A uniform, spherical, 1000.0-kg shell has a radius of 5.00 m. (a)...

Question

A uniform, spherical, 1000.0-kg shell has a radius of 5.00 m. (a) Find the gravitational force this shell exerts on a 2.00-kg point mass placed at the following distances from the center of the shell: (i) 5.01 m, (ii) 4.99 m, (iii) 2.72 m.

Accepted Answer

Welcome back everybody. We are looking at an imaginary planet that is in the shape of a uniform spherical shell. We have that. The mass of this planet is 3. times 10 to the 21st kilograms, we're told that the radius or distance from the center center being right here is kilometers. Now we are also looking at three little point charges or little masses In reference to the sphere. Two of them on the inside and one of them on the outside. Now the distances are we have this little guy, this little distance is only 200 m away from the center. This guy is 99 km away from the center. And then we have the outside guy Who is 101 km away from the center. Now we are told that the mass of all these little points are only 200 mg and we are asked to find the gravitational force of the planet on each of these masses. Well, we're gonna use this relationship right here. We know that for a uniform spherical shell, it has a potential energy. Well, sorry, pardon me for a uniform spherical shell in reference to a church or a point either inside or outside of that shell. The potential energy of that point is given by this formula, that U. Is equal to big G. Which is just some constant times the mass of our body or our planet times the mass of the little point times the distance of the point away from the center. Now, the force of gravity is what we're looking for And we actually know that the derivative of potential energy with respect to the distance away, the negative of that is equal to our force. So let me go ahead and take the derivative of this equation right here with respect to our Now, since we're looking at our, everything else is going to be treated as constant, the little imaginary negative one right there is going to move to the front. So the derivative or the force of gravity of the planet is equal to Big G times Big M times little M all divided by r squared. So let's take these points just one step at a time here away from the center and figure out the force of gravity of the planet on the point. So first we're gonna look at the point that is just right outside the shell, that is 100 and one kilometers away from the center. Let's just plug in our values like I said, Big G is just a constant equivalent to 6.67 times 10 to the negative 11 times the mass of our planet, which is 3.0 times 10 to 21 times the mass of our little point, which we know is 200 all divided by the distance of the point to the center, which is 100 and one kilometers or 101, m. This, when you plug it into your calculator, we get that the force of gravity on our 101, um, point away is 3.9 to times 10 to the third. Newton's great. So now let's look at our points B and C. Here. Apologies that didn't label point A. But that was our point A. Let's go ahead and look at B and C. Here at the same time. We also know by studying uniform spherical shells that any points on the inside of the cell shell. Sorry, the potential energy on the inside is just some constant, which when you take the derivative, we get that our force of gravity is zero for all points inside the shell, meaning that our force of gravity for the 1 99 kilometers away is equal to our force of gravity on the m point away equal to zero. This gives us our final answer of a thank you all so much for watching. Hope this video helped. We will see you all in the next one.

A uniform, spherical, 1000.0-kg shell has a radius of 5.00 m. (a) Find the gravitational force this shell exerts on a 2.00-kg point mass placed at the following distances from the center of the shell: (i) 5.01 m, (ii) 4.99 m, (iii) 2.72 m.

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