = $\frac{GM}{R^2}$ So we can modify the equation of velocity v = $\sqrt{\frac{2GM}{R} \times \frac{R}{R}}$ = $\sqrt{\frac{2RGM}{R^2}}$ Putting the value of g = $\frac{GM}{R^2}$ on the above equation of velocity we get ∴ Velocity v = $\sqrt{2Rg}$ That iis the velocity of a particle on reaching the earth if it falls from infinity is $\sqrt{2Rg}$