A satellite in a circular orbit of radius r has period T. A satellite in a nearby orbit with radius r + Δr, where Δr≪r , has the very slightly different period T + ΔT. (a) Show that
ΔT/T = (3/2) (Δr/r)

Question

A satellite in a circular orbit of radius r has period T. A satellite in a nearby orbit with radius r + Δr, where Δr≪r , has the very slightly different period T + ΔT. (a) Show that 
ΔT/T = (3/2) (Δr/r)

Accepted Answer

Welcome back. Everyone in this problem. Two drones are flying in circular orbits around a tall structure at a constant altitude. Drone A has a circular orbit with radius two R and takes time four to complete one full revolution around the structure. John B has a circular orbit with radius two R plus delta R and takes time four T plus delta T to complete one full revolution where delta R is much smaller than R and delta T is much smaller than T determine the relationship between the changes in the revolution times and path radius A says it's three halves of delta R divided by R minus one. B says it's two third of delta R divided by R plus one. C says delta T divided by T equals nine fourths of delta R divided by R minus one. And D says it's 4/9 of delta are divided by R plus one. Now, from our answer choices, we can tell that when we want to determine the relationship between the changes in the revolution times and path radi essentially, we're looking for an equation that relates delta T divided by T to delta R divided by R let's first try to visualize what's going on here. OK. So we have our structure and we have two drones orbiting our structure for zone A, its radius is two R and for zone B, its radius is two R plus delta R. It's just a bit bigger than, or a bit longer than the radius two R. OK. And for June A, it has a revolution time of 40. Well, for John B, it has a revolution of 40 plus delta T. Now, if we're going to figure out the relationship first, we have to ask ourselves, what do we know that relates to revolution to radius for an orbit? Well, recall OK, that the period of an orbit is equal to the square root of four pi squared divided by GM. OK? Multiplied by R raise to the power of three halves. OK. So if we can find expressions for both drone A and drone B, the, the periods for drone A and B, then we should be able to compare them to see if there's a relationship between the changes in the revolution times and the path radius. So let's go ahead and use this formula first for drone A. OK. Now, for June A, that means then that its period for is going to be equal to the square root of four pi squared divided by GM multiplied by R. Well, its radius which is two R raise to the power of three halves. No what this tells us then is that if we were to square both sides, OK. And that can help us to simplify our expression, getting rid of the fractional fractional power, then we're going to have 16 T squared being equal to four pi squared divided by GM multiplied by two RQ, which is going to be eight RQ because it's two RR QE. Now we could divide both sides by 16. OK? We are on the left hand side, 16 K 16 leaving us with T squared. OK. And on the right hand side, uh 16 goes in, sorry, it goes into 16, 2 times, two, goes into 42 times. And after we simplify, we get T squared to be two pi squared R cubed divided by GM. OK. No, we could go ahead and solve for T but we're just going to keep this expression for no, because it's very important you, you will see it soon again. OK. But this is a good expression for us to keep in mind. No, since it has simplified here for a, let's think about what happens for our formula for John B. Hm. So for B no, OK, our period is going to be 40 plus delta T four T plus delta T and that is going to be equal to the square root of four pi squared divided by GM multiplied by two R plus delta R raised to the power of three Hs. Now, if you compare it to what we had before. You'll probably notice that it will end up to something similar to the expression we have where we'll replace R with two R plus delta R and T with four T plus delta T. In other words, four T plus delta T all squared is going to be equal to two pi squared divided by GM multiplied by two R plus delta R QED. Now, at this point, we can go ahead, we can simplify by expanding both sides of our equation. OK. When we expand the left hand side, we end up with 16 T squared plus delta T squared. OK plus HT delta T. And on the right hand side, we have two pi squared divided by GM multiplied by eight R cubed, OK plus delta R cubed plus 12 R squared, delta R plus six R delta R squared. Now remember earlier, we noted that our delta terms just represented a small change in comparison to our, our, our quantities. In other words, delta R was a small change in R delta T, a small change in T. So we can neglect the delta terms since they're really small and squaring or cubing them will make them even smaller. So from our expression, we can neglect delta T squared. Likewise, delta R QED and six R delta R squared. On the other hand, notice here that we can factorize T squared on our left hand side and factorize for our cubed on our right hand side. So when we do that to our equation, then we're going to get T squared 16 plus eight delta T divided by T, which is the expression that we wanted being equal to two pi squared divided by GM multiplied by R cubed multiplied by eight plus 12 delta R divided by R. So we have the expressions that are the relationships that we want. But no, we can take this a step further because remember earlier, we said that T squared equals two pi squared R cubed divided by GM, which is the expression that we have right here. OK. So we can replace T squared with this expression. OK. Can replace T squared with two pi squared R cubed divided by GM. And now if we cancel it from both sides, then that tells us that 16 plus eight delta T divided by T is going to be equal to eight plus 12 delta R divided by R. Now we can simplify to find a relationship if we subtract it from both sides or rather if we subtract 16 from both sides, eight delta T divided by T is going to be equal to 12 delta R divided by R minus eight. OK. If we divide both sides by eight, then we get delta T divided by T to be equal to three halves of delta R divided by R minus one. Thus delta T divided by T equals three halves of delta. R divided by R minus one. That's the relationship between the change in period and the change in RADII. If we go back to our answer choices, OK, then that tells us, then that a is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

A satellite in a circular orbit of radius r has period T. A satellite in a nearby orbit with radius r + Δr, where Δr≪r , has the very slightly different period T + ΔT. (a) Show that ΔT/T = (3/2) (Δr/r)

Verified Solution

Key Concepts

Kepler's Third Law

Differential Calculus

Circular Motion