How can one derive an expression for \( w^2 / k \) using \( b, p, l \) only?

Click For Summary
The discussion focuses on deriving an expression for \( w^2 / k \) using the variables \( b, p, l \). It begins with the total energy of a particle, expressed as \( u^2 / 2 - k/R \), and simplifies to \( u^2 / 2 \) under certain conditions. By applying conservation of energy, the relationship \( \frac{u^2}{2} = \frac{w^2}{2} - \frac{k}{p} \) is established. The angular momentum equation \( l = bu \) allows for substitution of \( u \) in terms of \( b \) and \( l \). The problem is ultimately solved by recognizing the relationship \( l = pw \), linking all variables effectively.
lriuui0x0
Messages
101
Reaction score
25
Homework Statement
A particle with unit mass has distance ##R## from the origin and initial speed ##u##. It moves in the central potential ##\Phi(r) = -k/r##. If it doesn't move in the central field, it would move in a straight line whose shortest distance from the origin is ##b## (impact parameter). When it moves in the central field, it's closest distance from origin is ##p < b## with speed ##w##. Assume ##u^2 \gg 2k/R##, find ##w^2/k## in terms of ##b, p## only.
Relevant Equations
##l = bu##
The total energy of the particle is ##u^2 / 2 - k/R##. When ##u^2 \gg 2k/R##, we take the total energy to be ##u^2/2## only. By the conservation of energy, we have:

$$
\frac{u^2}{2} = \frac{w^2}{2} - \frac{k}{p}
$$

Take the angular momentum expression ##l = bu##, we can replace ##u## with ##b,l## thus getting an expression for ##w^2 / k## with ##b,p,l## only. But I don't know how to get an expression with ##b, p## only.
 
Last edited:
Physics news on Phys.org
You have used energy conservation. What else can you use?
 
Thanks. I solved the problem by noticing another relationship ##l = pw##.
 
  • Like
Likes hutchphd and bob012345
lriuui0x0 said:
Thanks. I solved the problem by noticing another relationship ##l = pw##.
Exactly, conservation of angular momentum ##l = bu = pw##.
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
3
Views
987
Replies
1
Views
2K
Replies
13
Views
2K
Replies
13
Views
2K
Replies
15
Views
2K
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K