(III) A 2800-kg space vehicle, initially at rest, falls vertic...

Question

(III) A 2800-kg space vehicle, initially at rest, falls vertically from a height of 2900 km above the Earth’s surface. Determine how much work is done by the force of gravity in bringing the vehicle to the Earth’s surface.

Accepted Answer

Welcome back. Everyone in this problem. A satellite of mass 200 kg was initially at rest in space at 14,000 kilometers above the Mars surface. Suddenly it fell vertically down to Mars. Calculate the work done by the force of gravity in moving the satellite to Mars surface from its initial position. Assume the mass and the radius of Mars are 6.4 multiplied by 10 to the 23rd kilograms and 3.4 multiplied by 10 to the sixth meters respectively. For our answer choices A says the work done is 6.2 multiplied by 10 to the seventh Juels B 1.2 multiplied by 10 to the 10th jewels. C 1.2 multiplied by 10 to the 32nd jewels and D is 1.1 multiplied by 10 to the 44th Js. Now, if we're going to figure this out, let's try to visualize what's really going on here. So we're talking about a satellite, OK, with a given mass that's falling toward a planet. Now, for that satellite, we know that it falls a distance of 14,000 kilometers. So that means the height it falls in si units is 1.4 multiplied by 10 to the seventh meters. OK. We also know that the radius of Mars, let's call that radius RM rm is equal to, OK. You make some space here and that, that radius IM equals 3.4 multiplied by 10 to the sixth meters. OK. And we're trying to figure out the work done by the force of gravity. If we make a note of all the other information we have, we also know that the mass of the satellite, I'm going to write that as a common M equals 1200 kg. OK. And we also know that the mass of Mars capital M equals 6.4 multiplied by 10 squared, sorry 10 to the 23rd kilograms. OK. So let's call that a capital M the mass of Mars. Now, if we're going to figure out the work done by the force of gravity right away, looking at the information we have, we don't necessarily know what that work is, but we're talking about the work done by the force of gravity. What do we know about gravitational forces? Well, recall, OK, that the gravitational force on the satellite F is going to be equal to negative G multiplied by capital M multiplied by a common M, both of those masses, the mass of mass and the mass of the satellite divided by R squared as a vector R. OK. Where are in our formula is the distance from the center of mass to the satellite. OK. No, the satellite is moved from above the surface of Mars to Mars surface. So in essence, then comparing it to the center of Mars, then the satellite went from RM plus H OK. That entire distance RM plus H down to RM. OK. Now the question is, what do we know about work and forces? Well, recall that work is the integral of force with respect to a displacement help. So, since that's true, that means then that the work done is going to be the integral of our force with respect to L OK, where it's going to be. Well, here we don't have our displacement but we know that LDL or change in L is equal to D was played by the vector art. OK. So what that means then if we substitute that into our integral, then let me put this in right here, read from Mars, right. Then this means the work done is going to be the integral of our force which is negative GMM divided by R squared. OK. R OK. RDR which is what we said was uh what we said was our, what was DL came Victor Dl where it's going to be between the boundaries of RM plus H and RM because that's the distance over satellite fell. Those are the bounds of satellite fell. And we're saying that R vector R points radially outward from the center. No, we can rewrite this, then we can rewrite our work. W as an integral between RM plus H and RM of a negative GMM tmm divided by R squared. Dr. So we're, we're integrating it with respect to RO and if we take O to our constants, that means we can write, I think I can make some space over here, we can rewrite our work. Then as negative GMM OK, multiplied by the integral between RM plus H and RM of one divided by R squared or rewriting that in terms of a power as R to the negative to D, now we're integrating purely with respect to R. So let's go ahead. So when we do that, we get our work to be equal to negative GMM multiplied by the integral of R ray to the power of negative two. We're integrating a term that's raised to a power, which means we're going to add one and write our new power as the denominator of our expression. So that's going to be R to the negative two plus one which is negative one divided by the new power, negative one. And it's still between the bonds RM plus H and RM. Now this is good because notice both terms in our product are negative. So we can divide negative GMM by a negative one and that leaves us then with the work GMM rw equals GMM multiplied by R to the negative one. T, the bunches of RM plus H and RM. Now, what does that mean for us? Well, if we expand it here, it's between these bonds. That means we're substituting RM and RM plus H into the result of our integral. So we can reri R to the negative one as one divided by R and no, that means the work done equals GMM multiplied by one divided by RM minus one divided by RM plus H. OK. So that is going to be the work done. Let's take a quick uh let's take a quick check here of what we know. Well, we know our gravitational constant G, we know the mass of Mars capital M. We know the mass of the satellite common M we know the distance or we know the radius of Mars RM and we can figure out the radius of Mars plus the height that our satellite dropped RM plus H. So now that we have all of this information, let's substitute to solve for the work done because now this tells us that the work done is going to be equal to G 6.67 multiplied by 10 to the negative level Newton meters squared per kilogram squared. OK. Multiplied by Armas of Mars which is 1200 kg multiplied by sorry, the mass of the satellite 1200 kg multiplied by the mass of Mars 6.4 multiplied by 10 to the 23 kg. OK? And all of this, I'm going to have to write this in two lines. All of this is multiplying one divided by the radius of Mars 3.4 multiplied by 10 to the sixth meters minus one divided by RM plus H. That is 3.4 multiplied by 10 to the sixth plus 1.4 multiplied by 10 to the seventh. OK. Which are all written in meters. So if we're going to wear a result, we can put our first expression into our calculator multiply it by our second expression. Or we can just write out uh the whole thing, whatever we do, when we multiply, we should find that we get our work to be equal to 1.212 multiplied by 10 to the 10th. Jus If we write it to two significant figures, we put it in standard form, it's going to be equal to approximately 1.2 multiplied by 10 to the 10th jewels. Therefore, the work done by the force of gravity in moving the satellite to Mars surface from its initial position is 1.2 multiplied by 10 to the 10th. Joes. If we look back on our answer choices, that means B is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

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(III) A 2800-kg space vehicle, initially at rest, falls vertically from a height of 2900 km above the Earth’s surface. Determine how much work is done by the force of gravity in bringing the vehicle to the Earth’s surface.

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