4.11 Projection in gravitational field  (Page 5/5)

Lesser massive star becomes neutron star instead of black hole. Even neutron star has very high gravitational field. We can realize this by calculating escape velocity for one such neutron star,

$$M=3X{10}^{30}kg$$

$$R=3X{10}^{4}m$$

$$\Rightarrow v_e=\sqrt{\left(\frac{2X6.67X{10}^{-11}X3X{10}^{30}}{3X{10}^4}\right)}$$

$$\Rightarrow v_e=1.1X{10}^{5}m/s$$

It is quite a speed comparable with that of light. Interstellar Black hole is suggested to be 5 times the mass of neutron star and 10 times the mass of sun! On the other hand, its dimension is in few kilometers. For this reason, following is possible :

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

where “c” is the speed of light. Hence even light will not escape the gravitational force of a black hole as the required velocity for escape is greater than speed of light.

Nature of trajectory

In this section, we shall attempt to analyze trajectory of a projectile for different speed range. We shall strive to get the qualitative assessment of the trajectory – not a quantitative one.

In order to have a clear picture of the trajectory of a projectile, let us assume that a projectile is projected from a height, in x-direction direction as shown. The point of projection is, though, close to the surface; but for visualization, we have shown the same at considerable distance in terms of the dimension of Earth.

Let “ $v_O$ ” be the speed of a satellite near Earth’s surface and “ $v_e$ ” be the escape velocity for Earth’s gravity. Then,

$$v_O=\sqrt{\left(gR\right)}$$

$$v_e=\sqrt{\left(2gR\right)}$$

Different possibilities are as following :

1 : $v=0$ : The gravity pulls the projectile back on the surface. The trajectory is a straight line (OA shown in the figure below).

2 : $v<v_C$ : We denote a projection velocity “ $v_C$ ” of the projectile such that it always clears Earth’s surface (OC shown in the figure below). A limiting trajectory will just clear Earth’s surface. If the projection velocity is less than this value then the trajectory of the projectile will intersect Earth and projectile will hit the surface (OB shown in the figure below).

3 : $v_C<v<v_O$ : Since projection velocity is greater than limiting velocity to clear Earth and less than the benchmark velocity of a satellite in circular orbit, the projectile will move along an elliptical orbit. The Earth will be at one of the foci of the elliptical trajectory (see figure above).

4 : $v=v_O$ : The projectile will move along a circular trajectory (see inner circle in the figure below).

5 : $v_O<v<v_e$ : The projection velocity is greater than orbital velocity for circular trajectory, the path of the projectile is not circular. On the other hand, since projection velocity is less than escape velocity, the projectile will not escape gravity either. It means that projectile will be bounded to the Earth. Hence, trajectory of the projectile is again elliptical with Earth at one of the foci (see outer ellipse in the figure above).

6 : $v=v_e$ : The projectile will escape gravity. In order to understand the nature of trajectory, we can think of force acting on the particle and resulting motion. The gravity pulls the projectile in the radial direction towards the center of Earth. Thus, projectile will have acceleration in radial direction all the time. The component of gravity along x-direction is opposite to the direction of horizontal component of velocity. As such, the particle will be retarded in x-direction. On the other hand, vertical component of gravity will accelerate projectile in the negative y – direction.

However, as the projection speed of the projectile is equal to escape velocity, the projectile will neither be intersected by Earth’s surface nor be bounded to the Earth. The resulting trajectory is parabola leading to the infinity. It is an open trajectory.

7 : $v>v_e$ : We can infer that projection velocity is just too great. The impact of gravity will be for a very short duration till the projectile is close to Earth. However, as distance increases quickly, the impact of gravitational force becomes almost negligible. The final path is parallel to x-direction.