Time to Reach the Origin in Central Force Attraction

In summary, the homework statement is that a particle of mass is attracted to the origin by a force -k/r^2. The equation to find the time t for the mass to reach the origin is \DeltaU= U-U0= \int\textbf{F}(r) dr. I found \DeltaU by \DeltaU= U-U0= \intF(r) dr and by following the convention r0=\infty U(\infty)=0, so \DeltaU= \frac{-k}{r} . Using the equation W=-k/r + c1, I found U to be k/r. I then used the expression \textbf{E
  • #1
littlehonda
10
0

Homework Statement



a particle of mass m is attracted to the origin by a force F=-k/r^2

find the time t for the mass to reach the origin

Homework Equations



[tex]\Delta[/tex]U= U-U0 = [tex]\int[/tex][tex]\textbf{F}[/tex](r) dr



The Attempt at a Solution




I found [tex]\Delta[/tex]U by [tex]\Delta[/tex]U= U-U0 = [tex]\int[/tex]F(r) dr and by following the convention r0=[itex]\infty[/itex] U([itex]\infty[/itex])=0

so [tex]\Delta[/tex]U = [tex]\frac{-k}{r}[/tex]

-[tex]\Delta[/tex]U=[tex]\textbf{W}[/tex]

not sure what to do now but i think I am just missing some easy steps. I know I have to equate this with 1/2mv^2 and then probably take a time derivative...
 
Physics news on Phys.org
  • #2
It seems to me that k is not a constant but is a function of mass. If it were otherwise, cricket balls would fall more quickly than cannonballs.
So if we calculate the work done by the fall and equate that to the kinetic energy gained we should be able to derive an expression for the time.
 
Last edited:
  • #3
K is given as a constant, analogous to Coulombs or the gravitational constant. The force could be expressed as

F = -k/r^2 = m[itex]\ddot{x}[/itex]

and we can see that acceleration is as always inversely proportional to the mass of the particle.

I believe you need to use U to solve this problem but I'm not sure how
 
  • #4
In that case, the method I already outlined will work but the time taken will be a function of both the mass and the distance.
 
  • #5
ok here's what I did

integrated F with respect to r and got W

W=-k/r + c1

using W = -[tex]\Delta[/tex]U I found U to be k/r

I then used the expression

[tex]\textbf{E}[/tex] = [tex]\frac{k}{r1}[/tex] +[tex]\frac{1}{2}[/tex]mv1[tex]^{2}[/tex] = [tex]\frac{k}{r2}[/tex] +[tex]\frac{1}{2}[/tex]mv2[tex]^{2}[/tex]

and since v1= 0 and r1=0

[tex]\textbf{E}[/tex] = [tex]\frac{k}{r1}[/tex] = [tex]\frac{1}{2}[/tex]mv2[tex]^{2}[/tex]

Now I solve for v

v = [tex]\frac{dL}{dt}[/tex] = [tex]\sqrt{\frac{2k}{mr}}[/tex]

[itex]\int[/itex]dL = [tex]\int[/tex] [tex]\sqrt{\frac{2k}{mr}}[/tex]dt

I end up with

t = [tex]\frac{L}{\sqrt{\frac{2k}{mr}}}[/tex]

is this the proper expression for time t it takes for the particle to move to the origin?
 
  • #6
Doesn't L = r ?
 

Related to Time to Reach the Origin in Central Force Attraction

1. What is central force attraction?

Central force attraction is a type of force that acts between two objects, with the force being directed towards the center of mass of the objects. It is a fundamental force in nature and is responsible for the motion of celestial bodies, such as planets orbiting around the sun.

2. How does central force attraction work?

Central force attraction is based on the law of gravitation, which states that any two objects with mass will attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. This force acts along the line connecting the two objects and is always directed towards the center of mass.

3. What are some examples of central force attraction?

The most well-known example of central force attraction is the gravitational force between the Earth and the moon. Other examples include the force between the sun and planets in our solar system, the force between stars in a galaxy, and the force between particles in an atom.

4. How is central force attraction related to Newton's laws of motion?

Central force attraction is directly related to Newton's laws of motion, specifically the second law which states that the force acting on an object is equal to its mass multiplied by its acceleration. In the case of central force attraction, the force acting on an object is the gravitational force, and the resulting acceleration is what causes the object to move in a circular or elliptical path around the other object.

5. Can central force attraction be repulsive?

No, central force attraction is always an attractive force. It is a fundamental force in nature and cannot be repulsive. However, there are other types of forces, such as electrostatic force, that can be either attractive or repulsive depending on the charges of the objects involved.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
19
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
250
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
254
  • Introductory Physics Homework Help
Replies
12
Views
244
  • Introductory Physics Homework Help
2
Replies
39
Views
2K
  • Introductory Physics Homework Help
Replies
11
Views
260
  • Introductory Physics Homework Help
Replies
5
Views
502
  • Introductory Physics Homework Help
Replies
16
Views
1K
Back
Top