# Gravitational Velocity of Mass ##M## at Distance ##d## from Gravitating Body

• ChrisVer
In summary: It is let's say \frac{v}{c} = \beta(r)... it shows the change of the velocity of a mass that began from infinity with 0 velocity under the "Newtonian" gravitational law of a point-like mass source M (in terms of "schwarzchild radii" of the source body- of course it should not be confused with schwarzchild radii since I don't work in GR). So for example, when the body is at a distance x=2 \times \frac{GM}{c^2} away from the source, it will have a velocity v \approx 0.75c...
ChrisVer
Gold Member
Starting from a locked thread I tried to work the gravity of a body of mass ##M## on another body starting from infinity to some distance ##d## from the gravitating body.
We have from the SR 2nd Newton law that:
$\gamma^3 a = \frac{GM}{r^2}$

writting $a= \frac{dv}{dt}= v \frac{dv}{dr} = \frac{1}{2} \frac{dv^2}{dr}$

Naming $v^2/c^2= x$ the above relation becomes:

$\frac{dx}{(1-x)^{3/2}}= \frac{2GM}{c^2 r^2}dr$

Integrating:

$\int_{0}^{x_d}\frac{dx}{(1-x)^{3/2}}= \int_{\infty}^{d} \frac{2GM}{c^2 r^2}dr$
$\frac{2}{\sqrt{1-x_d}}-2= - \frac{2GM}{c^2 d}$

$x_d=1- \Big( 1- \frac{GM}{c^2 d}\Big)^{-2}$

And so:

$v(d)= c \bigg[ 1- \Big( 1- \frac{GM}{c^2 d}\Big)^{-2} \bigg]^{1/2}$

I tried plotting this solution as $v(d)$, the good part is that $v<c$ for all distances however I don't understand why for $d=GM/c^2$ I'm obtaining an infinity (and worse- in the imaginary regime)? By the way, that's the Schwarzschild radius...

Even worse, if I set $\frac{GM}{c^2}=1$ the plot of $\beta = v/c = \bigg[ 1- \Big( 1- \frac{1}{d}\Big)^{-2} \bigg]^{1/2}$ is given in my attachment...and doesn't seem to have a real solution away from 1?

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• a1.jpg
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ChrisVer said:
I tried plotting this solution as v(d), the good part is that v<c for all distances however I don't understand why for d=GM/c2 I'm obtaining an infinity
Are you trying to compute the falling speed relative to a local hovering observer? You cannot have one hovering at the Schwarzschild radius.

A.T. said:
. You cannot have one hovering at the Schwarzschild radius.

So the information of Schwarzschild radius singularity is contained within SR? I solved the problem in a mechanical way...
But the main problem is that for larger radii from the source (distances ##d>R_s##), I am not getting a real solution for ##v##...

ChrisVer said:
Starting from a locked thread I tried to work the gravity of a body of mass ##M## on another body starting from infinity to some distance ##d## from the gravitating body.
We have from the SR 2nd Newton law that:
$\gamma^3 a = \frac{GM}{r^2}$

writting $a= \frac{dv}{dt}= v \frac{dv}{dr} = \frac{1}{2} \frac{dv^2}{dr}$

Naming $v^2/c^2= x$ the above relation becomes:
..
..

Typo here maybe

##\frac{1}{2} \frac{dv^2}{dr}##

Eh?
$\frac{1}{2} \frac{dv^2}{dr}= \frac{1}{2} \big( \frac{dv}{dr} v + v \frac{dv}{dr}\big) = v \frac{dv}{dr}= \frac{dr}{dt} \frac{dv}{dr} = \frac{dv}{dt}=a$

That looks like a GR conversation..

ChrisVer said:
That looks like a GR conversation..
Yes, for comparison of g under GR with the value you assume based on SR and Newton.

I think this is a mistake with the gravitational force I used...I think I should have used a minus sign: $\gamma^3 a = - GM/r^2$...
The main problem is that, as it is, there is no solution for $d > R_s$ ($v^2<0$) and this makes physically no sense, since the force I used could as well be some other $1/r^2$ type force (like electromagnetism), with the change $R_s \rightarrow D=\frac{k q_1 q_2}{c^2 m}$.

The result for $v$ is then:

$v= c \sqrt{1- (1+\frac{GM}{c^2 d} )^{-2}}$
With a plot (for $R_s= \frac{GM}{c^2}=1$) like the attached, which makes "sense".

#### Attachments

• a1.jpg
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ChrisVer said:
Eh?
$\frac{1}{2} \frac{dv^2}{dr}= \frac{1}{2} \big( \frac{dv}{dr} v + v \frac{dv}{dr}\big) = v \frac{dv}{dr}= \frac{dr}{dt} \frac{dv}{dr} = \frac{dv}{dt}=a$
Oh, you mean ##\frac{d}{dr}v^2##. Your notation confused me.

What is the plot of in your latest post ?

Mentz114 said:
What is the plot of in your latest post ?

It is let's say $\frac{v}{c} = \beta(r)$... it shows the change of the velocity of a mass that began from infinity with 0 velocity under the "Newtonian" gravitational law of a point-like mass source $M$ (in terms of "schwarzchild radii" of the source body- of course it should not be confused with schwarzchild radii since I don't work in GR). So for example, when the body is at a distance $x=2 \times \frac{GM}{c^2}$ away from the source, it will have a velocity $v \approx 0.75c$...
In a closed thread someone said that the velocity of the object would eventually reach a value larger than $c$ due to acceleration. I just wanted to counter this idea via special relativity alone...

I guess this could change for GR, but OK... It was more like a special relativistic application of $F= \frac{dp}{dt}$. Showing that by applying special relativity alone, it doesn't matter for how long your body has traveled (starting from infinity up to reaching the source at 0), its velocity won't exceed the speed of light.

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It saves the "classical"/non-relativistic prediction that:
$v= \sqrt{\frac{2GM}{r}}$
Which allows for $r$ such that $v>c$, eg $r= \frac{2GM}{9c^2}$ which gives $v=3c$

ChrisVer said:
It is let's say $\frac{v}{c} = \beta(r)$... it shows the change of the velocity of a mass that began from infinity with 0 velocity under the "Newtonian" gravitational law of a point-like mass source $M$ [...] So for example, when the body is at a distance $x=2 \times \frac{GM}{c^2}$ away from the source, it will have a velocity $v \approx 0.75c$...
In a closed thread someone said that the velocity of the object would eventually reach a value larger than $c$ due to acceleration. I just wanted to counter this idea via special relativity alone...
The curve certainly makes sense. You can see those ##\gamma##'s kicking in.

ChrisVer said:
We have from the SR 2nd Newton law

SR can't be used to model gravity.

ChrisVer said:
So the information of Schwarzschild radius singularity is contained within SR?

No. You need to use GR to model gravity.

ChrisVer said:
That looks like a GR conversation..

So should this thread be. You can't model gravity using SR.

ChrisVer said:
It was more like a special relativistic application of ##F= \frac{dp}{dt}##.

But for gravity, ##F = 0##. Gravity is not a force in relativity. A force in relativity is something that causes proper acceleration. Gravity doesn't. However the computations you are making work out formally, they are physically meaningless as far as gravity is concerned.

ChrisVer
ChrisVer said:

## 1. What is gravitational velocity of mass ##M## at distance ##d## from gravitating body?

The gravitational velocity of a mass ##M## at a distance ##d## from a gravitating body is the speed at which the mass is pulled towards the gravitating body due to the force of gravity. It is affected by the mass of the gravitating body, the mass of the object, and the distance between them.

## 2. How is gravitational velocity of a mass ##M## affected by the mass of the gravitating body?

The gravitational velocity of a mass ##M## is directly proportional to the mass of the gravitating body. This means that the larger the mass of the gravitating body, the greater the gravitational velocity of the object will be.

## 3. Is there a maximum gravitational velocity at a certain distance from a gravitating body?

Yes, there is a maximum gravitational velocity at a certain distance from a gravitating body. This is known as the escape velocity, and it is the minimum velocity required for an object to escape the gravitational pull of the body and not fall back down.

## 4. How does the distance between the mass ##M## and the gravitating body affect gravitational velocity?

The gravitational velocity of a mass ##M## is inversely proportional to the square of the distance between the mass and the gravitating body. This means that the farther the object is from the gravitating body, the weaker the gravitational pull will be and the lower the velocity will be.

## 5. Can the gravitational velocity of a mass ##M## be changed by altering the distance from the gravitating body?

Yes, the gravitational velocity of a mass ##M## can be changed by altering the distance from the gravitating body. As mentioned earlier, the gravitational velocity is inversely proportional to the distance, so by changing the distance, the velocity will also change accordingly.

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