Question

A body is projected vertically upward with a speed twice of the escape speed at earth's surface .calculate the speed of the body far away from the earth .the escape velocity of earth's surface is 11.2 km/s

Answer 1

Answer:

The speed of the body far away from the earth is 19.40 m/s

Explanation:

We know that,

  Escape velocity,    $\mathrm{v}_{\mathrm{e}}=\left(\frac{2 \mathrm{GM}}{\mathrm{R}}\right)^{\frac{1}{2} }$

Given,

             $\mathrm{V}=2 \mathrm{v}_{\mathrm{e}}=2\left(\frac{2 \mathrm{GM}}{\mathrm{R}}\right)^{\frac{1}{2} }=\left(\frac{8 \mathrm{GM}}{\mathrm{R}}\right)^{\frac{1}{2} }$

By conservation of Energy,

             $\frac{-\mathrm{GMm}}{\mathrm{R}}+\frac{8 \mathrm{GMm}}{2 \mathrm{R}}=0+\frac{\mathrm{mv}^{2}}{2}$

              $v=\left(\frac{3 \times 2 \mathrm{GM}}{\mathrm{R}}\right)^{\frac{1}{2} }=\sqrt{3} \mathrm{v_e}$

              $v=\sqrt{3}\times 11.2\ =\ 19.40\ m|s$