Physics Test

Gravitation and Rotation

  1. (2) At the surface of planet Vulcan, where Spock is from, g was equal to some constant we will call A. As a multiple of A, what wasg at the following locations:
  1. At a point in space above Vulcan where the distance to the center of the planet was 40 times the planet’s radius. A/1600 free points right here.
  1. On the surface of a planet whose radius is equal to Vulcan’s, but whose mass is triple that of Vulcan. 3A free points right here

True/False: Half a point each

  1. T/F: An orbiting moon in circular orbit has an acceleration vector that points tangent to the path of the orbit.
  2. T/F: The ancient Greeks and Roman astronomers were wrong when they said that we see Mars move backwards in its orbit around Earth from time-to-time.
  3. T/F: Johannes Kepler was wrong when he said that Mars and all the other planets move in a single direction on elliptical paths around the Sun.
  4. T/F: It is possible to use Newton’s equations to derive Kepler’s relationship that T2 is proportional to R3 for any orbit around a common body, where T is the period of orbit and R is the average radius.
  5. T/F: At any instant, an orbiting moon has a velocity that is not in the direction of its acceleration.
  6. T/F: Two satellites orbit a common planet with the same average orbital radius. The more massive satellite will have a greater speed than the less massive one.
  7. T/F: The force of gravity exerted on an object by a planet gets stronger as the object travels closer to the planet.
  8. T/F: g is a number which is universally constant.
  9. T/F: G is a number which is universally constant.
  10. T/F: Newton succeeded in showing mathematically how far the moon falls (i.e. deviates from a straight line path) in one second due to the force of gravity.
  11. T/F: The acceleration of the moon toward earth is g which is 9.8 m/s2.
  12. T/F: The force the earth exerts on the moon is Mg which is 9.8 m/s2divided by approximately 3600, and M is the mass of earth.
  1. You’ve likely seen images of astronauts making their tools float around with them in apparent weightlessness while in an orbiting satellite or space station. (You’ve also perhaps seen it in fiction films.) Meanwhile, we know that the thing we call g weakens as distance form earth’s center increases. So is the astronaut’s apparent weightlessness due to the fact that g has weakened substantially at this greater radius?Write a brief several sentences that definitively and quantitatively answers this question. Full credit will only be awarded to those who show a calculation of g that backs up his/her claim AND who end up saying clearly what is the reason for the apparent weightlessness. A typical satellite altitude is 200 km above the ground.

g = GM/R2 = (6.67 x 10-11Nm2/kg2)(6 x 1024kg)/(6.6 x 106m)2 = crank it out

This is probably the most time-consuming thing on this quiz. It should take a minute:

g = [(6.67 x 1013N/kg)(6)]/(6.62 x 1012) = [40/(6.62)] x 101N/kg still haven’t touched the calculator

g = 9.2 N/kg

is nowhere close to weak, so apparent weightlessness during orbit has nothing to do with g’s decrease with distance and instead is a result of the satellite being in a constant state of freefall.

That in bold italics is the only thing an A student should plan on writing for their essay, and 2 minutes total (including the math part) is all the time they should expect to spend. When people buy their AP test, but then never do any training with this in mind, I’m not sure what they are expecting in the test they purchased for $95.

Those who couldn’t answer, because they didn’t know, because they didn’t pay full attention in Chapter 7, should not waste their test time trying to polish an uninformed answer, trying to write lots of vague words to make it sound good. It will still be zero credit and a waste of test time. The answer is as short as what I just wrote in bold.

Those who calculated g as greater than 9 m/s2, but still said weightlessness is due to weak g, I don’t know what advice to give you, because you refuse to look at the facts on your own paper.

Jupiter problem solution on the next page.

  1. One of the niftiest events ever was when Galileo used a telescope to look at Jupiter. He quickly became the discoverer of the moons of Jupiter. It was the first time an object besides the Sun or Earth was found to have things orbiting it. Io and Europa are two well-known moons of Jupiter. Galileo’s data for the moons of Jupiter were quickly fed into Kepler’s equations, and they matched perfectly. That was the final nail in the coffin of the Geocentric Model of the solar system, at least among those who based their models on measured evidence (i.e. scientists).

Below are data for various objects:

Mass (kg) / Orbital Radius (AU) / Orbital Period (yr)
Jupiter / 1.9 x 1027 / 5.19 / 11.85
Io / 1.0 x 1023 / 2.81 x 10-3 / 4.85 x 10-3
Europa / 2.5 x 1023 / 5.62 x 10-3
Sun / 2 x 1030 /

N/A

/

N/A

  1. (3) Find the speed of Europa in its orbit. Use any valid units in your final answer.

v2/R = g so v2/R = GM/R2 so v = (GM/R)0.5

= [(1.96 x 10-29AU3kg-1yr-2)(1.9 x 1027kg)/(5.62 x 10-3yr)]0.5

= [(0.196)(1.9)/(5.62)]½ x (10-4)½AU/yr = 2.6 AU/yr

= 12.3 x 103m/s

  1. (1) Evaluate the quantity T2/R3 for Jupiter’s orbit around the sun in yr2/AU3.

It’s 1 yr2/AU3. Those who don’t know this will miss the question, and all the time in the world won’t change this. These are conceptual questions, not math. And when they look like math, they are conceptual questions disguised as math. Always.

  1. (2) If T is the period of Europa’s orbit around Jupiter and R is the average radius of Europa’s orbit, will the quantity T2/R3 = 1yr2/AU3 for Europa? Tell how you know.

It’s not. The Kepler constant depends on the mass of the central body. Jupiter ain’t sun. Be fast, be specific*, know your stuff, move on.

  1. (3) Solve for Europa’s period. Use any valid units.

T = 2πR/v = 2π(5.62 x 10-3)/(2.6 AU/yr) = 0.0137 yr = 4.33 x 105s

*In fact, those who really know their stuff can take full advantage of knowing that (thanks to Newton) we can now say: “because Sun’s Mass is 1050 times that of Jupiter, the conclusion is that the constant T2/R3 for the sun system is 1050 times LESS than what it is for the Jupiter system. This means that T2/R3 = 1050yr2/AU3for ALL things orbiting Jupiter. This could be used to solve for Europa’s period by plugging in R in AU.

16. The most important historical idea of this unit was the realization that Newton’s Gravitation confirms Kepler’s Laws and yields that M ~ K-1, where M is the mass of the thing being orbited. This means M is inversely proportional to the Kepler constant.

KStar = 4KSun so MStar = ¼MSun

MSun = 2 x 1030kg so MStar = 0.5 x 1030kg

But you can do it some tedious formulaic way if you want to.

  1. μs = 1 for both scenarios. The properties of the tire and the road are the determiners.
  2. 4.58 m/s, 6.53 N