A 55,000 kg space capsule is in a 28,000-km-diameter circular orbit around the moon. A brief but intense firing of its engine in the forward direction suddenly decreases its speed by 50%. This causes the space capsule to go into an elliptical orbit. What are the space capsule’s (a) maximum and (b) minimum distances from the center of the moon in its new orbit?

Hint: You will need to use two conservation laws.

Video transcript

Welcome back. Everyone in this problem, an artificial satellite having a mass of 100 kg orbits around the planet following a circular path, having a diameter of 700 kilometers. Due to some technical glitch, its velocity reduces suddenly by 30% leading it into an elliptical trajectory. What will be the maximum and the minimum distance from the center of the planet on this new trajectory, the mass of the planet is 3.5 multiplied by 10 to the 20th kilograms. A says the maximum distance is three multiplied by 10 square kilometers while the minimum distance is one multiplied by 10 squared kilometers. B says they are 3.5 multiplied by 10 squared and 1.1 multiplied by 10 squared kilometers respectively. C says there are 3.5 multiplied by 10 squared and 2.1 multiplied by 10 squared kilometers and D 4.5 multiplied by 10 squared and 2.5 multiplied by 10 squared kilometers respectively. Now let's first try to draw a diagram here and make note of some of the information we have. So we have a planet that's being orbited by a satellite in a circular path. OK. And no, for this orbit, we know it has a diameter of 700 kilometers. So that means its radius R is going to be 350 kilometers. No, in our problem, it says because of a technical glitch, the velocity reduces and it changes to an elliptical trajectory. In other words, if I were to represent that in red here, no, we could draw an ellipse. Let me draw the best lips that I can draw. OK. Let's say that this is the ellipse, not perfect, but this is the new elliptical path. OK? Where we have our distance R or maximum distance and our minimum distance. So let's call R one, let me put that here. Let's call R one or maximum distance. And let's call our two or minimum distance. OK? And we know that for our satellite it's moving at speeds for our trajectory V one, OK? At our max speed and V two, sorry at our max distance and V two for our minimum distance. And based on the information we have in our problem, we know that the mass of the satellite is 100 kg. We know the mass of the planet is 3.5 multiplied by 10 to the 20th kilograms. The radius here is equal to uh 350 kilometers. OK? So that's 350 multiplied by 10 to the third meters. And we know that our speed V one for this elliptical path is 30% of the speed for the circular path. In other words, V, one equals 0.7 V I. Now how can we use this information to figure out our maximum and minimum distances are one and R two? Well, let's think about the energies that are involved because the energy at the firing point is equal to the energy at the other end of the electrical trajectory. OK. So that means the energy is conserved. And by the law, by the principle, sorry, by the principle of conservation of energy, let's make a note of that here. OK. Then the energy before the initial energy which for our satellite would be its kinetic energy and its gravitational potential energy, both of which are acting against each other is equal to the final kinetic energy minus the final gravitational potential energy. Now, what do we know about both of those? Well, we know that the initial kinetic well, kinetic energy generally sorry is equal to half MV squared. OK. And our gravitational potential energy U is equal to the gravitational constant G multiplied by the product of the two masses divided by the distance between them. So if we apply the principle of conservation of energy here, and let me just move our formula for our gravitational energy up here. OK. Then if we apply this idea and this tells us OK, that a half of MV one squared OK. The initial kinetic energy minus GMM divided by R one, the initial potential gravitational energy equals half MV two squared. OK minus GMM divided by R two. Now, we have some good information here and let me just make some space by bringing this up here. We, we, we have an expression here to talk about or to illustrate what's going on. But the problem is that we don't know what V one is. OK. And we don't know what V two is and we need to use both of these to help us figure out our maximum and minimum distances which we can solve with R two. Now, before we get to that, let's first start by trying to solve for V one and V two. No solving for V one. OK. If we go back to our diagram, notice that at V one, our satellite is on its circular path. OK? And that means then that it has a centripetal force which is based on the gravitational force. The gravitational force provides the centripetal force needed to keep the satellite in circular orbit. So in solving for V one, OK, we know that our centripetal force equals the gravitational force which tells us that GMM divided by R squared, that's our gravitational force equals our centripetal force MV I squared divided by R one. Now, in this case, we can cancel M from both sides. OK? We can cancel R one from our right hand side, R one into R one squared leaves R one on our left hand side. And now if we square root both sides that tells us, then that V I equals the square root of GM divided by R. So now if we were to go ahead and solve for V one, OK. That means V one, which is zero is 70% of V I is going to be 0.7 multiplied by the square root of GM divided by R. Let's substitute those values because we know we're talking about R one. So that's 0.7 multiplied by the square root of the gravitational constant 6.674 multiplied by 10 to the negative 11th cubic meters per kilogram per second squared multiplied by the planet's mass. 3.5 multiplied by 10 to the 20th kilograms. OK? And now all of that is being divided by the radius for our circular path, which is equal to 350 kilometers. Now, when we go ahead and solve here for V one, we should get it to be equal to 180.8387 m per second. So that's our value for V one. Next, let's find our value for V two. OK. Now, for V two, let's ask ourselves what, what do we know that's going to help us? Let's look back on our diagram. OK. Now, if we think about it, the same, the angular momentum is conserved and the same at the minimum radius and the maximum radius. So by the conservation of momentum, we know that M one, V one, R one is going to be equal to M two V two R two. Our masses are the same. OK. So it's really just M so we can cancel M is the mass of the satellite. OK. And now solving for V two, this tells us then that V two can be written as an ex in, in terms of V one that is V two equals V one, R one divided by R two. Now we don't know R two yet. So we'll leave this expression for V two, at least we can use it to get rid of it in our formula or equation that we already have. So no substituting our values for V one and V two into our conservation of the equation we have from the conservation of energy. Then that tells us that a half of V one squared minus GM divided by R one is going to be equal to a half of V one, R one divided by R two, all squared. The expression for V two minus GM divided by R two and no, we don't have M in our formula anymore in this one. Because if you, if we go back to the first one, M was a common term in all of our expressions. So we could cancel M out. Now here, notice that if we want to solve A R two, we can multiply our en entire creation by R two squared. OK. We can get it out of our denominator because R two squared is going to be in this term. So now when we do that, when we multiply through by R two squared, let me put that in red, then we're going to have a half of V one squared, R two squared minus GMR two squared divided by R one equal to a half of V one squared, R one squared, R two squared is going to cancel from the denominator minus GM multiplied by R two. Now, if you notice here, we could rearrange our equation as R two squared. OK. Or rather, we could factor out R two squared from our terms here. So that's a half of V one squared minus GM divided by R one multiplied by R two squared. OK. We could bring Gmr two over to our left hand side. So now that's plus OK, GMR two, and we could also bring a half of V one squared R one squared to our left hand side. So that's gonna be minus OK. A half of a half of V one squared R one squared and equated to zero. Now, why would we do that? Well, no notice that we have a squared term for R. In other words, A X squared, OK, we have a linear term for R two which we can call BX and we have a constant. So no, our equation is in the quadratic form. And since it's in the quadratic form, we can solve it quadratic. Now, we know all of these values because of the time we have left, we can't solve for each of these. But if you go ahead and solve for them on your own, OK, you should find that you end up with the expression 50.3887 multiplied by 10 to the third R two squared minus 23.359 multiplied by 10 to the ninth, R two plus 2.003 multiplied by 10 to the 15th being equal to zero. This is our quadratic expression and know if we go ahead and solve quadratic, OK, using a calculator, then we'll get two values for R. Of course, we're ignoring our units for no, just to make it simpler to write. And you should find that you get R two to either be equal to 113.573 multiplied by 10 to the third meters. And you get it to be equal to 350.002 multiplied by 10 to the third meters. No, this is really 350 kilometers here. This second value. And that makes sense because if we go back to our diagram, our max is R one which is close to, which is approximately 350 kilometers. So then we can say from our second value that our max written to two significant figures is 3.5 multiplied by 10 square kilometers. Now 113.573 multiplied by 10 to the third meters in our minimum value. So when we write it to two significant figures, it equals 1.1 multiplied by 10 squared kilometers. Thus, the minimum distance is 1.1 multiplied by 10 square kilometers. And the maximum distance is 3.5 multiplied by 10 square kilometers for this elliptical trajectory. If we go back to our answer choices, that means B answer choice B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.