Verse from "A Brief History of Time"

[rudransh verma]

1.One can now see why all bodies fall at same rate: A body of twice the weight will have twice the force of gravity pulling it down, but it will also have twice the mass. According to Newton’s second law these two effects will exactly cancel each other, so the acceleration will be same in all cases.
2.Law of gravitation predicts the orbit of planets, moon ,earth with great accuracy. If the law were that the gravitational attraction of a star went down faster with distance, the orbits of the planets would not be elliptical, they would spiral into the sun. If it went down slower the gravitational force from distant star would dominate over that of the earth.

please explain the line in bold

 

[anuttarasammyak]

1. It seems all right to me.
2. Gravity follows law of inverse square of distance. Are you challenging it?

 

[rudransh verma]

1. It seems all right to me.
2. Gravity follows law of inverse square of distance. Are you challenging it?

How will it spiral into the sun?

 

[anuttarasammyak]

As I wrote as 2. , I am sorry to say I do not understand what you deduce in your 2.

 

[haushofer]

I guess that planets can only follow closed orbits iff gravity falls of as 1 over r-squared.

 

[Ibix]

You can write an effective potential assuming that the gravitational force $F$ depends on $r^{-p}$ where $p$ is not necessarily 2. If $p\neq 2$ orbits don't close. Outside a fairly narrow range of $p$ ($2\leq p\leq 4$ from somewhat distant memory) there turn out to be no stable orbits at all.

Edit: correction: $1\leq p\leq 3$ is the range for stable orbits

 

[ergospherical]

I guess that planets can only follow closed orbits iff gravity falls of as 1 over r-squared.

You can write an effective potential assuming that the gravitational force $F$ depends on $r^{-p}$ where $p$ is not necessarily 2. If $p\neq 2$ orbits don't close.

also $p=-1$ as well

 

[rudransh verma]

You can write an effective potential assuming that the gravitational force $F$ depends on $r^{-p}$ where $p$ is not necessarily 2. If $p\neq 2$ orbits don't close. Outside a fairly narrow range of $p$ ($2\leq p\leq 4$ from somewhat distant memory) there turn out to be no stable orbits at all.

I know it will be unstable. The book says it will spiral in. How will it spiral into the sun? Is it only a guess?

 

[Ibix]

The book says it will spiral in. How will it spiral into the sun?

I don't understand. Unstable orbits either spiral in or out. That's what it means to be an unstable orbit.

Is it only a guess?

No, it's a consequence of the effective potential calculation I mentioned above.

 

[rudransh verma]

I don't understand. Unstable orbits either spiral in or out. That's what it means to be an unstable orbit.

No, it's a consequence of the effective potential calculation I mentioned above

The book says if force falls faster with distance it will spiral in and ….
How will it spiral in if the force falls faster?

 

[ergospherical]

For some central force $f(r)$ the path equation of the trajectory is $\dfrac{d^2 u}{d\varphi^2} + u = -\dfrac{f(1/u)}{L^2 u^2}$ where $u=1/r$ and $L = r^2 \dot{\varphi}$. To investigate the stability, you imagine subjecting a particle in a stable circular orbit of radius $r_0$ to a perturbation $u = \dfrac{1}{r_0}(1 + \epsilon(\varphi))$.

As an exercise for you, make this change of variables from $u$ to $\epsilon$ in the path equation and - neglecting terms of $O(\epsilon^2)$ - what is the linearised equation satisfied by $\epsilon$?

 

[rudransh verma]

For some central force $f(r)$ the path equation of the trajectory is $\dfrac{d^2 u}{d\varphi^2} + u = -\dfrac{f(1/u)}{L^2 u^2}$ where $u=1/r$ and $L = r^2 \dot{\varphi}$. To investigate the stability, you imagine subjecting a particle in a stable circular orbit of radius $r_0$ to a perturbation $u = \dfrac{1}{r_0}(1 + \epsilon(\varphi))$.

As an exercise for you, make this change of variables from $u$ to $\epsilon$ in the path equation and - neglecting terms of $O(\epsilon^2)$ - what is the linearised equation satisfied by $\epsilon$?

Can you answer the first one in bold.

 

[Ibix]

The book says if force falls faster with distance it will spiral in and ….
How will it spiral in if the force falls faster?

It won't necessarily. There are closed circular orbits but they are unstable. If you perturb those circular orbits at all they will either fall into the central mass or escape.

 

[rudransh verma]

It won't necessarily. There are closed circular orbits but they are unstable. If you perturb those circular orbits at all they will either fall into the central mass or escape.

Does this mean force increases rapidly with decrease in distance.
Also please answer first one.

 

[Ibix]

Does this mean force increases rapidly with decrease in distance.

Does what mean force increases rapidly with decreasing distance?

Also please answer first one.

First what?

 

[rudransh verma]

Does what mean force increases rapidly with decreasing distance?

First what?

If force dercrease rapidly with distance then does it mean the other way around.
Answer the first question?

 

[Ibix]

If force dercrease rapidly with distance then does it mean the other way around.

Are you asking if "increases with decreasing distance" is the same as "decreases with increasing distance"? If so, yes.

Answer the first question?

Weight is proportional to mass. If you double the mass you double the weight.

 

[rudransh verma]

Are you asking if "increases with decreasing distance" is the same as "decreases with increasing distance"? If so, yes.

Weight is proportional to mass. If you double the mass you double the weight.

1.So According to the book somehow the distance decreases and the force increases rapidly making the planet spiral into the sun. Right?
2. F=mg. Double the weight means 2F. But it also means 2m. So 2F=2mg. So g=g Right?

 

[Ibix]

1.So According to the book somehow the distance decreases and the force increases rapidly making the planet spiral into the sun. Right?

If the distance decreases due to a micrometeorite impact or something (or the object isn't in a perfect circular orbit anyway) then the force increase as the satellite approaches it too much and the orbit spirals in, yes.

2. F=mg. Double the weight means 2F. But it also means 2m. So 2F=2mg. So a=g. Right?

More generally, the force due to gravity (i.e. the weight) is $F=GMm/r^2$ and the acceleration due to a force is $a=F/m$, so the acceleration due to gravity is $a=GM/r^2$ which doesn't depend on mass. The book is saying that same thing by arguing that if you replace a mass $m$ with a mass $2m$ you have to do it in both the first two equations, so the doubled force from the first equation is counteracted by the doubled mass in the second and hence the gravitational acceleration is independent of mass.

 

[vanhees71]

A very nice treatment of Bertrand's theorem can be found here:

https://aapt.scitation.org/doi/abs/10.1119/1.1430698

 

[Vanadium 50]

The book says

What book? What exactly does it say?

 

[rudransh verma]

Look up in question!

 

[Vanadium 50]

If that's what it says, it's wrong, or at least incomplete. If the law goes down faster with distance, planetary orbits stop being elliptical and become rosettes.