Question

The escape velocity of a body depends upon its mass as:

& \text{A}\text{. }{{\text{m}}^{\text{0}}} \\\ & \text{B}\text{.}{{\text{m}}^{\text{1}}} \\\ & \text{C}\text{.}{{\text{m}}^{\text{2}}} \\\ & \text{D}\text{.}{{\text{m}}^{\text{3}}} \\\ \end{aligned}$$

Answer 1

The escape velocity of an object like a satellite, is the minimum speed which the object must acquire in order to escape or move an infinite distance from the gravitational force of a body like earth, from where the object is projected.
Formula used: $v_{e}=\sqrt{\dfrac{2GM}{r}}$ where $G$ is the gravitational constant, $M$ is the mass of the body and $r$ is the distance of the object from the center of the body.

Complete step-by-step solution:
Escape velocity is also defined as the speed at which the magnitude kinetic energy of the object is equal to the gravitational potential energy of the body. The body which achieves this escape velocity in nowhere close to the orbit of the body. This velocity of the object keeps decreasing as it moves away from the body, it goes close to zero but doesn’t become zero.
It is given by $v_{e}=\sqrt{\dfrac{2GM}{r}}$where $G$ is the gravitational constant, $M$ is the mass of the body and $r$ is the distance of the object from the center of the body.
Thus escape velocity is a function of the mass of the body and the distance of the object from the center of the body.
Here, since the question says, the escape velocity of a body depends upon its mass as, i.e. the mass of the body which is projected. But we know that the escape velocity is independent of the mass of the body.
Hence the answer is $A.m^{0}$.

Note: The escape velocity is just enough to push needed by the object to continuously move away from the orbit of the body. No further force is needed to keep the object away from the orbit.
The escape velocity is a constant for the body from where the object is projected.
For example, the escape velocity of the earth’s surface is $11,186 m/s$.