Relating the universal law of gravitation and Newton's second law

Click For Summary

Homework Help Overview

The discussion revolves around relating the universal law of gravitation to Newton's second law, specifically in the context of two point particles attracting each other. The original poster attempts to analyze the accelerations of the particles and the implications for their motion as they approach each other, leading to questions about the applicability of kinematic equations and energy conservation principles.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between force, acceleration, and motion, questioning the use of kinematic equations under varying acceleration conditions. There is discussion about applying mechanical energy conservation and the correct formulation of velocity in relation to time versus position. Some participants suggest solving differential equations as an alternative approach.

Discussion Status

The conversation includes various attempts to clarify misunderstandings regarding the application of equations and the integration process. Participants are actively engaging with each other's suggestions, and some guidance has been offered regarding the correct forms of equations and the implications of attractive forces. There is an ongoing exploration of different methods to approach the problem.

Contextual Notes

Participants note that they have not yet covered gravitational energy in lectures, which may affect their understanding of the problem. There is also mention of the need to consider the one-dimensional nature of the motion and the relationship between the coordinates of the two particles.

hi im nimdA
Messages
7
Reaction score
0
Homework Statement
Two point particles, each of mass 100 kg, are initially at rest 1 m apart in outer space. (a) What is their initial acceleration? (b) What are their speeds when their separation is 0.5 m?
Relevant Equations
(1) F=G (m1m2)/r^2
(2) F=ma
First, I started with F_a = m_aa_a=G \frac{m_am_b}{r^2} and F_b = m_ba_b=G \frac{m_am_b}{r^2}. Solving for their respective accelerations, I got a_a=G \frac{m_b}{r^2} = 100G and a_b=G \frac{m_a}{r^2} = 100G, meaning that the initial acceleration of the two point particles are each 100G. When I tried to do part (b), I realized that their accelerations will keep changing as they get closer to each other, meaning that I can't use the kinematic equations, since they require a constant acceleration. I'm not too sure how to go about solving part (b).

I was thinking maybe v(r) = \int a(r) dr v(r) = \int_{1}^{0.5} \frac{100G}{r^2} dr Thanks for the help
 
Physics news on Phys.org
This$$v(r) = \int a(r) dr$$ is not correct. The correct expression is$$v(t) = \int a(t) dt.$$How about mechanical energy conservation?
 
kuruman said:
This$$v(r) = \int a(r) dr$$ is not correct. The correct expression is$$v(t) = \int a(t) dt.$$How about mechanical energy conservation?

Darn, I thought I could easily just apply it to its position instead of time. Mechanical energy conservation? Do you mean E_{mech, i} = E_{mech, f} K_i + U_i = K_f + U_f. We haven't covered energy for gravitation in lecture yet, but I found in the textbook that gravitational potential energy is U=-\frac{Gm_1m_2}{r}. Would kinetic energy still be K=\frac{1}{2} mv^2? Is there also a different way to solve this problem without energy? Thanks again
 
hi I am nimdA said:
Darn, I thought I could easily just apply it to its position instead of time.
Bear in mind that a(whatever).dt is multiplying an acceleration by a time, so gives dimension of velocity. a.dr would have the wrong dimension.
 
haruspex said:
Bear in mind that a(whatever).dt is multiplying an acceleration by a time, so gives dimension of velocity. a.dr would have the wrong dimension.
Ah. That's a good way to think about it. Thanks for correcting my misunderstanding.
 
If you want to do this without mechanical energy conservation, you need to solve a differential equation. Note that$$a=\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}=v\frac{dv}{dr}=\frac{Gm_2}{r^2}$$which gives$$v~dv=Gm_2\frac{dr}{r^2}$$Integrate, and you will have ##v(r)##. Note: This is known as a "first integral" which is equivalent to using energy conservation.
 
K_i+U_i=K_f+U_f Since it's initially at rest, -\frac{Gm_1m_2}{r_i} = \frac{1}{2}m_1v_f^2-\frac{Gm_1m_2}{r_f} -\frac{Gm_2}{r_i} +\frac{Gm_2}{r_f} = \frac{1}{2}v_f^2 v_f = \sqrt{2Gm_2(-\frac{1}{r_i}+\frac{1}{r_f})} v_f = \sqrt{2G(100)(-\frac{1}{1}+\frac{1}{0.5})} v_f=10\sqrt{2G} But when I do it with your integration method, \int_{v_o}^{v_f}v~dv = \int_{1}^{0.5} \frac{100G}{r^2} dr v_f^2 = -200G Is there something wrong with my bounds of integration? I did from 1 to 0.5 because the distance between the two particles goes from 1 to 0.5. If I swap the bounds, the answer will work out to be the same from both methods. Can someone help me fix my understanding. Thank you!
 
Actually, I was bit careless and I misled you, sorry. The correct equation to solve is $$v\frac{dv}{dr}=-\frac{Gm_2}{r^2}.$$ The negative sign indicates that the force is attractive.
 
kuruman said:
a=\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}=v\frac{dv}{dr}=\frac{Gm_2}{r^2}
kuruman said:
The correct equation to solve is v\frac{dv}{dr}=-\frac{Gm_2}{r^2}.
So just to clarify, we are still using the equation that I solved for in part (a) and then since the force is attractive, we tack on a negative sign? If so, thank you for all of your help!
 
  • #10
hi I am nimdA said:
So just to clarify, we are still using the equation that I solved for in part (a) and then since the force is attractive, we tack on a negative sign? If so, thank you for all of your help!
Yes.
 
  • #11
Keep in mind, that Newton's Second Law relates the acceleration of the particle with respect to a fixed point to the force acting on the particle. Here the force is inversely proportional to the square of the distance between the particles. You denoted both the coordinate and the distance with r.
As the particles are in rest initially, they will move toward each other along the straight line connecting them. So the problem is one-dimensional, you need a single coordinate to describe the motion of both particles, see picture.
243496


The distance between the particles is r=x2-x1
So the correct forms of Newton's Second Law are
##m_1\frac{d^2x_1}{dt^2} =G\frac{m_1m_2}{r}##
##m_2\frac{d^2x_2}{dt^2} = - G\frac{m_1m_2}{r}##
You can choose the mass centre of the system as the origin: it stays in rest. The two masses are equal, so the CoM is at the middle point between the particles: x1 = r/2 and x2 = r/2.
 
  • Like
Likes   Reactions: kuruman

Similar threads

  • · Replies 14 ·
Replies
14
Views
2K
Replies
6
Views
782
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 42 ·
2
Replies
42
Views
6K
Replies
11
Views
2K
Replies
2
Views
1K
Replies
8
Views
2K
Replies
15
Views
1K
Replies
3
Views
2K
Replies
35
Views
4K