Lab 8 Gravitational Force and Orbits-converted

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Lab 8 Gravitational Force and Orbits Resources The Virtual Lab https://phet.colorado.edu/sims/html/gravity-force-lab/latest/gravity-force-lab_en.html Paper. Pencil, Calculator Software Requirements The new HTML5 sims can run on iPads and Chromebooks, as well as PC, Mac, and Linux systems. iPad: iOS 11+ Safari iPad compatible sims Android: Not officially supported. If you are using the HTML5 sims on Android, we recommend using the latest version of Google Chrome. Chromebook: Latest version of Google Chrome The HTML5 and Flash PhET sims are supported on all Chromebooks. Chromebook compatible sims Windows Systems: Microsoft Edge and Internet Explorer 11, latest version of Firefox, latest version of Google Chrome. Macintosh Systems: macOS 10.9.5+, Safari 9+, latest version of Chrome. Linux Systems: Not officially supported. Please contact phethelp@colorado.edu with troubleshooting issues.

Isaac Newton's law of universal gravitation proposed that the gravitational attraction between any two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. In equation form, this is often expressed as follows: The constant of proportionality in this equation is G - the universal gravitation constant. The value of G was not experimentally determined until nearly a century later (1798) by Lord Henry Cavendish using a torsion balance. Cavendish's apparatus for experimentally determining the value of G involved a light, rigid rod about 2-feet long. Two small lead spheres were attached to the ends of the rod and the rod was suspended by a thin wire. When the rod becomes twisted, the torsion of the wire begins to exert a torsional force that is proportional to the angle of rotation of the rod. The more twist of the wire, the more the system pushes backwards to restore itself towards the original position. Cavendish had calibrated his instrument to determine the relationship between the angle of rotation and the amount of torsional force. A diagram of the apparatus is shown below. Cavendish then brought two large lead spheres near the smaller spheres attached to the rod. Since all masses attract, the large spheres exerted a gravitational force upon the smaller spheres and twisted the rod a measurable amount. Once the torsional force balanced the gravitational force, the rod and spheres came to rest and Cavendish was able to determine the gravitational force of attraction between the masses. By measuring m 1 , m 2 , d and F grav , the value of G could be determined.

Cavendish's measurements resulted in an experimentally determined value of 6.75 x 10 -11 N m 2 /kg 2 . Today, the currently accepted value is 6.67259 x 10 - 11 N m 2 /kg 2 . The value of G is an extremely small numerical value. Its smallness accounts for the fact that the force of gravitational attraction is only appreciable for objects with large mass. While two students will indeed exert gravitational forces upon each other, these forces are too small to be noticeable. Yet if one of the students is replaced with a planet, then the gravitational force between the other student and the planet becomes noticeable. School administrators generally do not allow physics classes to take field trips to other planets to study gravitational forces. And it is rather difficult to change the mass of the earth or the moon in an effort to see if such changes have an effect upon the gravitational force. But with this Interactive lab, changing the masses of planets and moons and the distance that separates them is a snap or click. And by doing so, we can discover the law of universal gravitation without ever leaving home or school.

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The Lab Environment Spend a few minutes to understand/ explore the functionalities of the different tabs/components. Move the slider(s) to change the values of the two masses Instructions 1. Hold and drag the two masses to adjust the distance between the two masses. Let it is r 2. Drag the sliders to adjust the two masses. Let these masses are m 1 and m 2 . 3. Note down the values of the forces exerted by the two masses on one another. Since these forces will be equal in magnitude and opposite in direction, select any one and let it is F 12 , that is force on mass 2 because of the mass 1. It can be noted that the force exerted by the mass 2 on mass 1 will be F 21 and is of the same magnitude. 4. We can make the G as subject of formula, that is

2 m m F 12 = G 1 2 ⟶ r 5. Using the above relation, find the value of the constant. Write your answer in scientific notation. 11 N . m 2 6. Compare the calculated value of G with G = 6.673 × 10 − kg 2 and calculate the %age error. Part 1 Determination of Value of Universal Gravitational Constant Observations and calculations No. of Obs Mass M 1 Mass m 2 r F 12 G F r 2 = m m 1 2 1 520 850 6 8.29 x 10^-7 6.75 x 10^-11 2 800 270 3 1.60 x 10^-6 6.67 x 10^-11 3 300 200 10 4.00 x 10^-8 6.67 x 10^-11 4 220 610 5 3.58 x 10^-7 6.67 x 10^-11 %age Error Calculated Value of Actual Value of G = 6.69x10^-11n xm^2/kg^2 G = 6.673x10^-11 nxm^2/kg^2 Average G =6.69x10^-11nxm^2/kg^2 %age Error Actual value of G − Calculated value of G = Actual value of G × 100 = 0.2547 % 1 2

Part 2: Gravity and Orbits Now go to the Gravity and Orbits simulation. https://phet.colorado.edu/sims/html/gravity-and-orbits/latest/gravity-and- orbits_en.html Use the Model . Make sure to click Gravity Force , Velocity and Path and make sure Gravity is checked On . Run several simulations, changing parameters as you go. Answer the following questions: 1. What direction is the gravitational force of the orbiting object? Towards the center of the earth. 2. What direction is the velocity of the orbiting object? West to east. 3. If you turn gravity off , what happens? Why does this happen? If you were to turn off gravity all things will float out of orbit because there will be no force forcing it to stay on its path. Gravity attracts all objects towards their center without that force there is no attraction between objects. 4. If you increase the mass of the Sun (or other object at the center of the orbit), provide an explanation of what happens to the Earth (or other satellite). If you were to increase the mass of the object in the center of the orbit, the outside object would start to orbit closer to the center object and the velocity of the orbit will increase. 5. If you decrease the mass of the Sun (or other object at the center of the orbit), provide an explanation of what happens to the Earth (or other satellite). When you decrease the mass of the object in the center the force of the gravitational pull decreases, the orbit of the outer object will be a lot farther away than the center object increase the orbit size.

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6. Move the satellite (the orbiting object) closer to the object it is orbiting and further from the object it is orbiting. Describe what happens and explain why this happens. As the satellite is moved closer to the object it is orbiting, it will experience a stronger gravitational pull. This will increase its orbital speed because the gravitational force is directly proportional to the distance from the center of mass of the central object. As the satellite is moved further from the object it is orbiting, the gravitational force it experiences will decrease. This will result in a decrease in its orbital speed because the gravitational force is inversely proportional to the square of the distance. 7. Change the mass of the satellite only. Will this change the orbit of the satellite? Why or why not? Explain what happens. Changing the mass of the satellite will not change the orbit. The central object's mass is larger than that of the satellite, so the satellite's mass is insignificant in the gravitational equation. 8. The mass of the Moon is 7.35x10 22 kg and the mass of the Earth is 5.97x10 24 kg. The Moon’s orbit distance is 3.84x10 8 m. What is the gravitational force on the Moon by the Earth? What is the gravitational force of the Earth on the Moon? Explain the difference between the forces. (either type out your work below or do your work on paper and take a photo and insert it here). Gravitational F = !.!#×%& !"" ’( # /*+ # (#.-.×%&//*+ 1 ..2#×%&/3*+) (-.53×%&56) # = 1.99 x 10 20 N The gravitational attraction between the Earth and the Moon is an illustration of Newton's third law of motion, which asserts that there is an equal and opposite response to every action. The Earth's gravitational force on the Moon is the action in this example, while the Moon's gravitational pull on the Earth is the response. They have the same magnitude but act in opposing directions, keeping both objects in orbit around their shared center of mass.

9. The mass of the Moon is 7.35x10 22 kg and the radius of the Moon is only 1737.1 km. The value of g on earth is 9.8 m/s 2 . What would g be on the moon? G Moon = !#×%& !"" ’( # /*+ # (#.-.×%& ## *+) (%.#-#%×%& $ () # = 1.625 m/s 2 10. If the mass of the Earth is … increased by a factor of 2, then the F grav is increased by a factor of 2 . ... increased by a factor of 3, then the F grav is increased by a factor of 3 . ... decreased by a factor of 4, then the F grav is decreased by a factor of 4. 11. If the separation distance between the moon and the planet is … ... increased by a factor of 2, then the F grav is decreased by a factor of 4 . ... increased by a factor of 3, then the F grav is decreased by a factor of 9 . ... increased by a factor of 4, then the F grav is decreased by a factor of 16.

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