4.11 Projection in gravitational field  (Page 3/5)

Now, it is clear from the definition of binding energy itself that the initial kinetic energy of the projection should be equal to the binding energy of the body in order that it moves out of the gravitational influence. Now, the body to escape is at rest before being initiated in projection. Thus, its binding energy is equal to potential energy only.

Therefore, kinetic energy of the projection should be equal to the magnitude of potential energy on the surface of Earth,

$$\frac{1}{2}m{v}_e^2=\frac{GMm}{R}$$

where “ $v_e$ ” is the escape velocity. Note that we have used “R” to denote Earth’s radius, which is the distance between the center of Earth and projectile on the surface. Solving above equation for escape velocity, we have :

$$\Rightarrow v_e=\sqrt{\left(\frac{2GM}{R}\right)}$$

2. conservation of mechanical energy :

The act of putting a body into interstellar space is equivalent to taking the body to infinity i.e. at a very large distance. Infinity, as we know, has been used as zero potential energy reference. The reference is also said to represent zero kinetic energy.

From conservation of mechanical energy, it follows that total mechanical energy on Earth’s should be equal to mechanical energy at infinity i.e. should be equal to zero. But, we know that potential energy at the surface is given by :

$$U=-\frac{GMm}{R}$$

On the other hand, for body to escape gravitational field,

$$K+U=0$$

Therefore, kinetic energy required by the projectile to escape is :

$$\Rightarrow K=-U=\frac{GMm}{R}$$

Now, putting expression for kinetic energy and proceeding as in the earlier derivation :

$$\Rightarrow v_e=\sqrt{\left(\frac{2GM}{R}\right)}$$

3: final velocity is not zero :

We again use conservation of mechanical energy, but with a difference. Let us consider that projected body of mass, “m” has initial velocity “u” and an intermediate velocity, “v”, at a height “h”. The idea here is to find condition for which intermediate velocity ,”v”, never becomes zero and hence escape Earth’s influence. Applying conservation of mechanical energy, we have :

$$E_i=E_f$$

$$K_i+U_i=K_f+U_f$$

$$\frac{1}{2}m{u}^2-\frac{GMm}{R}=\frac{1}{2}m{v}^2-\frac{GMm}{R+h}$$

Rearranging,

$$\Rightarrow \frac{1}{2}m{v}^2=\frac{1}{2}m{u}^2-\frac{GMm}{R}+\frac{GMm}{R+h}$$

In order that, final velocity (“v”) is positive, the expressions on the right should evaluate to a positive value. For this,

$$\Rightarrow \frac{1}{2}m{u}^2\ge \frac{GMm}{R}$$

For the limiting case, u = $v_e$ ,

$$\Rightarrow v_e=\sqrt{\left(\frac{2GM}{R}\right)}$$

Interpreting escape velocity

These three approaches to determine escape velocity illustrates how we can analyze a given motion in gravitational field in many different ways. We should be aware that we have determined the minimum velocity required to escape Earth’s gravity. It is so because we have used the expression of potential energy, which is defined for work by external force slowly.

However, it is found that the velocity so calculated is good enough for escaping gravitational field. Once projected body achieves considerable height, the gravitational attraction due to other celestial bodies also facilitates escape from Earth's gravity.

Further, we can write the expression of escape velocity in terms of gravitational acceleration (consider $g=g_0$ ),

$$g=\frac{GM}{r^2}$$

$$\Rightarrow \frac{GM}{r}=gr$$

Putting in the expression of escape velocity, we have :

$$\Rightarrow v_e=\sqrt{\left(\frac{2GM}{R}\right)}=\sqrt{\left(2gR\right)}$$

Escape velocity of earth

In the case of Earth,