4.12 Projection in gravitational field (application)

Questions and their answers are presented here in the module text format as if it were an extension of the treatment of the topic. The idea is to provide a verbose explanation, detailing the application of theory. Solution presented is, therefore, treated as the part of the understanding process – not merely a Q/A session. The emphasis is to enforce ideas and concepts, which can not be completely absorbed unless they are put to real time situation.

Representative problems and their solutions

We discuss problems, which highlight certain aspects of the study leading to the projection in gravitational field. The questions are categorized in terms of the characterizing features of the subject matter :

• Satellite
• Vertical projection
• Escape velocity

Satellite

Problem 1: A satellite of mass “m” is to be launched into an orbit around Earth of mass “M” and radius “R” at a distance “2R” from the surface. Find the minimum energy required to launch the satellite in the orbit.

Solution : The energy to launch the satellite should equal to difference of total mechanical energy of the system in the orbit and at Earth’s surface. The mechanical energy of the satellite at the surface is only its potential energy. It is given by,

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

On the other hand, satellite is placed at a total distance of R + 2R = 3R. The total mechanical energy of the satellite in the orbit is,

$$E_O=-\frac{GMm}{2r}=-\frac{GMm}{2X3R}=-\frac{GMm}{6R}$$

Hence, energy required to launch the satellite is :

$$E=E_O-E_S$$

$$\Rightarrow E=-\frac{GMm}{6R}-\left(-\frac{GMm}{R}\right)=\frac{5GMm}{6R}$$

Problem 2: A satellite revolves around Earth of radius “R” with a speed “v”. If rockets are fired to stop the satellite to make it standstill, then find the speed with which satellite will strike the Earth. Take g = 10 $m/s^2$ .

Solution : When the seed of satellite is reduced to zero, it starts falling towards center of Earth. Let the velocity with which it strikes the surface be “v’” and distance between center of Earth and satellite be “r”.

Applying conservation of energy :

$$\frac{1}{2}mv{\prime }^2-\frac{GMm}{R}=0-\frac{GMm}{r}$$

For satellite, we know that kinetic energy is :

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

Also, we can express "GM" in terms of acceleration at the surface,

$$g{R}^2=GM$$

Substituting these expressions in the equation of law of conservation of mechanical energy and rearranging,

$$\Rightarrow \frac{1}{2}mv{\prime }^2=gRm-m{v}^2$$

$$\Rightarrow \frac{1}{2}v{\prime }^2=gR-v^2$$

$$\Rightarrow v\prime =\sqrt{2\left(gR-v^2\right)}$$

$$\Rightarrow v\prime =\sqrt{\left(20R-2v^2\right)}$$

Vertical projection

Problem 3: A particle is projected with initial speed equal to the orbital speed of a satellite near Earth’s surface. If the radius of Earth is “R”, then find the height to which the particle rises.

Solution : It is given that speed of projection is equal to orbital speed of a satellite near Earth’s surface. The orbital speed of the satellite near Earth's surface is given by putting "r = R" in the expression of orbital velocity :

$$v=\sqrt{\left(\frac{GM}{R}\right)}$$

$$v^2=\frac{GM}{R}$$

Since orbital velocity is less than escape velocity, the particle is returned to the surface after attaining a certain maximum height, “h”. Applying conservation of energy, the height attained by projectile is obtained as :

$$h=\frac{v^2}{2g-\frac{v^2}{R}}$$

Substituting for “v” and “g”, we have :

$$\Rightarrow h=\frac{\frac{GM}{R}}{\frac{2GM}{R^2}-\frac{GM}{R}}$$

$$\Rightarrow h=R$$

Escape velocity

Problem 4: A particle is fired with a velocity 16 km/s from the surface Earth. Find its velocity with which it moves in the interstellar space. Consider Earth’s escape velocity as 11.2 km/s and neglect friction.

Solution : We observe here that initial velocity of the particle is greater than Earth’s escape velocity. We can visualize this situation in terms of energy. The kinetic energy of the particle is used to (i) overcome the mechanical energy binding it to the gravitational influence of Earth and (ii) to move into interstellar space with a certain velocity.

Let “v”, “ $v_e$ ” and “ $v_i$ ” be velocity of projection, escape velocity and velocity in the interstellar space respectively. Then, applying law of conservation of energy :

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

$$\Rightarrow \frac{1}{2}m{v}^2=\frac{GMm}{R}+\frac{1}{2}m{v}_i^2$$

Here, we have considered gravitational potential energy in the interstellar space as zero. Also, we know that kinetic energy corresponding to escape velocity is equal to the magnitude of gravitational potential energy of the particle on the surface. Hence,

$$\Rightarrow \frac{1}{2}m{v}^2=\frac{1}{2}m{v}_e^2+\frac{1}{2}m{v}_i^2$$

$$\Rightarrow v_i^2=v^2-v_e^2$$

The escape velocity for Earth is 11.2 km/s. Putting values in the equation, we have :

$$\Rightarrow v_i^2={\left(16\right)}^2-{\left(11.2\right)}^2$$

$$\Rightarrow v_i^2=256-125.44=130.56$$

$$\Rightarrow v_i=11.43km/s$$

Problem 5: A satellite is orbiting near surface with a speed “v”. What additional velocity is required to be imparted to the satellite so that it escapes Earth’s gravitation. Consider, g = 10 $m/s^2$ and R = 6400 km.

Solution : The orbital speed of the satellite near Earth’s surface is given by :

$$v=\sqrt{\left(\frac{GM}{R}\right)}$$

We can write this expression in terms of acceleration at the surface (g),

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

On the other hand, escape velocity is given by :

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

Hence, additional velocity to be imparted is difference of two speeds,

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

$$\Rightarrow v_e-v=\left(\sqrt{2}-1\right)\sqrt{\left(gR\right)}$$

$$\Rightarrow v_e-v=\left(\sqrt{2}-1\right)\sqrt{\left(10X6.4X{10}^6\right)}$$

$$v_e-v=3.31X{10}^{3}m/s$$