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Assessment
https://cms.psu.edu/Section/Assessment/Delivery/AssessmentAl...
1.
Open the simulation "gravity and orbits" found here:
http://phet.colorado.edu/en/simulation/gravity-and-orbits
Start by exploring the simulation. Turn on the gravity force arrow, the velocity arrow and click to show path as well. Note that you can move
the Earth and the Sun (or moon or ISS) by simple click and drag. Try all sort of things and describe a strange or fun situation that you were
able to simulate. Explain what you had to do to get to that situation.
many possible answers here. It is very easy to crash things.
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2.
Push reset and go back to the Earth and Sun (in either cartoon or to scale view). What does the length of the gravity arrow represents?
Strength of gravity force
3.
Looking at the gravity arrow. The gravitational pull from the Earth on the Sun is
A) lower than the gravitational pull from the Sun on the Earth.
B) greater than the gravitational pull from the Sun on the Earth.
C) equal to the gravitational pull from the Sun on the Earth.
D) sometimes greater and sometimes smaller than the gravitational pull from the Sun on the Earth.
4.
Can you explain how your previous choice make physical sense?
The exchange in momentum is equal. What momentum the Earth loses or gains, the Sun gains or loses.
So the force of gravity on the Sun from the Earth is equal (but opposite direction) to the force of
gravity from the Earth on the Sun).
You can see this from Newton's universal law of gravity which depends on both masses.
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5.
Go to Earth-Sun in the "to scale" mode. Click to show the masses of each object. At this point the Earth's period should be 365 days (if it is
not, just hit the reset button). Double the Sun’s mass. What is approximately the new period? Don't forget your units!
140 days
6.
Hit reset to go back to a 365 days period. Keep the Sun’s mass at its original value but now double the mass of the Earth. What is the new
orbiting period?
365 days
7.
From Newton’s law of gravity, we can work out the following formula for the orbital velocity:
v2 = G M/r
where v is the orbiting speed, r is the orbital distance and G is Newton's constant (see the course page for more on this).
The orbital period is just the distance traveled in one revolution divided by the speed. So the period T is inversely proportional to the
velocity (T~1/v).
In the formula for the velocity (or period), the mass M is the mass of which object?
mass of the sun
8.
According to Newton's theory of gravity, if I double the mass of the Sun, by how much should the velocity of the Earth change? How much
should the period change?. Does this agree with what you found in question 5? If not, why not?
If I double the mass the velocity should change by square root of 2. Or about 1.4. The period which
goes like 1/v will therefore decreases by 1.4. So Newton's theory predict that it the period should
be 365/1.4 = 260 days.
In question 5, I found 140 days. This is very different. The reason is that he orbit is now an
ellipse and not a circle!
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9.
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If I double the radial distance between the Earth and the Sun what happens to the orbital speed?
A) It increases by square root of 2 (about 1.4)
B) It decreases by square root of 2 (about 1.4)
C) It doubles
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Assessment
https://cms.psu.edu/Section/Assessment/Delivery/AssessmentAl...
D) It is reduced by 2
10.
Now imagine that instead of orbiting around an object you are orbiting inside of an object of uniform density (this is possible if you are
orbting inside of a thin gas with no drag for example). Newton showed that the only mass that matter for your orbiting motion is the mass
inside of the radius you are orbiting as the figure below shows.
The mass inside of a radius r increases as the volume increases. As I move the satellite further away from the center, there is more and
more mass inside the orbit and the force of gravity will be bigger not smaller.
Imagine I now the double the radius (still saying inside the gas). What will happen to the velocity?
Hint: Newton's formula for v is still good, its just that the mass affecting the orbital motion now increases as you go move out (see figure).
A) The orbiting velocity increases by square root of 2 (about 1.4)
B) The orbiting velocity decreases by square root of 2 (about 1.4)
C) The orbiting velocity doubles
D) The orbiting velocity is reduced by 2
11.
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The figure below shows the measured speed of stars (in km/s) orbiting a neighbor galaxy as a function of distance from the center of that
galaxy. Kpc stands for kilo parsec and it is an astronomical unit to measure distance. On the figure, we added the image of the galaxy.
Amost all the visible matter ends at 4 kilo parsec. Can you explain the graph between 1 kpc and 4 kpc? What do you think is going on
between 4 kpc and 6 kpc?
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Assessment
https://cms.psu.edu/Section/Assessment/Delivery/AssessmentAl...
Between 1 kpc and 4 kpc. It looks like the speed decreases with radius. This makes sense with
Newton's theory. You get farther away, the force of gravity is weaker.
From 4 to 6 kpc, the speed start increasing again. This looks like the case of Q10. It is as if the
stars were orbiting inside of a gas of something. The further you are the more gas is pulling you
in. ****This is dark matter and this is how it was discovered.
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Done
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11/25/13 10:46 AM