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Step 1: On a separate sheet of paper; describe method to determine the gravitational acceleraton on Planet X using Spring and the gold cylinder You may conduct expe...

Question

Step 1: On a separate sheet of paper; describe method to determine the gravitational acceleraton on Planet X using Spring and the gold cylinder You may conduct experiments on both worlds; ana you may Use knowledge 'gained in previous steps But you may not use any other masses or springs Step 2: Also on that sheet; record the data and any calculations needed to determine the gravitational acceleration on Planet X Organize your data neatly and show calculations completely;the mas by mass ineq

Step 1: On a separate sheet of paper; describe method to determine the gravitational acceleraton on Planet X using Spring and the gold cylinder You may conduct experiments on both worlds; ana you may Use knowledge 'gained in previous steps But you may not use any other masses or springs Step 2: Also on that sheet; record the data and any calculations needed to determine the gravitational acceleration on Planet X Organize your data neatly and show calculations completely; the mas by mass inequall Tean B, and



Answers

Refer to the formula $F=\frac{G m_{1} m_{2}}{d^{2}}$. This gives the gravitational force $F$ (in Newtons, $N$ ) between two masses $m_{1}$ and $m_{2}$ (each measured in kg) that are a distance of $d$ meters apart. In the formula, $G=6.6726 \times 10^{-11} \mathrm{~N}-\mathrm{m}^{2} / \mathrm{kg}^{2}$. Determine the gravitational force between the Earth $\left(\mathrm{mass}=5.98 \times 10^{24} \mathrm{~kg}\right)$ and Jupiter $\left(\right.$ mass $\left.=1.901 \times 10^{27} \mathrm{~kg}\right)$ if at one point in their orbits, the distance between them is $7.0 \times 10^{11} \mathrm{~m}$.

So here we have the tabulated values on the uh just turned into a graph on the y axis. We have the force in newtons and on the x axis we have one over R squared. Of course. This would be meters uh to the negative race. The native second power, we know that the force due to gravity would be equal to the gravitational constant, multiplied by the mass of planet X times the at times M, which would be the mass of the space probe divided by R squared are are being the distance between the center of the mass, rather the center of planet X. And the space probe. So given that we have this expression and we have the tabulated values graft. Now we can say that from the slope, we see that. Then the slope of this graph is equaling to 7.0 times 10 to the 15th. And the units would be newton meters squared. Of course, considering that this would be multiplied by one over R squared, Leaving us with simply newton's, which is of course the units for force. So we can compare this slope to our expression here. This intercept here is very small. We're going to say it's a negligible and we can say that then comparing the slope to um the expression for the force of gravity. We can say that then the gravitational constant multiplied by the, multiplied by the massive planet X. Multiplied by the mass of the space probe. Should be equal to 7.0 times 10 to the 15th Newton meters squared. So to find the massive planet X, we would simply divide the gravitational constant and the mass of the space probe. So we have 7.0 times 10 to the 15th. Newton meters squared. This would be divided Bye 6.67 times 10 to the negative 11th Newton meters squared per kilogram squared. And this would be multiplied by 10 kg the mass of the space probe. And so we find that then the mass of planet X would be equal to approximately 1.5 times 10 to the 25th kilograms. This would be our final answer. That is the end of the solution. Thank you for watching.

Calculate the gravitational acceleration on the surface of, ah, mercury and on Saturn's moon. So, um, the acceleration according to the ah, laws of universal gravitation, is G times the mass of the planet or moon in this case, um, over r squared. So we're going to have to go to Appendix E, um, and put in information about mercury and information about Titan. All right, so I have appendix here, Here, here, Appendix E here. Um, the acceleration from mercury is, um, first of all, the universal gravitational constant, which is 6.67 times 10 to the negative 11th power. I'm gonna use all s I based units, so I'm not gonna write my units. I know my answer will be in meters per second squared. OK, so there's the universal gravitational constant. The mass of mercury is 0.330 times 10 to the 24th power point 330 times 10 to the 24th power, and that's in kilograms. So that's an S I based unit and, um, the radius. Um um, Mercury. So this is on the surface on the surface. So that's I just need the radius of mercury. The mean radius for mercury is 2.44 times 10 to the sixth power and that is in meters good. That's in meters and at Sybase unit to the second power. So we put this into a calculator and I'll be right back with you, all right. My calculator tells me 3.70 meters per second squared, So now we need to move on to tighten. The acceleration is same. Universal gravitation, constant. That's why we call it universal, because it's always the same. Um, I need the mass of Titan. So let's get back to the appendix. Saturn Titan, Mass. Okay, the mass is the first thing. Um, it is point one 35 times 10 to the 24th power. So it has less mass than mercury over 90. The radius of Titan check here Mean radius is times 10 to the sixth, power 2.58 to the sixth power second power. So that's interesting. Even though Titan has a lower mass, its diameter or radius in this case is larger. Yeah, must have a lot lower density. So we'll put that into the calculator, and my calculator tells me one 0.35 meters per second squared

Question. Number 74 wants us to find an expression for a planet's mass using its radius, its surface gravity and the gravitational constant G. We can use Newton's law of universal gravitation to find the force of gravity on an object surface that is F super G is equal to G times the mass of the planet times the mess of an object on the planet or around the planet, divided by the distance between the two r squared. Now that's just one expression for the force of gravity on a planet. There is another expression for the force of gravity, which is useful when objects air on the surface of that planet. That is, F sub G is equal to the mass of the object times G, where g is thea acceleration due to gravity on the surface of that planet. Since both of these expressions air useful in determining the force of gravity on a planet, let's go ahead and set them equal to each other and then solve for mass. We will have the gravitational constant g times the mass of the planet, times the mass of an object on the planet or around the planet divided by r squared is equal to little m the mass of the object times thea acceleration due to gravity tree run that planet. We can cancel out the little AM on both sides of this expression and then multiply both sides by r squared over g gravitational constant. When we do that, we find that the mess of our planet is equal to G For that planet. Times are squared over the gravitational constant and that is our expression for the mass of a planet using only the gravitational acceleration on that planet, the planet's radius and the gravitational constant G.

And again we have a gravitational force problem. We know that the gravitational constant is given here and we want to know the gravitational force between the earth and a human standing on the earth given that the diane the distance from the center of the earth to um the human is Um six and 371000s times 10 to the 60 m. So we're going to take the gravitational constant six 6726 times 10 to the negative 11 times the mass of the first body which is Earth. Yeah, comes the math mass of the second body which is the human. And we're going to divide that by The distance between the two squared 2637. 1 Times 10 to the six. That would be the radius of the earth. Mhm And multiply 6.67- six. I'm just multiplying this fabulous about this Times 5.28 tons 80. That gives me 3192.17184. And that's going to be times 10 to the -11 plus 24. About that by 6.371 squared just 40 0.59 Times 10 to the 12. So this is three. Mhm She's 10 to the 13. Yeah, So then this is going to be 3.19 times 10 to the 16th because I have to move this decimal three places divided by 4.06 times 10 to the 13. So it would be okay 0 6. So I divide these two And get 78 seven And then divide these 2 10 to the 1613. So that's gonna be seven Point Paint Times. This would be to 3 -3 positive three. Then I go one to the right Height that becomes Thomas 10. Mhm. Yeah. Yeah. To the fourth. Yes to the second. Yeah. So 7.8 times 10.7857 Times 10 to the 3rd. His 785.7. And then If I write it in scientific notation at seven .85 times 10 to the second. And again, depending on how your instructor or your directions tell you around you may need to keep them more exact but I'm rounding these sort of in scientific notation so it would be 7.8 times 10 to the second. Mhm.


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