The Near Earth Asteroid Rendezvous (NEAR) spacecraft, after tr...

Question

The Near Earth Asteroid Rendezvous (NEAR) spacecraft, after traveling 2.1 billion km, orbited the asteroid Eros with an orbital radius of about 20 km. Eros is roughly 40km x 6km x 6km. Assume Eros has a density (mass/volume) of about 2.3 x 10³ kg/m³ .

(c) Estimate the orbital period of NEAR around Eros, as if Eros were a sphere.

(c) Estimate the orbital period of NEAR around Eros, as if Eros were a sphere.

Accepted Answer

Welcome back. Everyone. In this problem. A satellite is orbiting a strange space rock at a distance of 3.5 billion kilometers from earth. The space rock has a mass of 1.5 multiplied by 10 to the 14th kilograms and can be modeled as a sphere with a radius of 2.5 multiplied by 10 to the third meters. The satellite orbits at an altitude of four kilometers above the surface of the space rock, find the orbital period of the satellite above the space rock. A says it's 2.2 hours. B 4.4 hours C 9.1 hours and d 1100 hours. Now, if we're going to find the orbital period of our satellite, let's try to first draw a diagram and then make note of all the information we have. So let's say that this is our satellite. OK. And it's above our space rock and our space rock has a radius of R the satellite is above our space rock by a height of H. And now we know that first, the mass of the space rock is 1.5 multiplied by 10 to the 14th kilograms. The radius of the space rock is 2.5 multiplied by 10 to the third meters. We know the satellite orbits at an altitude of four kilometers above the space rock. So H equals four kilometers, four multiplied by 10 to the third meters. And thus the distance from the center of the space rock to the satellite. This distance here is going to be R plus H OK. And R plus H R plus H equals four, multiplied by 10 cubed meters plus 2.5 multiplied by 10 cub meters, which is 6.5 multiplied by 10 cubed meters. So we have this information, we also know its distance from earth. And now we want to find the orbital period of the satellite about the spacer. Now, what do we know about the orbital period that can be related to all this information that we have? Well, first, OK, recall that the orbital period of our satellite can be calculated as T equals two pi multiplied by the distance from the center of the object. It's orbiting around to the satellite. OK. So that's going to be R plus H divided by the speed V. Now, we know how far it is. We know R plus H but we don't know the speed that our object or satellite is orbiting at. So we'll need to figure that out. What else do we know? Well, we also know that our speed V is equal to the square root of GM divided by R plus H. OK. So now since we have that information, then if we substitute what we know about V, then that tells us our period T is equal to two pi multiplied by R plus H divided by the square root of GM divided by R plus H. No, we can do a bit of work here. OK? Because no, if I were to rewrite this or if I were to distribute my square root across the numerator and the denominator based on what we know about uh radicals, then we could rewrite this as two pi multiplied by R plus H multiplied by the square root of R plus H all divided by the square root of GM. OK? No, we can let me keep my two pi outside here. No, we can even write this. We can combine our terms for R plus H and the square root of R plus H. Because if we write, if we rewrite R plus H with a square root, that's gonna be two pi multiplied by R plus hr squared, let me, let me write that a little better here can multiplied by the square root of R plus H. And that's being divided by the square root of GM. I know all of these terms are under the square root. So we can combine them. In other words, the numerator is going to be R plus H squared multiplied by R plus H divided by GM all under the square root, which we can simplify to be two pi multiplied by the square root of R plus HQ divided by GM. Know that we have that and we know R plus H, we know our gravitational constant G and we know the mass of our, uh we know the mass of our space rock, then we can go ahead and substitute. So that's going to be two pi multiplied by the square root of R plus 86.5 multiplied by 10 to the third meters cube. OK? And all of that is being divided by the gravitational constant 6.67 multiplied by 10 to the negative 11th, Newton square meters per square kilogram multiplied by the mass of the rock. 1.5 multiplied by 10 to the 14th kilograms. Now, when we work the salt, OK, then we should get our orbital period. Pt sorry to be 32,918 0.667 seconds. So we know what the period is. But if we look back at our answers, they were written in hours. So we'll have to convert this to hours and we can do that by multiplying by one hour divided by 3600 seconds because 3600 seconds are in an hour. Now, when we multiply, then in hours, we get our period to be equal to 9.14 hours or to two significant figures. 9.1 hours. Therefore, if we go back to our answer choices, that means c is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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