 So, today's session what we will do is, we will first start with problem practice, problems on whatever we have done last, problems on whatever we have done till now in gravitation we will take up those problems and then we will discuss few of the concepts which are beyond your school curriculum, but they are you know they are there in J syllabus okay. So, we will start with the problem solving all of you ready should we start or you have any specific doubt in this chapter which you want to discuss first okay. So, I think let us start fine I will start taking up your questions from your school book okay first once we are done with your school level questions then only we will go to the next level okay. So, you can see this question right in front of your screen question number 8.5 you see this please solve this anyways please attempt this question number 8.5. So, in this chapter in this chapter you will see that concepts are straightforward when it comes to the school level questions, but calculation is bigger. So, you should be good at calculation and let me tell you a fact in 2019 in 2019 the common feedback is that common feedback in January J means is that the paper was calculation intensive okay. So, many students could attempt only 20 out of 30 questions fine the attempt was less because they could not calculate very fast getting it. So, you must not use calculator in these type of calculations you must do the hand calculation otherwise you know the speed will take a beating your speed will be very less and because of that even though you know how to solve it you will not be able to attempt the full question paper okay. So, don't make that error which you know I have again told the same thing to your seniors also but few listened few chose not to listen. So, no surprises who suffered at the end okay. Anyone is able to get the answer? R is not the radius of Phobos R could be the orbital radius of the Phobos. What concept will you use here? Which concept will you use? You will use T square is proportional to R cube isn't it? So, T1 by T2 whole square will be equal to R1 by R2 whole cube fine alright. So, Phobos has a period of 7 hours 39 minutes and orbital radius of Phobos is given. So, let's say T1 is for Phobos okay we need to find the okay this is I think this is not used right now the first one the first part is something else Phobos has time period given and the orbital radius is given we need to find mass of the Phobos okay. So, we know that the when you equate mass of Phobos into orbital velocity square divided by R this should be equal to G times G times mass of mass into mass of Phobos divided by R square. So, from here you will see that mass of Phobos get cancelled out alright. So, velocity will come out to be a root over G mass of mass divided by R okay this is the velocity okay time period is 2 pi R divided by velocity okay. So, you will get 2 pi root GM G mass of mass okay into R raise to power 3 by 2 okay. So, we need to find the mass of the mass right. So, just rearrange the term first you should get what is mass of what is the mass of the mass in terms of other variables okay. So, root over mass of mass is equal to 2 pi divided by TG R raise to power 3 by 2 right then square both sides. So, mass of mass will become equal to 2 pi divided by GT square into R cube alright. Once you get this then you substitute the values okay then you substitute the values to get the answer okay T must be in seconds. So, convert 7 hours 39 minutes into seconds how will you convert that 7 hours 60 minutes 60 seconds okay plus 39 minutes. So, plus 39 into 60 seconds this is the value of T which you should substitute over here R cube you convert this in meters okay and then cube it and the value of G is known okay yeah 2 pi is squared. So, this fine 7 calculated is it no Shavan we are calculating mass we are not calculating the time period right now mass will be 5.97 into transfer 24 kg all of you clear about this one. Now, the second part of this question is assume that earth and mass move in a circular orbit around the sun Martian orbit is 1.52 times the orbit radius of the earth what is the length of Martian year in days one year should be what one year should be the time taken by the planet to complete one revolution around the sun that is how one year for every planet is defined okay. So, here you need to use this Kepler's law time period of earth around this sun which is one year is given as 365 days okay. So, let's say T1 is for the earth okay R1 is radius of revolution for the earth and R2 is for the mass okay. So, I can say that T2 is equal to T1 times R2 by R1 times 3 by 2 okay R2 by R1 is 1.52 okay 1.52 raised to power 3 by 2 you have to do it here. Now, you can you can't find out 1.52 raised to power 3 by 2 using hand calculation. So, you must be very much familiar with the lock tables okay. So, try using lock tables if you don't have a lock table please get a copy of that. So, using lock tables you'll get 1.52 raised to power 3 by 2 and then multiply that with T1. T1 is what 365 days it's just been clear any doubt please ask quickly okay fine great this is 8.5 8.6 mass is this 6.4 into transport 23 this one 8.6 okay can you get the expression like how will you find forget about the final answer okay can you tell me using what all two different methods can you get the mass of earth okay can you tell me the two different methods in which you can get the mass of earth don't worry about the final answer tell me the concept really what all two concepts will you be using to get the mass of earth yes this is one method this is one method the value of accession due to gravity G is given right. So, since G is given I can use the value of G to find the mass of earth because G must be equal to G into mass of earth divided by radius of earth square radius of earth is also given okay what is the other method other method is time period okay but time period revolution of moons orbit is given it's not the earth's time period that is given so how will you do that this is suppose moon that is orbiting the earth okay this is the earth fine so if moon has a velocity of let's say u okay and this is the orbit of the moon distance of the moon is given as this so this is the orbit of moon also right so that is r and let's say this is moon mass of earth is m e and mass of moon is m m what did I say last class any situation in which something is moving in a circle the first formula you should write is this m v square by r should be equal to g times m m divided by r square okay so from here you'll get this velocity velocity of the orbit okay and once you get the velocity of orbit okay time period will be 2 pi r divided by v getting it okay and you can see that the velocity of orbit depends on the mass of the earth all right so when you substitute v over here mass of earth will come in the expression okay so that time period is also given 27.3 days that you convert in seconds okay how do you convert in second 27.3 into 24 into 60 into 60 okay now it is a good idea to remember how many seconds one day is how many second one day is it is a good idea to remember this and also one year has how many seconds so if you remember this it will really help you so I hope all of you are clear any doubts this question 8.8 the hint is you need to use conservation of energy conservation of energy okay if final energy e2 is more than e1 so how much energy you need to give you need to give e2 minus e1 right energy cannot be created you have to give the energy anyone got the answer just do the first part okay just till here should I do it all right so let us see the potential energy is what potential energy is minus of g m1 m2 by r this is a potential energy right and we have already derived that kinetic energy is minus of potential energy by 2 right so total energy is what total energy is potential energy plus kinetic energy which is u by 2 right so always remember if radius of revolution is r total energy will be always g m1 m2 divided by 2 r this is the total energy you don't need to find out a kinetic energy and potential energy separately so it's a good idea to remember that if radius of revolution of object is r okay then the total energy will be minus g m1 m2 by 2 r okay now don't calculate right now so so initial energy initial energy when the this satellite is at that orbit will be equal to what minus of g mass of earth into mass of satellite divided by two times of 2 re that is 4 re this is your initial energy initial total energy and final total energy will be minus of g mass of earth into mass of satellite divided by two times of the new orbit radius which is 4 re so that is 8 times re okay now tell me is t2 greater than or less than t1 is t2 more or less than t2 t1 t2 is more than t1 or not right t2 is more right it is closer to i mean the absolute value of t2 is less but when you put a negative sign with it okay then t2 becomes more than t1 all right so if t2 is more than t1 then t2 minus t1 extra energy is required to have t2 okay because one thing is very uh uh you know you should note a point that if if a satellite is revolving around a radius of r then it's potential energy it's kinetic energy and total energy everything can be written in terms of r only so if r is fixed its energy is fixed okay if if its energy increases automatically its radius will change fine so suppose you have a let's say a satellite which is revolving around the radius okay and if you want to change its uh if you want to change its orbital radius then what you do you burn some fuel put you can put it like a rocket you can increase its velocity so if you increase its velocity its kinetic energy automatically increases okay so if energy increases automatically it will spiral into the larger radius having more energy getting it so if you want to decrease its radius you decrease its velocity velocity of satellite if you as soon as you decrease the velocity of satellite its orbital radius will orbital radius will change because its total energy is changing fine so total energy is a direct function of radius of revolution all of you understood this one once you know what is the uh change in a total energy you can find the change in kinetic and change in potential energy as well clear all of you okay so let's try to solve few of your school questions can you solve 8.12 okay you're doing it or should i should i attempt it yes yes all of you get the value okay learn from your seniors mistake don't avoid calculation in fact you should welcome it you should feel good that okay fine it should be like a challenge right that let me quickly find the correct answer then only you'll improve on calculation check from your back of the book is it correct whatever answers you're getting you can check the back of your book 8.12 question is that the correct answer all of you tried right now i can solve should i solve now ashutosh got 0.1 into 10 is for 8 meters deviation which is like 10 is for 7 meters 10 is for 7 meter is like 10 is for 4 kilometers that's like 10 000 kilometers more he got okay others are also getting the similar thing all right let me solve this a rocket is fired from the earth towards the sun at what distance from the earth center is the gravitation force on the rocket 0 mass of sun is given mass of earth is given and so this is earth the sun is vast it will be very very big okay but compared to the distance between earth and sun it will size of the sun and earth will be like a point mass okay so rocket is fired like this you have to find distance where the gravitation force on the rocket is zero so i think this is a straightforward question so let's say this is the distance at which the gravitational force due to earth equals the gravitational force due to the sun so they will be equal and opposite right so all you have to do is equate them so g times mass of earth into mass of rocket divided by mass of sun is given mass of earth orbital radius is this okay so let's say this distance is d distance from the earth is as right so let's say this is d and that distance center to this point distance will become r minus d okay so this should be equal to g times mass of sun into mass of rocket divided by r minus d whole square right so all you do equate this and mass of rocket will get cancelled out fine so from here you get the value of d even g is gone fine so d will be equal to r minus okay r minus d under root of me by ms so from here just simplify it further you get the value of d okay all of you clear i think this should be something just straight forward