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

[hi im nimdA]

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

 

[kuruman]

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?

 

[hi im nimdA]

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

 

[haruspex]

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.

 

[hi im nimdA]

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.

 

[kuruman]

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.

 

[hi im nimdA]

$$K_i+U_i=K_f+U_f$$ 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!

 

[kuruman]

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.

 

[hi im nimdA]

$$a=\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}=v\frac{dv}{dr}=\frac{Gm_2}{r^2}$$

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!

 

[kuruman]

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.

 

[ehild]

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.

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.