Kepler's Law for finding velocity

Click For Summary
SUMMARY

This discussion focuses on applying Kepler's Laws, specifically Kepler's Second Law, to determine the velocity of a planet in orbit. The key equations involved are the velocity equation \(v = \frac{2\pi}{T}\) and the area swept out by the planet, represented as \(\frac{dA}{dt} = 0.5 \cdot r \cdot v \cdot \sin(\theta)\). The solution involves using geometric approximations of triangles to represent the areas swept out in equal time intervals, leading to the conclusion that the velocity at periapsis can be expressed as \(v_{periastrom} = \frac{r_A}{r_P}v_0\). The discussion emphasizes the importance of small time intervals for accurate calculations.

PREREQUISITES
  • Understanding of Kepler's Laws, particularly Kepler's Second Law
  • Familiarity with basic geometry and calculus concepts
  • Knowledge of angular momentum conservation principles
  • Ability to manipulate algebraic equations involving velocity and time
NEXT STEPS
  • Study the derivation of Kepler's Second Law and its implications for orbital mechanics
  • Explore geometric methods for approximating areas in calculus
  • Learn about angular momentum in the context of celestial mechanics
  • Investigate the relationship between orbital radius and velocity in elliptical orbits
USEFUL FOR

Students and educators in physics, particularly those focusing on celestial mechanics and orbital dynamics, as well as anyone interested in the mathematical applications of Kepler's Laws.

Carpetfizz
Messages
13
Reaction score
0

Homework Statement


[/B]
I need to help solving part a)

HAoejbq.png


Homework Equations


[/B]
$$v= \frac{2\pi}{T}$$

$$(\frac{T_1}{T_2})^2 = (\frac{r_1}{r_2})^3$$

The Attempt at a Solution



I'm not sure where to begin really. One approach I tried was getting T_1 in terms of v_0 and plugging it into Kepler's Third Law to get T_2, and plugging T_2 back into the velocity equation to get the new velocity. However, this did not match the given answer of:

$$v_{periastrom} = \frac{r_A}{r_P}v_0$$
 
Physics news on Phys.org
There's only one orbit, so only one period. That rules out Kepler III as being of use here. What's left?
 
Kepler's second law? That each planet sweeps equal areas for equal periods of time? Not sure how to use that here.
 
Carpetfizz said:
Kepler's second law? That each planet sweeps equal areas for equal periods of time? Not sure how to use that here.
Yes. Kepler II is essentially a statement about the conservation of angular momentum. Consider a small time interval Δt and the areas swept out at the two locations over that small time interval. You might want to make a sketch using a bit of geometry to construct the approximations.
 
Okay so I have that $$\frac{dA}{dt} = 0.5*rvsin(\theta)$$ but I don't know how to use it to solve for v. I know that $$\frac{dA}{dt}$$ is constant though.
 
Go much simpler. Basic geometry. Draw the triangles that result assuming tangential velocities at the two locations. The areas of the triangles will approximate the areas swept out. If the Δt is small enough then these triangles will be a good approximation. In fact, as the Δt → 0, it's essentially the calculus approximation for the differential area element. What are the expressions for the areas of the two triangles?
 
Okay, so I have this so far:

$$A = \frac{1}{2}(t_1*v_0)r_A$$
$$A = \frac{1}{2}{t_2*v}r_P$$

but since I don't know the times I can't solve for v by setting them equal to each other.
 
Kepler II states equal areas in equal times. The times are identical.
 
Oh okay great, thank you so much!
 
  • #10
When you write this up, use Δt to represent the time interval and make it clear that Δt is a small time interval (theoretically approaching zero). The geometric approximation using triangles to represent the areas swept out will fail if the time interval is large, as the triangle will not approximate the curvature of the ellipse.
 
  • Like
Likes   Reactions: Carpetfizz

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 32 ·
2
Replies
32
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 40 ·
2
Replies
40
Views
5K
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K