A uniform, solid, 1000.0-kg sphere has a radius of 5.00 m. Fin...

Question

A uniform, solid, 1000.0-kg sphere has a radius of 5.00 m. Find the gravitational force this sphere exerts on a 2.00-kg point mass placed at the following distances from the center of the sphere: (i) 5.01 m, (ii) 2.50 m.

Accepted Answer

Welcome back. Everyone. In this problem, a spherical asteroid of mass 1500 kg has a radius of 15 m. Calculate how much gravitational force the asteroid exerts on a nearby tiny space rock of 1.5 kg. If the space rock is kept at a distance of 1st 15.5 m and 2nd 10 m from the center of the asteroid. For simplicity, assume that the space rock is a point particle. For our answer choices A says when the distance is 15.5 m, the force is 2.15 multiplied by 10 to the negative 10th newtons. While when it is 10 m, the force is 4.45 multiplied by 10 to the negative 10th newtons. B says they are 2.15 multiplied by 10 to the negative 10th and 1.25 multiplied by 10 to the negative 10th newtons respectively. C says they are 6.25 multiplied by 10 to the negative 10th and 4.45 multiplied by 10 to the negative 10th newtons respectively. And the D says they are 6.25 multiplied by 10 to the negative 10th and 1.25 multiplied by 10 to the negative 10th newtons respectively. Now, in this problem, we're basically trying to find how much gravitational force or asteroid exerts for different distances. Ok. And so far one thing that's constant is the mass of the asteroid, let's call that capital M of 1500 kg. OK. It's a radius of 15 m. OK. And it's the, the, the mass of the time is space rock. Let's give that the value of common M of 1.5 kg. OK. So the question is, how are we going to find a gravitational force? What do we know about gravitational force that's related to distance? Well, recall that the gravitational force F is going to be equal to the gravitational constant G multiplied by the product of the two masses divided by the distance between the masses squared. OK. So I'm going to represent R here, the radius of the asteroid as R not so that we don't confuse it with R from our formula. But to figure out the gravitational force, we can use this formula in both scenarios. So let's start with part one and in part one, we're at a distance at a distance of 15.5 m. OK. Now note that the distance of 15.5 m from the center of the asteroid to the space rock is greater than the radius of the asteroid R equals 15 m. So for spherical masses, it can be assumed that all of the mass is concentrated at a point at the center of the spherical mass. And since the space rock is tiny, we assume it's a point particle. So now that means we can substitute R here in our problem for 15.5 m. So this force let's call it F one is going to be equal to the gravitational constant 6.7674 multiplied by 10 to the negative 11th Newton square meters per square kilogram multiplied by or the masa asteroid 1500 kg multiplied by the masa for space rock 1.5 kg, which is all going to be divided by that distance of 15.5 m squared. Now, when we go ahead and solve here and run our answer to three significant figures like like the rest of our answer choices, then our first force when it's at a distance of 15.5 m is approximately equal to 6.25 multiplied by 10 to the negative 10th newtons. OK. Now let's move on to part two of the question. And in part two, remember we're trying to find it at a distance of 10 m from the center of the asteroid. So let's make a note here, let's say at 10 m away. Now, in this one, it's a bit different from part one because here we can tell that the distance from the center of the asteroid 10 m to the space rock is less than the radius of the asteroid. So that means the space rock is placed inside the asteroid. And in such cases, only the inner mass of the asteroid that is at a distance less than 10 m from the center of the asteroid contributes to the gravitational acceler gravitational attraction. Sorry, the outer mass plays no part. OK. So hence, we're going to consider the attraction between all the mass of the asteroid at a distance of less than 10 m and the space rock and the space rock at all distance for 10 m. In other words, F two, OK, is going to be equal to g multiplied by our mass or inner mass m multiplied by the mass of the space rock divided by our inner radius of 10 m. OK. Now, what do we know about that inner mass? We don't know what it is yet. OK. We only know the mass of the asteroid. How can we find the inner mass? Well, our inner mass, if we think about it is going to be equal to the asteroids density multiplied by that inner volume. OK. And we know that the density of the asteroid is going to be constant here if I can just make a quick note, remember that density is equal to mass over volume. So we can use the asteroids mass and the asteroids overall volume to find that constant density and multiply it by the inner volume. Now, what is that overall volume going to be. We remember we're assuming our asteroid is feral. So its mass is going to be M and its volume is going to be four thirds of Pi Rnq. OK. We remember we said RN is the radius of the asteroid and we're going to now be multiplying that by our inner volume. OK. Which as we said, let me write it here is going to be four thirds of pi multiplied by R I cubed for our inner radius of 10 m. Now you probably notice we have some common terms here. We can cancel four thirds, we can cancel pi and that leaves us then with an inner mass of M multiplied by R I cubed divided by R not cubed. OK? Where R not again is the mass of our inner or, or sorry, it's the mass of or asteroid. So basically, we've been able to find our inner mass using this formula now that we have our inner mass, we can substitute it into our formula for our force. OK? Because now that tells us then that F two is going to be equal to GM I. Well, what do we know? M is M is MRI cubed divided by, are not cubed. OK. Multiplied by M divided by R I squared. OK. Now notice we have some terms that we can cancel here. OK. We can say, well, you know what, let's um let's go ahead and cancel, say R I squared into R I cubed. That goes, that will leave us with R I OK. And no, our formula then is going to be GM MRI divided by R not cubed. Since we know what all of those values are, we can go ahead and substitute. So that's 6.674 multiplied by 10 to the negative 11th, Newton square meters per square kilogram multiplied by our mass of 1500 kg, multiplied by the mass of the tiny space rock, 1.5 kg multiplied by the inner radius of 10 m. And that's all going to be divided by the radius of the asteroid 15 m cubed. Now, when we go ahead and solve here and write it to three significant figures, then we get our answer to be approximately 4.45 multiplied by 10 to the negative 10th new times. Thus, at a distance of 10 m, the gravitational force is 4.45 multiplied by 10 to the negative 10th newtons. While at 15.5 m, the force is 6.25 multiplied by 10 to the negative 10th newtons. That means C is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

A uniform, solid, 1000.0-kg sphere has a radius of 5.00 m. Find the gravitational force this sphere exerts on a 2.00-kg point mass placed at the following distances from the center of the sphere: (i) 5.01 m, (ii) 2.50 m.

Key Concepts

Newton's Law of Universal Gravitation

Shell Theorem

Gravitational Field Inside a Sphere

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