4. Radar measurements give the following data for a...

Transcript

00:01 So here on this problem we are given that satellite is rotating around the earth and the mass of the satellite we are given as 220 kg and its altitude is 640 kilometer that means this distance we are given as 640 kilometer now if you calculate the total distance means this total distance are so this total distance are would be the radius of the earth plus the altitude so radius of earth we know it is 6 point its radius of earth are one we have 6 .37 times 10 to the power 6 meter and this altitude we are given as 640 times 10 to the power 3 meter if we calculate the total radius are you to add this these two terms that is the radius of earth plus the altitude so it would be 6 .3 three seven times 10 to the power six plus six forty times ten to the power three and we'll get this as seven point zero one times ten to the power six meter and there eight parts in the equation we'll be solving the a part first now in the a part we have to calculate the speed of the satellite so b we have to calculate now if you see the force acting on this satellite the magnitude would be if equals to gm m over r square where capital m is the mass of earth and it would be acting the force would be acting like this and if you see the acceleration of this satellite the acceleration would be v square over r where v is the velocity of satellite so by using the newton's second law of motion here we can write that f is equals to mass times acceleration and acceleration we know it's v square over r so v square over r and this f is equals to this value so we can write gm times m over r square is equals to m times v square over r so from here the expression for we comes out to be as under root gm over r now you can put the values g is the constant and its value we have six point six seven times 10 to the power minus 11 it's meter cube over second square kg times m we have this m is the mass of earth and mass of earth we know it is 5 .98 times 10 to the power 24 kg divide by r and this total radius r we have 7 .01 times 10 to the power 6 meter and it's under root now we're going to solve this and we'll get the velocity or the speed of satellite as 7 .54 times 10 to the power 3 meter per second so this is the answer for the a part now next we'll be solving the b part of the problem now in the b part we have to calculate the time period that means the time taken by satellite to complete one revolution so time would be equals to 2 pi r over the velocity and 2 by r we have 7 .01 times 10 to the power 6 meter divide by velocity we have calculated earlier as 7 .54 times 10 to the power 3 meter per second now we're going to solve this and we'll get the time period t as 5 .84 times 10 to the power 3 second and if you convert this thing into minute we know that one second is equals to 1 over 60 minute so we'll get the time period t in minutes as approximately 97 minute so this is the answer for the b part now next we'll be solving the c part now next they are saying that the satellite is losing mechanical energy at an average rate of 1 .4 times 10 to the 5 jule per orbit revolution and we have to calculate the altitude and then in the d part we have to calculate the speed and in the a part we have to calculate the period time period t after the 15th 1 ,500 revolution.

05:18 Now first of all to solve these three parts we have to calculate its initial energy that is a not is equals to minus g m over 2 r then this the initial mechanical energy we're going to put the values to calculate its initial energy so we have minus g we have 6 .67 times 10 to the power 11 multiplied with mass of the earth we have 5 .98 times 10 to the power 24 times the mass of the satellite which is 220 kg divide by 2 times r and r we have 7 .01 times 10 to the power 6 meter from here we get the answer in rule in a jule because we have a substitute the values in si unit so a not would be minus 6 .26 times 10 to the power 9 jule so this is the initial mechanical energy now if we calculate the energy left after 1500 revolution a it would be equals to e0 minus n times c so n we have 1 500 and c we have this much we plug it here to calculate the energy lift so a not we have 6 .626 sorry 6 .26 times 10 to the power 9 jule minus n we have 1 ,500 multiplied with c which is 1 .4 times 10 to the power 5 jule joule per orbit so from here we'll get the value of e as minus 6 .47 times 10 to the power 9 jule so this is the energy left with the satellite after 1500 revolution now the satellite would come here after losing energy and now it is orbiting in this orbit now we have to calculate the altitude in the c part basically this h this height h we have to calculate so for that we have to calculate this radius and this radius r we can calculate by using the equation that we used earlier that is equals to minus gm times m over 2 r now if we have this much so we can plug it here this is minus 6 .47 times 10 to the power 9 jule is equals to g we have 6 .67 times 10 to the power minus 11 and m we have 5 .98 times 10 to the power 24 multiplied with the mass of this satellite which we have 220 kg divide with two times r and r we have to calculate we're going to simplify this and from here we get this radius r as 6 .78 times 10 to the power 6 meter now this altitude h would be equals to the radius r minus the radius of earth so radius r we have 6 .78 times 10 to the power 6 minus r1 that is the radius of earth we have 6 .37 times 10 to the power 6 meter now we're going to simplify this and we'll get our result as 4 .1 times 10 to the power 5 meter so this is the answer for the c part now next we'll be solving the d part where we have to calculate the speed of the satellite now the speed of the satellite we have v equals to under root gm over r this expression we have derived earlier now can put the values so under root g we have 6 .67 times 10 to the power minus 11 multiplied with m that is a mass of earth 5 .98 multiplied with 10 to the power 24 kg divide with the radius r now this time we'll take this radius are that we have calculated earlier to be 6 .78 times 10 to the power 6 meter.

09:53 So from here, velocity of the satellite in this orbit would be 7 .67 times 10 to the power 3 meter per second...

Question

Please give Ace some feedback