Gravity of Moon & Roller Coaster Proof Help

In summary, the conversation discusses two physics problems: the first being about the difference in gravity between the moon and earth, and the second being about the apparent weight on a roller coaster with a circular loop. The conversation includes discussions on the use of kinematic equations and the calculations for gravity on the moon and earth.
  • #1
Klinger
4
0
I would appreciate any hints or help in solving the following two physics problems. At this point I am stuck after trying to solve the problems for a while.

1. Apollo astronauts hit a golf ball on the moon 180 meters and 30 meters on earth. Assume that the swing, launch angle, etc are the same on the moon and earth. Assume no air resistance.

We know that Vxo and Vyo is the same on both the moon and Earth (Vxo is the initial velocity in the x direction). We also know that tmoon = tearth *(180/30).

I try substituting these values into the kinematic equations for projectile motion but get stuck. Vy are different for Earth and moon and thus there are too many unknowns.

2. Show that on a roller coaster with a circular vertical loop the difference in your apparent weight at the top of the loop and the bottom of the loop is 6g's. Ignore friction. Show also that as long as your speed is above the minimum needed, this answer doesn't depend on the size of the loop or how fast you go through it. Assume that roller coaster starts traveling from height h and the radius of the loop is R.

My approach was to calculate the ratio of the centripidal acceleration as it exits the loop divided by the centripidal acceleration at the top of loop. I calculate the decease in velocity due to the elevation gain of 2R. But I don't seem to get the expected answer.

Thanks for any help.

:bugeye:
 
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  • #2
for your first question, are you asked for the gravity on the moon?
then you can use this:
On earth:
[tex] V_{earth}^{2} = V_{o\ earth}^{2} + 2ad [/tex]
[tex] V_{earth}^{2} - V_{o\ earth}^{2} = 2a(1) [/tex]
[tex] V_{earth}^{2} = V_{o\ earth}^{2} = 2a [/tex]

On moon:
[tex] V_{moon}^{2} = V_{o\ moon}^{2} + 2ad [/tex]
[tex] V_{moon}^{2} - V_{o\ moon}^{2} = 2a(\frac{30m}{180m}) [/tex]
[tex] V_{moon}^{2} = V_{o\ moon}^{2} = \frac{1}{3}a [/tex]

Since the initial and final speed are the same, we can eqate the two:
[tex]2a = \frac{1}{3}a [/tex]
[tex]a_{moon} = \frac{1}{6}a_{earth} [/tex]
[tex]g_{moon} = \frac{1}{6}g_{earth} [/tex]

I think you can do the second one yourself.
 
  • #3
A Couple of Clarification Questions

I have a couple of questions:
-- in the 2nd "earth" equation why did you set d=1?
-- in the 2nd "moon" equation why did you set d=30/180?

Nenad said:
for your first question, are you asked for the gravity on the moon?
then you can use this:
On earth:
[tex] V_{earth}^{2} = V_{o\ earth}^{2} + 2ad [/tex]
[tex] V_{earth}^{2} - V_{o\ earth}^{2} = 2a(1) [/tex]
[tex] V_{earth}^{2} = V_{o\ earth}^{2} = 2a [/tex]

On moon:
[tex] V_{moon}^{2} = V_{o\ moon}^{2} + 2ad [/tex]
[tex] V_{moon}^{2} - V_{o\ moon}^{2} = 2a(\frac{30m}{180m}) [/tex]
[tex] V_{moon}^{2} = V_{o\ moon}^{2} = \frac{1}{3}a [/tex]

Since the initial and final speed are the same, we can eqate the two:
[tex]2a = \frac{1}{3}a [/tex]
[tex]a_{moon} = \frac{1}{6}a_{earth} [/tex]
[tex]g_{moon} = \frac{1}{6}g_{earth} [/tex]

I think you can do the second one yourself.
 

1. How does the gravity of the moon differ from that of Earth?

The gravity of the moon is about one-sixth of the gravity on Earth. This means that a person or object on the moon would weigh about one-sixth of what they weigh on Earth. This is because the moon is much smaller and has less mass than Earth, so it has a weaker gravitational pull.

2. Can the gravity of the moon affect a roller coaster ride?

Yes, the gravity of the moon can affect a roller coaster ride, but not significantly. Since the moon's gravity is much weaker than Earth's, the roller coaster would experience slightly less gravitational force. However, the effects would be minimal and likely not noticeable to riders.

3. How does the gravity of the moon impact the tides on Earth?

The gravity of the moon plays a major role in creating the tides on Earth. The moon's gravitational pull causes the ocean water closest to the moon to bulge, creating a high tide. As the moon orbits around Earth, it also creates a bulge on the opposite side, resulting in a second high tide. The areas in between these bulges experience low tides.

4. Can the gravity of the moon affect human behavior or health?

There is no scientific evidence that the gravity of the moon has any direct impact on human behavior or health. Some studies have shown a slight increase in hospital visits during full moons, but this is likely due to other factors such as increased outdoor activities or psychological beliefs.

5. Is there any evidence to support the myth that the gravity of the moon can make objects lighter or heavier?

No, there is no evidence to support this myth. As mentioned earlier, the gravity of the moon is much weaker than Earth's and would not have a significant effect on the weight of objects. The only way an object would feel lighter on the moon is because it is experiencing less gravitational pull, not because the object itself has changed weight.

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