Error in my units, and I don't understand why

[warfreak131]

Homework Statement

Let's consider the a piece of a meteor falling to earth. The gravitational acceleration is experiences if GM/r^2. If I integrate acceleration as a function of r, i should get velocity as a function of r, right?

$$\int \frac{GM}{r^2}dr=GM\int \frac{1}{r^2}dr=\frac{-GM}{r}+const.$$

The problem is that G has the units of m^3 kg^-1 s^2. If you plug this in, you get the units of m^2 / s^2.

I know that orbital velocity of a body is the square root of GM/R, I just don't know how they got the square root, and why I didn't

Homework Equations

The Attempt at a Solution

 

[jhae2.718]

That is gravitational potential.

Velocity is the integral of acceleration w.r.t. time, not position.

 

[warfreak131]

but if you know acceleration as a function of r, why CANT you calculate velocity as a function of r?

 

[gneill]

but if you know acceleration as a function of r, why CANT you calculate velocity as a function of r?

You can. Calculate the WORK done on the body by the gravitational field. In other words, integrate F.d and use that as the change in KE --> m*v²/2 .

 

[warfreak131]

You can. Calculate the WORK done on the body by the gravitational field. In other words, integrate F.d and use that as the change in KE --> m*v²/2 .

Lets say I am 10,000 meters away from the Earth and I find F(r). Then work would be equal $$\int_0^{10000} F(r) dr$$?

And then that equals 1/2 mv^2. So if I assume that I start at rest, and since you mentioned that it is the change in KE, my final KE minus initial KE would just be my final KE?

 

[gneill]

You should be clear on what you mean by "10,000 meters away from the earth". The force formula in its simple algebraic form expects to see radial distance (distance from the Earth's center).

But essentially this is correct. Integrate the force*dr over the change in radial distance and you get the work done. Work done tells you how the KE changes.

 

[warfreak131]

You should be clear on what you mean by "10,000 meters away from the earth". The force formula in its simple algebraic form expects to see radial distance (distance from the Earth's center).

But essentially this is correct. Integrate the force*dr over the change in radial distance and you get the work done. Work done tells you how the KE changes.

Nice, now onto my second question. Let's say I am at one edge of the observable universe (and for sake of ease, we will say that the edge is static), and a proton is at the other edge, so we're about 100 billion light years apart. Also assume that we are the only things in the universe, and that I am stationary, relative to the edge of the observable universe. I want to know how long it will take that proton to reach me from my gravity alone. I weigh 77 kilograms.

I know the equation d=vit+1/2at^2 can be solved for time, but that only works for a constant acceleration. How would this equation change for variable acceleration? And since initial velocity is 0, and the distance traveled is 100 BLY (of course adjusted to meters), we can simplify.

 

[HallsofIvy]

The force on the proton is, as you said before,
$$-\frac{GmM}{r^2}$$
The accelation, force over mass, is given by
$$-\frac{GM}{r^2}$$

so the velocity is given by
$$\frac{dv}{dt}= -\frac{GM}{r^2}$$

It is also true that $v= dr/dt$ so that, by the chain rule
$$a= \frac{dv}{dt}= \frac{dv}{dr}\frac{dr}{dt}= v\frac{dv}{dr}= -\frac{GM}{r^2}$$

That is a separable equation:
$$dv= -\frac{GM}{r^2}dr$$
and integrating both sides,
$$v= \frac{dr}{dt}= \frac{GM}{r}+ v_0$$
and, assuming the initial speed is 0, $v_0= 0$.

Again, that is separable.
$$\frac{r}{GM}dr= dt$$

 

[gneill]

If it's time your after, you'll need to write the differential equation for the motion and solve it. Keep in mind that you'll be accelerating towards the proton as it accelerates towards you. You'll have something like:

$$\frac{d^2 r}{dt^2} = -\frac{G(M + m)}{r^2}$$

Take a look at the thread here:

https://www.physicsforums.com/showthread.php?t=246833