Question

The escape velocity for a body projected vertically upwards from the surface of the earth is $11.2km{s^{ - 1}}.$ if the body is projected in a direction making an angle ${45^0}$ with the vertical, the escape velocity will be
(A). $\dfrac{{11.2}}{{\sqrt 2 }}km{s^{ - 1}}$
(B). $11.2 \times \sqrt 2 km{s^{ - 1}}$
(C). $11.2 \times 2km{s^{ - 1}}$
(D). $11.2km{s^{ - 1}}$

Answer 1

Hint: To solve this question, we will use the concept of gravitational potential energy and kinetic energy. As we know If constituents of a system are separated by infinite distance then their interaction gravitational potential energy is zero. If the total mechanical energy of a system is greater than or equal to zero then the system may go to infinity and thus the system is unstable.

Formula used:
$K.E = \dfrac{1}{2}m{v^2} \\\ P.E = - \dfrac{{GrMm}}{R} \\\$
Where
m is the mass of object
v is the velocity of the object
R is the radius of the earth
G is the gravitational constant.
M is the mass of the earth

Complete step-by-step solution -
Escape velocity-It is defined corresponding to a planet or satellite as the minimum speed given to a body from the surface of the planet so that the body goes beyond the gravitational field of the planet.
For calculating escape velocity, we will equate the total mechanical energy equal to zero.
The total mechanical energy or TME = 0
$\Rightarrow K.E + P.E = 0 \\\ \Rightarrow \dfrac{1}{2}m{v^2} + \left( { - \dfrac{{GrM}}{R}m} \right) = 0 \\\ \Rightarrow v = \sqrt {\dfrac{{2GM}}{R}} \\\$
Substituting the values of mass of earth = $5.972{\text{ }} \times {10^{24}}kg$ , radius of the earth = $6.37 \times {10^6}m$ and gravitational constant $= 6.673 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$ , we will get the escape velocity of earth.

$\Rightarrow v = \sqrt {\dfrac{{2 \times 5.972 \times {{10}^{24}}kg \times 6.673 \times {{10}^{ - 11}}N{m^2}k{g^{ - 2}}}}{{6.37 \times {{10}^6}m}}} \\\ \Rightarrow v = 11.2\sqrt 2 km{s^{ - 1}} \\\$

Since it’s mainly related to energy in the form of kinetic energy and gravitational potential energy .Thus it is a scalar quantity which does not depend upon the angle of projection.
Hence, the correct option is “B”.

Note: If we throw the object with this escape velocity with any angle of projection then possibly it will not return back again to us. The value of escape velocity depends on the distance to the center of the source of gravity. When we talk about the “escape velocity of the Earth”, we generally mean from the surface of the Earth, about 6400 km from the center of the Earth.