Question

The escape velocity for a body projected vertically upwards from the surface of the earth is 11.2 km/s. If the body is projected in a direction making an angle of 45$^\circ$ with vertical, the escape velocity will be
${\text{A}}{\text{. 11}}{\text{.2}}km/s \\\ {\text{B}}{\text{. 11}}{\text{.2}} \times \sqrt 2 km/s \\\ {\text{C}}{\text{. 11}}{\text{.2}} \times {\text{2}}km/s \\\ {\text{D}}{\text{. 11}}{\text{.2/}}\sqrt 2 km/s \\\$

Answer 1

- Hint: The escape velocity for a body given as the square of the ratio of two times the product of the universal gravitational constant and the mass of earth to the distance of the body from the centre of the earth. There is no dependence on the angle of projection here.
Formula used:
The expression for escape velocity is given as
$V = \sqrt {\dfrac{{2GM}}{r}}$
where V is used to represent the escape velocity of the earth
G is the universal gravitational constant whose value is given as
$G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$
r represents the distance of the object from the center of the earth

Complete step-by-step answer:
The escape velocity of earth is defined as the minimum velocity with an object that must be projected from the surface of the earth so that the object escapes the gravitational force of earth enough to start revolving in a closed orbit around earth.
We are given that the escape velocity for a body projected vertically upwards from the surface of the earth is 11.2 km/s. We need to find the escape velocity if the body is projected in a direction making an angle of 45$^\circ$ with vertical.
The escape velocity is calculated using the following relation that the kinetic energy of the body should be equal to the gravitational potential energy of the earth. Therefore, we can write
$\dfrac{1}{2}m{\text{v}}_e^2 = \dfrac{{GMm}}{r} \\\ \Rightarrow {{\text{v}}_e} = \sqrt {\dfrac{{2GM}}{r}} \\\$
where ${{\text{v}}_e}$ is the escape velocity and m is the mass of the object. We notice that escape velocity does not depend on the angle of projection and is constant for every object irrespective of the angle of projection equal to 11.2 km/s. The correct answer is option A

Note: The student should also remember that the gravitational force is a conservative in nature. This means that work done in a gravitational field does not depend on the path followed. This is also the reason why the escape velocity does not depend on the angle of projection which represents a different path in the gravitational field.