The escape velocity for a body projected vertically upwards from the surface of the earth is 11...

Question:

The escape velocity for a body projected vertically upwards from the surface of the earth is 11 km/s. If the body is projected at an angle of 45 degrees with the vertical, then what is the escape velocity?

Escape Velocity:

All the objects on the surface of the earth are under the influence of the earth's gravitational field. They are all bounded to the gravitational force due to the Earth. In order to escape this gravitational field, the object must have greater energy than its gravitational binding energy.

The escape velocity V is obtained by

{eq}V=\sqrt{\dfrac{2GM}{R}}\\ \rm Here:\\ \,\,\,\, \, \bullet \, G(=6.67\times10^{-11} N m^2/kg^2)\text{: gravitational constant}\\ \,\,\,\, \, \bullet \,M(=5.98\times 10^{24}\, kg)\text{: mass of the Earth}\\ \,\,\,\, \, \bullet \,R(=6378\, km) \text{: radius of the Earth} {/eq}

The escape velocity on the earth for a body is dependent on the mass and the size of the Earth, not the physical property of the body. The projected angle does not affect the speed. Escape velocity is not dependent on direction. In addition, the formula is the result when the kinetic energy of the body at launch is balanced with the potential energy change on leaving earth's gravity. Kinetic energy and potential energy are not vectors. The escape speed of the body with 45 degrees angle is 11.2 km/s.