Question

29.An object of mass 'm' at rest on earths surface.Escape speed of this object is V_e
same object is orbiting earth with h=r then escape speed is V¹_e. then----

Answer 1

$\begin{array}{l}\text{Escape velocity for a spherically symmetric object at a distance r from its}\\ \text{center is given by}\\ V_c=\sqrt{\frac{2GM}{r}}\\ \text{At surface of the earth thus, the escape velocity becomes}\\ V_c=\sqrt{\frac{2GM}{R}}\text{where R is the radius of the earth}\\ \text{Now if the height from the surface of the earth h = R,}\\ \text{distance from the center becomes = R+R = 2R}\\ \text{Thus, escape velocity becomes}\\ V_c^1=\sqrt{\frac{2GM}{2R}}=\frac{1}{\sqrt{2}}\sqrt{\frac{2GM}{R}}=\frac{1}{\sqrt{2}}{V}_c\end{array}$