FIGURE P13.57 shows two planets of mass m orbiting a star of mass M. The planets are in the same orbit, with radius r, but are always at opposite ends of a diameter. Find an exact expression for the orbital period T. Hint: Each planet feels two forces.

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Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Two satellites with an identical mass of M are revolving around earth that has a mass of capital M in the same circular path with radius of R but is always positioned at opposite ends of a diameter line as shown in the figure below what would be their orbital period T value. So that's our angle, our angle, we're trying to figure out what the orbital T value is. So we're trying to figure out what an expression for T is. So what is T equal to? So with that in mind now that we know that we're ultimately solving for, let's look at the figure that's provided to us by the prom itself. So in the center of our circular orbital radius or not orbital radius or circular orbital path, we have earth which has a massive capital M and we also directly above and below earth pointing towards the center of earth, we have our satellites that are both of a mass of some value lowercase M mass. And they all, they both have some radius value R. So from the center of earth going up to where this satellite is, is some value R. Awesome. And then we have our black circle going around earth denoting our circular orbit. Awesome. So with that in mind, let's read off our multiple choice answers to see what our final answer might be. And let us note that they're all for T equals some expression. So A is equal to 16 multiplied by pi multiplied by the square root of two multiplied by art. The power of three divided by capital G multiplied by capital M plus lowercase M divided by four B is four multiplied by pi multiplied by the square root of R to the power of three divided by capital G multiplied by capital M plus. Lower case M divided by four C is 16 multiplied by pi multiplied by the square root of R to the power of three divided by capital G multiplied by four multiplied by capital M plus. Lower case M and D is four multiplied by pi multiplied by the square root of R to the power of three divided by capital G multiplied by four multiplied by capital M plus lowercase M OK. So now, first off, we need to write down all of our known variables. So let us note that lowercase M is the mass of our satellite. So this is just to keep things simple. I'm just gonna write S for satellites. So M is the mass of our satellites. Capital M is equal to the mass of earth, which I'm gonna use a capital E for earth. And lowercase R is equal to the distance between earth and each satellite. So I'm just gonna put a lowercase D to represent distance. So once again, R is equal to the distance between earth and each satellite. So now we need to note that from these conditions that are provided to us by the pro itself, we can write and say that the gravitational force between the earth and the satellite plus the gravitational force between the two satellites is equal to the centripetal force between the earth and the satellites. Which as we should be able to recall, we can write this equation as capital G which is the gravitational constant multiplied by capital M multiplied by lowercase M divided by R squared plus capital G multiplied by lower case M multiplied by lower case M divided by two multiplied by R to the power of two is all equal to M multiplied by V squared divided by R. And once again, we need to note that capital G is the gravitational constant. And let's make a little side note here off to the side in blue. So the numerical value of G is equal to 6.6 74 multiplied by 10 to the power of negative 11 in its units are newtons multiplied by meters squared per kilogram squared. Awesome. And let us also note that V, the speed or the velocity is equal to two multiplied by pi multiplied by R all divided by capital T where capital T is our orbital period, which what we're ultimately trying to solve for. So we do not know what the value of T is at this particular moment. So now at this point, we can go ahead and write, we could take our equation and we can write that capital G multiplied by capital M multiplied by lower case M divided by R squared plus capital G multiplied by lowercase M multiplied by lowercase M divided by two multiplied by R squared is equal to M multiplied by or multiplied by pi squared multiplied by R squared all divided by R multiplied by T squared. So when we cancel out the M variable, we will be able to write. And specifically when we cancel out the lower case M variable, we can go ahead and write that capital G multiplied by capital M divided by R squared plus capital G multiplied by lower case M divided by four multiplied by R squared is equal to four multiplied by pi squared multiplied by R divided by T square. So now at this point, we need to start rearranging this equation to isolate and solve for T, which is our final answer that we're ultimately trying to solve for. So when we start to rearrange and simplify, we will get 1/4 multiplied by capital G divided by R square multiplied by four multiplied by capital M plus. So we need to add plus lowercase M is all equal to four multiplied by pi squared multiplied by R divided by capital T squared. So getting TS by itself, we will say that T squared is equal to 16 multiplied by pi squared multiplied by R to the power of three divided by capital G multiplied by four multiplied by capital M plus lowercase M. So to get rid of the squared on the period, we can square root both sides. So when we do, we will find that capital T is equal to 16 multiplied by pi squared multiplied by art, the power of three divided by capital G multiplied by four multiplied by capital M plus lowercase M. So when we simplify this expression, we will find that T, the period is equal to four multiplied by pi multiplied by the square root of R to the power of three divided by capital G multiplied by four multiplied by capital M plus lowercase M. And that's it we've solved for this problem. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter D which states that the period T is equal to four multiplied by pi multiplied by the square root of R to the power of three divided by capital G multiplied by four multiplied by capital M plus lowercase M and that's it. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

FIGURE P13.57 shows two planets of mass m orbiting a star of mass M. The planets are in the same orbit, with radius r, but are always at opposite ends of a diameter. Find an exact expression for the orbital period T. <IMAGE> Hint: Each planet feels two forces.

Verified Solution

Key Concepts

Gravitational Force

Centripetal Force

Orbital Period