 Hello and welcome. In the last lecture, we had started the discussion on the gravity turn trajectory and in that context, we had looked at one specific case of constant pitch rate based gravity turn trajectory solutions and we found that these were quite easily obtained as well as useful. In this lecture, we will look at the next concept of gravity turn trajectory where we achieve a constant velocity along the trajectory and still achieve a reasonable amount of turning of the trajectory through appropriate modulation of the thrust and the mass. So, let us begin. So, let us look at the various aspects of a constant velocity solution which will give us a reasonable amount of trajectory curvature and will also provide useful information. So, let us first understand why we would need such a scenario. While we see that gravity turn is a mechanism by which we can achieve the trajectory curvature and inclination right after the lift off, there is a phase of the trajectory called the terminal phase where the vehicle is out of the atmosphere and also has achieved a very high velocity, but it is possible that it may not have the desired inclination for injection. So, in such a case, it is generally possible to employ a constant velocity gravity turn trajectory which helps us to achieve the desired inclination while maintaining the velocity at the desired level. Of course, as we will see, these solutions are simpler to obtain as well as implement. Let us now look at the basic formulation and the applicable equations for the constant V case. So, let us go back to our gravity turn equations and put the constraint that B equal to B naught a constant. We immediately see that our theta dot equilibrium now has B naught which is a constant value in the denominator and we can now convert this directly into an equation between theta and t as 1 by sin theta d theta equal to g theta by V naught into dt. We will see that integration of this equation will directly give theta as an explicit solution for time. So, this is one benefit that we get for constant velocity constraint. Further, when we look at the velocity equation, the V dot becomes 0 because it is a constant and we now have the equation that minus m dot g naught i sp by m minus g tilde cos theta equal to 0. This equation now we can rewrite as an equation in terms of mass and theta. So, now you will find that we can obtain mass as an explicit function of theta through this differential equation as minus d m by m equal to V naught cot theta by g naught i sp t theta. So, you realize that the original two equations under the constraint that V is V naught are modified into the two equations one an explicit equation for theta in terms of time and the other equation for mass explicitly in terms of theta. We realize that in this context for the mass solution theta becomes the primary parameter and that is an important benefit particularly in the context where we have a requirement on inclination to be achieved at a given velocity. By using the value of those inclinations, we can directly estimate the amount of mass or the propellant that will be required for that particular inclination to be achieved and that is a great design benefit. Now, of course, we can solve these two differential equations. So, first let us solve theta as a function of t. So, it is a straight forward trigonometric integral equation and we can show that the time will be V naught by g theta ln of tan theta by 2 plus of course, a set of integration to be determined from initial condition which can be transformed into the delta t solution as given below. Now, this solution of course as we see is bidirectional. So, it can be worked in two ways. If you provide an inclination that is desired, this particular equation will tell you how long it will take at that velocity to complete the maneuver or the relation can be reversed and can be said that at that velocity if the maneuver is to be achieved in a certain time, what is the angle that can be obtained and if the final angle is specified, the same relation can also be used to determine the starting theta naught from which this maneuver should be carried out. And you will realize that this becomes an extremely useful design information. Of course, we also see that a higher delta t would generally provide a higher theta. But again, that will depend upon V naught. So, if you have a larger V naught, then we can appropriately modify this relation. Next, let us now look at the solution of mass as a function of theta. So, we go back to the integral equation dm by m equal to minus V naught by g naught isp cos theta by sin theta d theta, which is another simple trigonometric integral which can be obtained by substitution of sin theta as x and cos theta d theta is dx and because of which we can directly solve for this as ln m equal to minus V naught by g naught isp ln sin theta plus the constant of integration. We can do a little bit of algebraic manipulation on this expression and we can show that mass as a function of theta is the constant of integration k into sin theta to the power minus V naught by g naught isp. When we substitute the initial condition that at t equal to 0 theta equal to theta naught, we can obtain k as this expression so that we cannot talk about the mass fraction m by m naught as sin theta by sin theta naught to the power minus V naught by g naught isp. Now, we clearly see that for a given V naught and isp, which means a given velocity and a given propellant, if you want a higher correction in theta, you will need to have a higher propellant because you will get a smaller m by m naught indicating that there is a larger propellant mass that will be necessary to achieve that theta. Let us now look at the trajectory solutions in terms of the h and x profiles. So, we go back to the kinematic equations that we have derived earlier that is the h by dt as V naught cos theta. Again, it is an extremely simple trigonometric integral. So, we get a solution for h as V naught square by g theta ln sin theta plus constant of integration. And the dx by dt which is V naught sin theta is directly a function of V naught square by g into delta theta. So, you realize that the expressions for h and x are quite simple. And of course, for higher delta theta, you will travel a longer distance delta x and same thing will happen for a higher velocity. Let us now examine the implication of these relations for the problem that we have been considering. So, let us consider a rocket with following specifications. So, we have a lift off mass of 80 tons with the propellant mass of 60 tons of got ISP 240 seconds, theta naught is 2 degrees and our V naught is 300 meters per second. At the requirement is that we should become parallel to local horizon assuming flat earth approximation. So, theta B is 90 degree. Let us try and look at the solutions in terms of your mass profile, altitude profile and the horizontal distance profile. So, the time taken for this trajectory the first solution because we have given both theta naught and theta B at this V naught it would take about 123.8 seconds to complete the mission. And then we go to the mass profile that is m by m naught or m b by m naught expression. And it can be shown that the burnout mass in this case will be 52.17 tons. And if we start with 80 tons then effectively about 28 tons of propellant is going to be consumed during this. So, you will consume about 28 tons of propellant in the time interval of about 123 seconds to achieve an inclination from 2 degree to 90 degree at a constant velocity of 300 meters per second. So, which means that you can actually achieve a significantly larger inclination by just burning a reasonably smaller amount of propellant as long as your velocity is constant. Here it is worth noting that this particular feature is due to the fact that you are not really using your propellant to accelerate the vehicle, but only to turn the vehicle. And this is directly responsible for the normal acceleration component which is coming because of gravity. So, you would realize that we can achieve large angular changes by burning a smaller amount of propellant. And the results also justify our original hypothesis that constant velocity solutions manage the propellant better from a practical perspective. Of course, because the trajectory is for a long duration of about 124 seconds, the altitude during this will be of the order of around 31 kilometers and it will travel about 14 kilometers on the surface of the earth. As the distance traveled over the surface of earth is not very large during this period, a flatter approximation with which we have started becomes reasonably applicable. So, the results obtained in this case are fairly accurate and can be used for initial design and sizing exercise for such a mission. So, to summarize, we know that a constant velocity solution is much simpler to obtain and obviously, as the expressions are simpler, their implementation also would become lot easier. And there is an important aspect which comes out very clearly is that there is now an explicit control over the propellant mass for a desired velocity and inclination which means if we specify a velocity and an inclination, then it directly tells you how much of propellant you are going to need to complete the mission, where a time would then come out as a natural consequence of the solution. Hi, so we have seen that a constant velocity solution for gravity turn is an extremely useful solution for minor trajectory corrections in the context of errors in inclination when the vehicle has reached its desired terminal velocity. And the solution also indicates that the propellant that we are going to consume is kept to a minimum because we are not accelerating the vehicle. With this, we will now look at in the next lecture the last of the simplified gravity turn solution trajectory which will be driven by a constant specific thrust or what is called constant forward acceleration which also has its own practical value. So, bye, see you in the next lecture and thank you.