Three satellites orbit a planet of radius R, as shown in FIGURE EX13.24. Satellites S₁ and S₃ have mass m. Satellite S₂ has mass 2m. Satellite S₁ orbits in 250 minutes and the force on S₁ is 10,000 N.

(c) What is the kinetic-energy ratio for K₁ / K₃ for S₁ and S₃?

Question

Three satellites orbit a planet of radius R, as shown in FIGURE EX13.24. Satellites S₁ and S₃ have mass m. Satellite S₂ has mass 2m. Satellite S₁ orbits in 250 minutes and the force on S₁ is 10,000 N.

(c) What is the kinetic-energy ratio for K₁ / K₃ for S₁ and S₃?

Accepted Answer

Hey, everyone in this problem, we have a figure that shows a star with a radius of R and three comets, C one, C two and C three orbiting that star. OK. So we have C one on the left, OK? With a mass of W, a distance of two R from the center of the star, we have C two with a mass of four W at a distance of five R from the center of the star. And finally, we have C three with a mass of five W at a distance six R from the center of that star. OK. So we're told that the time taken by C one is 320 minutes in order to traverse the entire orbit. And we're asked to determine the ratios of kinetic energies, OK? Of C one and C three, if C one experiences a force of 12,000 nodes, now we have five answer choices here or sorry, four, only four answer choices. Option A 56, option B five thirds, option, option C 3/5 and option D six fifth. So let's start with what we're trying to find here. We're looking for a ratio of kinetic energy. That means we're gonna need to calculate the kinetic energy for those two comments of interest. So C one and C three, now we're called the kinetic energy K is given by one half MV squared. All right. So we know the mass of our comets. We're given that in the problem. What about the speed? So if we look at our diagram, we can see that we have circular orbits. OK? So the speed of a comment in a circular orbit. Well, we know that call that that's given by is equal to the square root of G M two divided by. So what that told us if we substitute V into our kinetic energy equation is that the kinetic energy K is gonna be equal to one half multiplied I am the map and this should be little M not big M. OK? This is the mass of our comet multiplied by V squared. So it's gonna be multiplied by G capital and mass of the star. Good bye bye. All right. So we have this equation for K, the kinetic energy. So we're gonna do is go through and calculate the kinetic energy for comment one and comment three and then we can write the ratio that we're interested in. So let's start with comment one. OK. So for comment one, the one we're given that the mass little M is equal to W and the distance between the center of the star And this comment is two R. So little R is gonna be equal to two multiplied by bigger. That means that our kinetic energy K one is going to be equal to one half multiplied by W multiply by G multiplied by M divided by two. We can kind of combine these constant coefficients and write this as W G big M divided by four R. All right. So that's our kinetic energy for comment one, it looks a little messy. We have a lot of um variables in here that we don't have values for. But that's OK. OK. Remember that we're looking for the ratio of these kinetic energies. So let's look at the kinetic energy for C three and we may be able to cancel some of these values. So for comet C three, we're told that the mass is gonna be equal to five W in the radius little R, the distance between the center of the star and our comet that's orbiting is six multiplied by A R. So our kinetic energy K three again, using our equation in this green box, one half little M multiplied by G multiplied by big M divided by R. Yeah, I get that. This is gonna be five WGM divided by 12 R. And we just skipped one step. We showed it with K one where we have our one half and then we have two R and we simplify to four R. We've just skipped that step. Here we have one half and then we had multiplied by R in the denominator. Ok? So that gives us that denominator of 12 R. All righty. So we have our K one, we have our K three. Let's go ahead and look at the ratio, we want the ratio of K 12 K 3. So K one divided by K three going to be equal to WG M divided by two R. Oops not two R four R. Here we go divided by five WGM divided by 12 R. And now you'll start to see these terms canceled. OK? We can rewrite this. When we divide by a fraction, we can multiply by the reciprocal. So we can write this as WGM divided by four R multiplied by 12 R divided by five WGF. And you'll see that the WGM and R term all divides it. OK? There's one in the numerator and one in the denominator everywhere. What we're left with is just these numbers. We have 12 in the numerator, we have four multiplied by five in the denominator which gives us 12 divided by 20. We can simplify both the numerator and denominator. Denominator are divisible by four. So let's go ahead and divide by four. We get that this simplifies to three bits. And so there was a lot of values in this problem. We didn't know and we didn't fill in the value for W or big G or big M or R but we were still able to find this ratio of kinetic energies of 3/5. OK. If we go up and compare this to our answer choices, we can see that this corresponds with answer choice. C Thanks everyone for watching. I hope this video helped see you in the next one.

Three satellites orbit a planet of radius R, as shown in FIGURE EX13.24. Satellites S₁ and S₃ have mass m. Satellite S₂ has mass 2m. Satellite S₁ orbits in 250 minutes and the force on S₁ is 10,000 N. (c) What is the kinetic-energy ratio for K₁ / K₃ for S₁ and S₃?

Verified Solution

Key Concepts

Gravitational Force

Kinetic Energy in Orbital Motion

Orbital Period and Velocity Relationship