AP Physics MC Questions: Orbits, Momentum

  • #1

Homework Statement



[Broken]
Answer is D.
[Broken]
Answer is E.

Homework Equations



14. v=√(Gm/r)
17. p=mv
τ=rxF
L=Iω

The Attempt at a Solution



Number 14 made perfect sense until I saw the answer, which is D. If the velocity of the spacecraft increases, according to v=√(Gm/r), the radius of the orbit decreases. So, I chose A. Why would the answer be D, where the earth becomes the focus of an elliptical orbit?

Number 17 also puzzles me. I thought that the answer was A, because the linear momentum decreases from its original value, while angular momentum is increased from 0 to a larger value. Why then, is the answer that linear momentum and angular momentum are both conserved?

Thanks in advance!
 
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Answers and Replies

  • #2
SammyS
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Homework Statement



[ IMG]https://imageshack.us/scaled/large/89/172004u.jpg [Broken]
Answer is D.
[ IMG]https://imageshack.us/scaled/large/17/142004h.jpg [Broken]
Answer is E.

Homework Equations



14. v=√(Gm/r)
17. p=mv
τ=rxF
L=Iω

The Attempt at a Solution

Number 14 made perfect sense until I saw the answer, which is D. If the velocity of the spacecraft increases, according to v=√(Gm/r), the radius of the orbit decreases. So, I chose A. Why would the answer be D, where the earth becomes the focus of an elliptical orbit?
v2 = Gm/r is only valid for a particular kind of orbit.

For elliptical and circular orbits the total energy of the orbit is negative. An increase in speed, corresponds to an increase in total energy, so the total energy becomes 'less negative' so to speak.

Number 17 also puzzles me. I thought that the answer was A, because the linear momentum decreases from its original value, while angular momentum is increased from 0 to a larger value. Why then, is the answer that linear momentum and angular momentum are both conserved?

Thanks in advance!
For #17, they're referring to the whole system, not to the individual components.
 
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  • #3
Wait, for what kind of orbits is v^2=Gm/r valid? Also, I tried the energy approach but it doesn't seem to follow what you said (I probably did it wrong). According to v^2=Gm/r, the velocity is inversely proportional to r^.5. But with the energy approach, where KE=-E(total), which equals GMm/2r, if KE is increased, shouldn't the radius also decrease?

For #17, even if I consider the whole system, I can't quite erase the thought that angular momentum, at the very least, is not conserved. Before the collision, neither the skater nor the board are rotating, so isn't the angular momentum 0? After the collision, the two are rotating together, indicating an increase in angular momentum?
 
  • #4
SammyS
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Wait, for what kind of orbits is v^2=Gm/r valid? Also, I tried the energy approach but it doesn't seem to follow what you said (I probably did it wrong). According to v^2=Gm/r, the velocity is inversely proportional to r^.5. But with the energy approach, where KE=-E(total), which equals GMm/2r, if KE is increased, shouldn't the radius also decrease?
Equating centripetal force to gravitational force gives [itex]\displaystyle m\frac{v^2}{r}=m\frac{GM}{r^2}\ .[/itex]

Multiplying by r/m gives v2 = GM/r . So this is true (only) for circular orbits.

If you increase the speed, from that appropriate for a circular orbit, then the gravitational force will not be adequate to maintain a circular orbit.

If you increase the KE while maintaining a circular orbit, the yes this must be done in such a manner so that the orbital radius decreases at the same time.

For #17, even if I consider the whole system, I can't quite erase the thought that angular momentum, at the very least, is not conserved. Before the collision, neither the skater nor the board are rotating, so isn't the angular momentum 0? After the collision, the two are rotating together, indicating an increase in angular momentum?

Angular momentum is always calculated relative to some point. No rotation is necessary for there to be angular momentum.
 
  • #5
Okay, that makes sense! Thank you! I should probably brush up on angular momentum, haha!
 
  • #6
Redbelly98
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Good luck on Monday!

Regarding #17, remember that a change in linear momentum requires that an external force be applied to the system.

Similarly, a change in angular momentum requires that an external torque be applied to the system.

The lack of friction is a clue that there is no external force or torque.

Regarding #14, remember Kepler's laws. Orbits are elliptical, with the massive body located at a focus of the ellipse.
 
  • #7
Good luck on Monday!

Regarding #17, remember that a change in linear momentum requires that an external force be applied to the system.

Similarly, a change in angular momentum requires that an external torque be applied to the system.

The lack of friction is a clue that there is no external force or torque.

Regarding #14, remember Kepler's laws. Orbits are elliptical, with the massive body located at a focus of the ellipse.

Great, thank you for the clarifications! Also, I'm not sure if I'm allowed to link another help post, but I was having a bit of trouble on the 2008 AP test here https://www.physicsforums.com/showthread.php?p=4378951 If you could take a look, that would be awesome!
 

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