(III) Two identical particles, each of mass m, are located on ...

Question

(III) Two identical particles, each of mass m, are located on the x axis at x= +x₀ and x = -x₀. At what point (or points) on the y axis is the magnitude of g a maximum value, and what is its value there? [Hint: Take the derivative dg/dy.]

Accepted Answer

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 spheres each with a massive M are positioned on the X axis at X equals positive D and X equals negative D. The gravitational field at any point on the Y axis is given by the expression lower case G is equal to two multiplied by capital G multiplied by M multiplied by Y all divided by D squared plus Y squared all to the power of three halves where capital G is the gravitational constant, identify the points on the Y axis where the magnitude of lowercase G is maximized and determine the maximum value. So that's our end goal. So ultimately, we're trying to solve for two separate answers. Our first answer is we're trying to figure out the points. We need to identify sulfur the points on the Y axis where the magnitude of lower case G is maximized. And then our second answer, we're trying to figure out what the maximum value is. So with that in mind, looking at our multiple choice answers, our first answer, which is the points on the Y axis where the magnitude of lowercase G is maximized will be ye some value. And the maximum value will be denoted as G max, G subscript max is equal to sum value. So let's read off our multiple choice answers to see which answer pair might be. Our final answer. A is plus or minus D divided by the square root of two and three multiplied by capital G multiplied by M all divided by two, multiplied by D squared. B is plus or minus D divided by the square root of two and four multiplied by capital G multiplied by M all divided by three multiplied by the square root of three multiplied by D squared C is plus or minus D and capital G multiplied by M all divided by three multiplied by D squared and D is zero and two multiplied by the square root. Sorry, two multiplied by capital G multiplied by M all divided by two multiplied by D squared. OK. So first off, let us note that sense, the gravitational field at any point on the Y axis is as given to us in the problem itself, G equals two multiplied by capital G multiplied by M multiplied by Y all divided by D squared plus Y squared all to the power of three halves. We can identify the points of the greatest magnitude by setting the first derivative to zero. So since the expression is always non negative, any extremo found will be considered to be a maximum value. Awesome. Therefore, with that in mind, we could write that DG of DT is equal to two multiplied by capital G multiplied by lowercase M multiplied by bracket D squared. So once again, quickly, a quick reminder here is we're taking the first derivative and we're setting it equal to zero. So when you take D squared plus Y squared, all to the power of three halves minus Y to the power of three halves multiplied by D squared plus Y squared to the power of one half multiplied by two multiplied by Y all divided by D squared plus Y squared. And this is all to the power of three. Don't forget your bracket and don't forget the set or derivative equal to zero. So now let me simplify for the first time, we will get D squared plus Y squared is al to the power of three hacks minus why? To the power of three halves multiplied by D squared plus Y squared, all to the power of one half multiplied by two, multiplied by Y all equal to zero. So now what we're trying to do is we're gonna keep simplifying this expression down to the simple, most simple form. And what we're ultimately gonna be trying to do is we're trying to simplify and then we're gonna have to rearrange this equation to isolate and solve for Y because that's our first answers, we're trying to figure out the points on the Y axis where the magnitude of lowercase G is maximized. So we're trying to isolate and solve for Y. So with that in mind, we need to just keep simplifying until we can't simplify anymore. So we need to take D squared plus Y squared to the power of three halves minus three multiplied by Y squared, multiplied by D squared plus Y squared. And this will be all to the power of one half is equal to zero. So we need to keep simplifying. So three multiplied by Y squared multiplied by D squared plus Y squared to the power of one half is all equal to D squared plus Y squared. Do the power of three halves. So now keep simplifying three multiplied by Y squared is equal to D squared plus Y squared, which we can simplify. And we could write that and don't forget your parentheses. We can write that three multiplied by Y squared minus Y squared is equal to D squared. Awesome. And we can simplify again and we can write that two multiplied by Y squared is equal to D squared. And finally, we can rearrange and solve for Y by itself. And when we do, we'll get that Y is equal to plus or minus D divided by the square root of two. And that's it. We did it, we solve for our first answer. Hooray. Now, I need a sol for G max. So we need to note that for G max, we need to know that G max is equal to two multiplied by capital G multiplied by M multiplied by Y all divided by D squared plus Y squared all to the power of three halves. OK. So this is also equal to two multiplied by capital G multiplied by M. And now we can plug in our value for Y that we just found together. So that's D divided by the square root of two, all divided by C squared plus D divided by the square root of two all to the power of two. And this whole expression is to the power of three halves which this is all equal to two multiplied by capital G Awesome. And this is multiplied by N and this is multiplied by D divided by the square root of two, all divided by three halves multiplied by the square root of three halves multiplied by D to the power of three. And finally, this is all equal to two multiplied by capital G multiplied by M multiplied by D divided by the square root of two, all divided by three halves multiplied by the square root of three halves multiplied by D squared multiplied by D divided by two. And when we simplify one last time, you will get four multiplied by capital G multiplied by M all divided by three multiplied by the square root of three multiplied by D squared. And that's it we've solved for G max hooray, we did it. So therefore, we need to also note that the same magnitude of G max can be taken at Y equals negative D divided by the square root of two. So with that in mind, looking at our multiple choice answers, the correct answer has to be the letter by equals plus or minus D divided by the square root of two. And G max is equal to four multiplied by capital G multiplied by M all divided by three multiplied by the square root of three multiplied by D squared. Thank you so much for watching. Hopefully that helped and I can't wait to see in the next video. Bye.

(III) Two identical particles, each of mass m, are located on the x axis at x= +x₀ and x = -x₀. At what point (or points) on the y axis is the magnitude of g a maximum value, and what is its value there? [Hint: Take the derivative dg/dy.]

Key Concepts

Gravitational Field Strength (g)

Superposition Principle

Maximization using Derivatives

Watch next

(III) Two identical particles, each of mass m, are located on the x axis at x= +x0 and x = -x0. At what point (or points) on the y axis is the magnitude of g a maximum value, and what is its value there? [Hint: Take the derivative dg/dy.]

Key Concepts

Gravitational Field Strength (g)

Superposition Principle

Maximization using Derivatives

Watch next

(III) Two identical particles, each of mass m, are located on the x axis at x= +x₀ and x = -x₀. At what point (or points) on the y axis is the magnitude of g a maximum value, and what is its value there? [Hint: Take the derivative dg/dy.]