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3.8. Taking p. = 10-31 g cm ', calculate the radius of the Einstein universe and its total mass in spherical space_...

Question

3.8. Taking p. = 10-31 g cm ', calculate the radius of the Einstein universe and its total mass in spherical space_

3.8. Taking p. = 10-31 g cm ', calculate the radius of the Einstein universe and its total mass in spherical space_



Answers

Suppose all the mass of the Earth were compacted into a small
spherical ball. What radius must the sphere have so that the
acceleration due to gravity at the Earth's new surface was equal
to the acceleration due to gravity at the surface of the Sun?

Hey, everyone, this is question number eighty seven from chapter five. This problem is a little different. It makes you think Suppose all the mass of the earth were compacted into a small spherical ball. What radius must have spirit sphere have on sphere have so that the acceleration due to gravity on the new Earth's surface would equal the acceleration due to gravity at the surface of the sun. Okay, so, over this problem, we're just going to set the gravity of the Earth new equal to gravity of the sun. And we know our gravitational equation is GM earth over our and then we're going to do the same liquid expression for the sun gm son over our son squared and we can rearrange and solve for our of the new earth, which is what we want. Multiply this over and then you can divide and then we take the square root. So our earth knew is equal to our son times the square root of M earth over m son. And we can solve that. That is equal to raise to the sun. Six point nine six times ten to the eighth meters, times the square root most of the earth, five point nine eight times ten to the twenty fourth meters kilogram. Excuse me, master kilograms divided by mass of the sun. One point nine nine times ten to the thirty kilograms. Plug that into the calculator and you get the new radius equal to one point two one times ten to the sixth meters, which is about a fifth the actual Earth's radius.

Were asked to use spherical coordinates to find the mass and center of mass of a solid hemisphere of radius A. The density at any point is proportional to its distance from the base. So for this problem, let's place the center of the base at the origin 000 and then if we have the hemisphere facing upwards So it's above the X Y plane in the density is given by the function row of X y Z equals okay, times the height above the X y plane z for some constant K. Then we have the total mass of this hemisphere is integral from Fadi equals 0 to 2 pi, integral from five equals zero to pi over two. Since it's a hemisphere integral from zero equals zero to row equals a the radius of our density, which is K Times E, which in spherical coordinates is K times row cosine phi times the differential rose squared sine phi dear oh defy d theta and taking anti derivatives and using for beans theorem weaken right. This is a product of inter grills and so we get factoring are constants que times and then from the Fada integral two pi times Ian to grow from zero to pi over two. Then what we have left over. We have the answer. Go from zero to a of co sign Phi Sine Phi Times wrote Cubed Let me take the interim. But if we get 1/4 row to the fourth so you get 1/4 eight to the fourth. Do you find and taking the anti derivative taking out Constance we get to over 1/4 is one half pi times k times in the anti derivative, both also times eight of the fourth. Then we get negative. Let's see a few different ways to do this. But if you use double angle identities, this is negative. 1/4 co sign of to fi from zero to pi over two. By the way, if you evaluate went up with pi over four K eight to the fourth to really be a one half, then sorry now by symmetry. Well, we know that our solid of symmetric about the wise e and the X seaplanes. So the moments about those planes are zero and we have that the moment about the X Y plane is integral from 0 to 2 pi integral from Phi equals zero pi over two integral from row equals zero to a of our density que times wrote times cosine phi times z which is again row times cosine phi times the differential rose squared sine phi the road If I d theta So this is inside K times wrote to the fourth times cosine squared fi times sine phi And so pulling out constants and taking into derivatives we get well, we have 1/5 from the rose times two pi is 2/5 times pi times k times a to the fifth I'm Santa girl from zero to pi over two of cosine squared Phi sine phi defy taking into derivatives We get to fifth spy Que eight to the fifth times and this is negative one third doing many use institution cosign cubed fi from zero to pi over two evaluating this is 2/5 times one third, which is to 15th. Hi. Okay, eight to the fifth. This is our moment about the X y plane and therefore the center of mass X bar y Barzee bar. Well, this is going to be 00 And then the moment about the x y plane to 15th spike a eight to the fifth over the total mass. However, four k a to the fourth. And so if we divide these, we get eight over 15 times A. This is our center of mass.

And this question were us to find the density of a steel ball. We know the steel ball has a mass of 3.475 Grand's and a diameter of 9.4 millimeters were given that the volume of a sphere is 4/3. Pie are cute where R is the radius, so let's go ahead and calculate the volume. The radius is going to be equal to 4.70 millimeters, and the way we confined this was we just take the the diameter and divided by two because the radio's was half a diameter. Now let's go ahead and change this millimeters two centimeters just to make it a little easier on ourselves when we're calculating the volume. So we know that there are 10 millimeters in one centimeter, so if we divide this by 10 will find the radius and centimeters so 4.7 divided by 10 is equal to 4.47 centimeters. Now we can go ahead and plug this into our equation, and we can use a calculator to find the volume, which we find to be 0.4 35 cubic centimeters now. We're also fine, whereas to find the density which we know is mass over volume. So we have the mass and we just calculated the volume. So we just take these two together and we simplify. We find a density is equal to 7.99 grams per cubic centimeter.

In this question we're told that we have a steel spear which has a radius Equal to 158". For us to find the mass of the sphere given the density of steel, let's convert this radius into centimetres to 1.58 inches. Conversion here one inches equivalent to 2.54 centimeters. That would give us 4.1 centimeters. So that would be the radius of the sphere in centimeters, Calculate the volume of the sphere that's equal to 4/3 pie. Are cubes, four thirds pi as 3.1416 radius is 4.1 centimeters cubed. And we find that the volume of the sphere is equal to 272 3 sig figs centimeters cubed. And then the mass of our sphere is equal to the density of the steel times the volume of the sphere. So the density of Steel is given to us at 788 grounds for centimeters cubed and we saw for the volume of the sphere, which is 270 centimeters cubed. Therefore, the mass of the sphere works out to 2.13 Times 10 to be three grams.


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